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\def\book{In book.}
\def\exindent{\dimen0 = \hsize\advance\dimen0 by -61.75pt
\noindent\parshape = 2 0pt\hsize 61.75pt\dimen0}
\def \E #1{\itemitem{#1.\hskip 10pt}}
\def \EE{\itemitem{}}
\def \e #1{\itemitem{#1.\hskip 10pt}}
\def \ee{\itemitem{}}
\def\NE{\vskip 6pt}
\def\b{In book.}
\def\IND #1{\exindent #1 \hskip .5em}
\def \D{\displaystyle}
\def \I{\vskip 5pt \itemitem{}}
\def \IN{\itemitem{}\IND}
\def \\{\bf}
\def \o{\omega}

%% ==============================================
\def\endNE{\relax}
%% ===============================================
%% Dana Stuff
\newif\ifprintoddonly
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\long\def\NE \e#1#2\endNE{\ifprintoddonly\ifodd #1
\nextex \e {#1} #2\fi\else\nextex\e{#1}
#2\fi}


%||*****||*****||*****||%sec1.1
 


\hskip60pt \bf Charles M. Grinstead and J. Laurie Snell:
\vskip 5pt

\hskip60pt \bf INTRODUCTION to PROBABILITY
\vskip 5pt

\hskip60pt \bf Published by AMS
\vskip 20pt

\indent\bf Solutions to the exercises
\vskip 20pt

%\input mydefinitions
\indent\bf SECTION 1.1\rm
\NE
\e{1}%exer 1.1.1
As $n$ increases, the {\it proportion} of heads gets closer to 1/2,
but the {\it difference} between the number of heads and half the
number of flips tends to increase (although it will occasionally be
0).
\endNE\NE      
\e{2}%exer 1.1.2  
$n$ must be approximately 100.
\endNE\NE
\e{3}%exer 1.1.3
(b) If one simulates a sufficiently large number of rolls, one
should be able to conclude that the gamblers were correct.
\endNE\NE
\e{4}%exer 1.1.4
Player one has a probability of about
.83 of winning.
\endNE\NE
\e{5}%exer 1.1.5
The smallest $n$ should be about 150.
\endNE\NE
\e{7}%exer 1.1.7
The graph of winnings for betting on a color is much smoother (i.e. has
smaller fluctuations) than the graph for betting on a number.
\endNE\NE
\e{8}%exer 1.1.8
For two tosses both probabilities are
$1/2$.  For four tosses they are both $6/16$.  They
are, in fact, the same for any even number of tosses.  (This is not 
at all obvious; see Chapter 12 for a discussion of this and related
topics.)
\endNE\NE
\e{9}%exer 1.1.9
Each time you win, you either
win an amount that you have already lost or one of the
original numbers 1,2,3,4, and hence your net winning is
just the sum of these four numbers.  This is not a
foolproof system, since you may reach a point where you
have to bet more money than you have. If you and the bank
had unlimited resources it would be foolproof.
\endNE\NE
\e{10}%exer 1.1.10
You are very likely to win 5 dollars, but we shall see that this is 
still an unfair game, so we might say that Thackeray was right.
\endNE\NE
\e{11}%exer 1.1.11
For two tosses, the probabilities that Peter wins 0 and 2 are 1/2 and
1/4, respectively.  For four tosses, the probabilities that Peter wins
0, 2, and 4 are 3/8, 1/4, and 1/16, respectively.
\endNE\NE 
\e{13}%exer 1.1.13
Your simulation should result in about 25 days in a year
having more than 60 percent boys in the large hospital
and about 55 days in a year having more than 60
percent boys in the small hospital.
\endNE\NE 
\e{14}%exer 1.1.14
About 1/2 the time you win 2, 1/4 of the time you win 4, 1/8 of
the time you win 8, etc. If you add up all of these potential 
winnings, weighted by their probabilities, you get $\infty$, so
it would seem that you should be willing to pay quite a lot to play 
this game.  Few are willing to pay more than \$10.
\endNE\NE
\e{15}%exer 1.1.15
In about 25 percent of the games the player will have
a streak of five. 
\endNE\NE 
\e{16}%exer 1.1.16
In the case of having children until they have a boy, they
should have about 200{,}000 children.  In the case that they
have children until they have both a boy and a girl, they
should have about 300{,}000 children, or about 100{,}000 more.
\endNE

\vskip 20pt
%||*****||*****||*****||%sec1.2
\vskip 20pt

%Modified on 6/21/96
%\input mydefinitions
\indent\bf SECTION 1.2\rm 

\NE
\e{1}%exer 1.2.1
$P(\{a, b, c\}) = 1\qquad P(\{a\}) = 1/2$
\par
\hskip 19pt$P(\{a, b\}) = 5/6\qquad P(\{b\}) = 1/3$
\par
\hskip 19pt$P(\{b, c\}) = 1/2\qquad P(\{c\}) = 1/6$
\par
\hskip 19pt$P(\{a, c\}) = 2/3\qquad P(\phi) = 0$
\endNE\NE
\e{2}%exer 1.2.2
(a)\ \ $\Omega= \ \{A$ elected,
$B$ elected$\}$.

\itemitem{}(b)\ \ $\Omega= \ \{$Head,Tail$\}.$

\itemitem{}(c)\ \ $\Omega= \ \{$(Jan., Mon.), (Jan.,
Tue.),\dots,(Jan., Sun.),\dots,(Dec.,
 Sun.)$\}.$
\itemitem{}(d)\ \ $\Omega=\ \{$Student\ 
1,\dots,Student\ 10$\}.$
\itemitem{}(e)\ \ $\Omega=\ \{A, B, C, D, F \}.$ 
\endNE\NE
\e{3}%exer 1.2.3
(b), (d)
\endNE\NE
\e{4}%exer 1.2.4
(a) In three tosses of a
coin the first outcome is a head.
\vskip 5pt
\itemitem{}(b) In three tosses of a coin the same
side turns up on each toss.
\vskip 5pt
\itemitem{}(c) In three tosses of a coin exactly one tail
turns up.
\vskip 5pt
\itemitem{}(d) In three tosses of a coin at least one
tail turns up.
\endNE\NE
\e{5}%exer 1.2.5
(a) 1/2
\itemitem{}(b) 1/4
\itemitem{}(c) 3/8
\itemitem{}(d) 7/8
\endNE\NE
\e{6}%exer 1.2.6 
$4\over 7$.
\endNE\NE
\e{7}%exer 1.2.7
11/12
\endNE\NE
\e{8}%exer 1.2.8
Art $1\over4$, Psychology
$1\over2$, Geology $1\over4$.
\endNE\NE
\e{9}%exer 1.2.9
$3/4,\ 1$
\endNE\NE
\e{10}%exer 1.2.10
 $1\over2$.
\endNE\NE
\e{11}%exer 1.2.11
$1:12,\ 1:3,\ 1:35$
\endNE\NE
\e{12}%exer 1.2.12
$3\over4$.
\endNE\NE
\e{13}%exer 1.2.13
11:4
\endNE\NE
\e{14}%exer 1.2.13.1
(a) $m_Y(2) = 1/5,\ m_Y(3) = 1/5,\ m_Y(4) = 2/5,\ m_Y(5) = 1/5$
\itemitem{}(b) $m_Z(0) = 1/5,\ m_Z(1) = 3/5,\ m_Z(4) = 1/5$
\endNE\NE
\e{15}%exer 1.2.14
Let the sample space be: 
 \vskip 5pt
\itemitem{} $\omega_1 = \{A,A\}\qquad \omega_4 =
\{B,A\}\qquad\omega_7 = \{C,A\}$ \vskip 5pt
\itemitem{}$\omega_2 = \{A,B\}\qquad\omega_5 =
\{B,B\}\qquad \omega_8 = \{C,B\}$
\vskip 5pt 
\itemitem{} $\omega_3= \{A,C\}\qquad \omega_6 =
\{B,C\}\qquad \omega_9 = \{C,C\}$
\vskip 5pt
\itemitem{} where the first grade is John's and the second is
Mary's. 
You are given
that $$P(\omega_4)+P(\omega_5)+P(\omega_6)=.3,$$
$$P(\omega_2)+P(\omega_5)+P(\omega_8)=.4,$$
$$P(\omega_5)+P(\omega_6)+P(\omega_8)=.1.$$
\vskip 2pt
\itemitem{} Adding the first two equations and subtracting the
third, we obtain the desired probability as
\vskip 2pt
$$P(\omega_2)+P(\omega_4)+P(\omega_5)=.6.$$
\endNE\NE
\e{16}%exer 1.2.15
10 per cent.  An example: 10 lost eye, ear, hand, and leg; 15 eye, ear, and
hand; 
20 eye, ear, and leg; 25 eye, hand, and leg; 30 ear, hand, and leg.
\endNE\NE
\e{17}%exer 1.2.16
The sample space for a sequence of $m$ experiments is the set of $m$-tuples of
$S$'s and $F$'s, where $S$ represents a success and $F$ a failure.  The
probability
assigned to a sample point with $k$ successes and $m-k$ failures is
$$\Bigl({1\over n}\Bigr)^k\Bigl({{n-1}\over n}\Bigr)^{m-k}\ .$$
(a)   Let $k = 0$ in the above expression.
\vskip 5pt
\itemitem{} (b)  If $m = n\log 2$, then
$$\eqalign{\lim_{n \rightarrow \infty} \Bigl(1 - {1\over n}\Bigr)^m &=
\lim_{n \rightarrow \infty} \biggl(\Bigl(1 - {1\over n}\Bigr)^n\biggr)^{\log 2}
\cr
&= \biggl(\lim_{n \rightarrow \infty} (\Bigl(1 - {1\over n}\Bigr)^n
\biggr)^{\log 2}
\cr
&= \Bigl(e^{-1}\Bigr)^{\log 2}\cr
&= {1\over 2}\ .\cr}
$$
\vskip 5pt
\itemitem{}(c)  Probably, since $6\log 2 \approx 4.159$ and $36\log 2 \approx
24.953$.

\endNE\NE
\e{18}%exer 1.2.17
\IND{(a)}   The
right-hand side is the sum of the probabilities of
 all outcomes occurring in the left-hand side plus some
more
 because of duplication.
\vskip 5pt 
\itemitem{} (b)
$$1\ge P(A\cup B) = P(A) + P(B) - P(A\cap B).$$
\endNE\NE
\e{19}%exer 1.2.18
The left-side is the sum of
the probabilities of all elements in one of the three sets.
For the right side, if an outcome is in all three sets its
probability is added three times, then subtracted three
times,\ then added once,\ so in the final sum it is counted
just once.  An element that is in exactly two sets is
added twice, then subtracted once, and so it is counted
correctly.  Finally, an element in exactly one set is
counted only once by the right side.
\endNE\NE
\e{20}%exer 1.2.19
We would have to have the
same probability assigned to all outcomes.  If this
probability is 0, the sum of the probabilities would be 0
so that $P(\Omega) = 0$ instead of 1 as it should be.  If
this common probability is $a > 0$, then the sum of all the
probabilities of the first $n$ outcomes would be $na$ and for
large enough $n$ this would be greater than 1, contradicting
the requirement that the sum of the probabilities for all
possible outcomes should be 1.
\endNE\NE 
\e{21}%exer 1.2.20
$7/2^{12}$
\endNE\NE
\e{22}%exer 1.2.21
$\Omega = \{1,2,3,\dots\}$ and the
distribution is  $m(n)= (5/6)^{n-1}(1/6).$
Now if $0 < x < 1$, then 
$$\sum_{n=0}^\infty x^j = {1\over{1-x}}\ .$$
Hence
$$(1/6)\sum_{n=1}^\infty {(5/6)^{n-1}} =
(1/6)\cdot{1\over{1-5/6}} = 1.$$
\endNE\NE
\e{23}%exer 1.2.22
We have 
$$\sum_{n=0}^\infty
m(\omega_n) = \sum_{n=0}^\infty r (1-r)^n = {r\over{1-(1-r)}} 
= 1\ .$$ 
\endNE\NE
\e{24}%exer 1.2.23
He just meant that if you pick a
month at random within a complete 400-year cycle of the
calendar the thirteenth of the month is more likely to fall on
Friday than on any other day.
\endNE\NE 
\e{25}%exer 1.2.24
They call it a fallacy
because if the subjects are thinking about probabilities
they should realize that $$P(\hbox{Linda is bank teller
and in feminist movement)} \le P\hbox{(Linda is bank
teller)}.$$
 One 
explanation is that the subjects are not thinking about
probability as a measure of likelihood. 
For another explanation see Exercise 52 of Section 4.1.
\endNE\NE
\e{26}%exer 1.2.25
 The probability that the 
two cards are of the same rank is ${{52 \cdot 3}\over{52 \cdot 51}} =
{1\over17}$. Thus
$2x + {1\over 17} = 1$ giving $x = {8\over17}$
\endNE\NE
\e{27}%exer 1.2.26
$$P_x = P\hbox{(male lives to age}
\ x)\ =\ {{\hbox{number of male survivors at age}
\ x}\over100,000}\ .$$ 
$$Q_x = P\hbox{(female lives to age}\  x)\ =\ 
{{\hbox{number of female survivors at age}\ x} 
\over100,000}\ .$$
\endNE\NE
\e{28}%exer 1.2.27
(a) $1\over3$ \vskip 5pt 
\itemitem{\hskip 10pt}(b){ $P_3(N) =$}
${{[{N\over 3}]}\over N}$, where
$[{N\over3}]$ is the greatest integer in
 ${N\over 3}$.  Note that $${N\over 3} - 1
\le \biggl[{N\over 3}\biggr] \le {N\over 3}\ .$$

\itemitem{}From this we see that $$P_3 = \lim_{N\to\infty}
P_3(N) = {1\over 3}\ .$$

\itemitem{}(c)  If $A$ is a finite set with  $K$ 
elements then $${A(N)\over N}\le
{K\over N}\ ,$$
\noindent so $$\lim_{N\to\infty}{A(N)\over N} = 0\ .$$


\itemitem{}On the other hand, if $A$ is the set of all positive
integers, then $$\lim_{N\to\infty} {A(N)\over N} =
\lim_{N\to\infty} {N\over N} = 1\ .$$
\itemitem{}(d)  Let $N_k = 10^k - 1$.  Then the integers between $N_{k-1}+1$ and
$N_k$ have exactly $k$ digits.  Thus, if $k$ is odd, then
$$A(N_k) = (N_k - N_{k-1}) + (N_{k-2} - N_{k-3}) + \ldots\ ,$$
while if $k$ is even, then
$$A(N_k) = (N_{k-1} - N_{k-2}) + (N_{k-3} - N_{k-4}) + \ldots\ .$$
Thus, if $k$ is odd, then 
$$A(N_k) \ge (N_k - N_{k-1}) = 9\cdot 10^{k-1}\ ,$$
while if $k$ is even, then
$$A(N_k) \le N_{k-1} < 10^{k-1}\ .$$
So, if $k$ is odd, then
$${{A(N_k)}\over{N_k}} \ge {{9\cdot 10^{k-1}}\over{10^k - 1}} > {{9}\over {10}}\
,$$
while if $k$ is even, then
$${{A(N_k)}\over{N_k}} < {{10^{k-1}}\over{10^k - 1}} < {{2}\over{10}}\ .$$
Therefore, 
$$\lim_{N \rightarrow \infty} {{A(N)}\over N}$$
does not exist.
\vskip 20pt

\endNE\NE
\e{29}%exer 1.2.28
(Solution by Richard Beigel) 
\itemitem{}(a) In order to emerge from the interchange going west, the car must
go straight at the first point of decision, then make $4n+1$ right turns, and
finally go
straight a second time.  The probability $P(r)$ of this occurring is
$$P(r) = \sum_{n = 0}^\infty (1-r)^2r^{4n+1} = {{r(1-r)^2}\over{1-r^4}} =
{1\over{1+r^2}} -
{1\over{1+r}}\ ,$$
if $0 \le r < 1$, but $P(1) = 0$.  So $P(1/2) = 2/15$.
\vskip 5pt
\itemitem{}(b) Using standard methods from calculus, one can show that $P(r)$
attains a
maximum at the value
$$r = {{1 + \sqrt 5}\over 2} - \sqrt {{{1 + \sqrt 5}\over 2}} \approx .346\ .$$
At this value of $r$, $P(r) \approx .15$.

\endNE\NE
\e{30}%\exer 1.2.29
In order to depart to the east, one must make $4n+3$ right-hand turns in
succession, and then
go straight.  The probability is
$$P(r) = \sum_{n = 0}^\infty (1-r)r^{4n+3} = {{r^3}\over{(1+r)(1+r^2)}}\ ,$$
if $0 \le r < 1$.  This function increases on the interval $[0, 1)$, so the
maximum value of
$P(r)$, if a maximum exists, must occur at $r = 1$.  Unfortunately, if $r = 1$,
then the car
never leaves the interchange, so no maximum exists.

\endNE\NE
\e{31}%\exer 1.2.30
(a) Assuming that each student gives any given tire as an answer with
probability 1/4, then
probability that they both give the same answer is 1/4.
\vskip 5pt
\itemitem{}(b) In this case, they will both answer `right front' with
probability $(.58)^2$,
etc.  Thus, the probability that they both give the same answer is $39.8\%$.
\endNE

\vskip 20pt
%||*****||*****||*****||%sec2.1
\vskip 20pt

\indent\bf SECTION 2.1\rm
\vskip 5pt
\item{} The problems in this section are all computer
programs.

\vskip 20pt
%||*****||*****||*****||%sec2.2
\vskip 20pt

%\input mydefinitions
\indent\bf SECTION 2.2\rm
\vskip 5pt
\NE
\e{1}%exer 2.2.1
(a)\ \ \ $f(\omega) = 1/8$ on $[2, 10]$
\vskip 5pt
\itemitem{}(b)\ \ \ $P([a, b]) = {{b-a}\over 8}\ .$
\endNE\NE
\e{2}%exer 2.2.2 
(a)\ \ \  $c = 1/48.$\
\vskip 6pt
\itemitem {\hskip 10pt} (b)\ \  $P(E) =
{1\over96}(b^2-a^2).$ \vskip 6pt
\itemitem{\hskip 10pt} (c)\ \  $P(X > 5)={75\over96},\  P(x <
7) = {45\over96}\ .$ 
\vskip 6pt
\itemitem{\hskip 10pt}(d) 
\vskip -20pt$$\eqalign{\ \ \ \ P(x^2-12x+35>
0)&= P(x-5>0,x-7>0) + P(x-5 < 0,x-7<0)\cr &=P(x>7) + P(x <
5) ={3\over4}\ .\cr}$$
\endNE\NE
\e{3}%exer 2.2.100
(a)\ \ \ $C = {1\over{\log 5}} \approx .621$
\vskip 5pt
\itemitem{}(b)\ \ \ $P([a,b]) = (.621)\log(b/a)$
\vskip 5pt
\itemitem{}(c)$$\eqalign{P(x > 5) &= {{\log 2}\over{\log 5}} \approx .431\cr
P(x < 7) &= {{\log(7/2)}\over{\log 5}} \approx .778\cr
P(x^2 - 12x + 35 > 0) &= {{\log(25/7)}\over{\log 5}} \approx .791\ .\cr}
$$
\endNE\NE
\e{4}%exer 2.2.4
(a) .04,\ \   (b) .36,\ \ (c)\ \ 
.25, \ \ (d) .09. 
\endNE\NE
\e{5}%exer 2.2.5
(a)\ \ \ $1 - {1\over{e^1}} \approx .632$
\vskip 5pt
\itemitem{}(b)\ \ \ $1 - {1\over{e^3}} \approx .950$
\vskip 5pt
\itemitem{}(c)\ \ \ $1 - {1\over{e^1}} \approx .632$
\vskip 5pt
\itemitem{}(d)\ \ \ $1$
\endNE\NE
\e{6}%exer 2.2.6
 (a) $e^{-.01T}$, (b)\  $T=
100 \hbox{log}(2) = 69.3.$
\endNE\NE
\e{7}%exer 2.2.7
(a) 1/3,\ \ (b) 1/2,\ \ (c) 1/2,\ \ (d) 1/3
\endNE\NE
\e{8}%exer 2.2.8
(a) \ \  $1/8$,\ \   (b)
${1\over2}(1+\hbox{log}(2))$, \ \  (c) .75,
  \ \ (d) .25,
\vskip 6pt 
\itemitem{}(e)\ \  $3/4$,
\ \ 
(f)$\ \ 1/4$,  \ \  (g) $1/8$,  \ \ 
(h) $\pi/8$,  \ \  (i) $\pi/4$.
\endNE\NE
\e{12}%exer 2.2.13
$1/4$.
\endNE\NE
\e{13}%exer 2.2.14
$2\log 2 - 1$.
\endNE\NE
\e{14}%ex:cr 
(a) $13/24$, \ \ (b)
$1/48$. 
\endNE\NE
\e{15}%ex:cs 
Yes.
\endNE\NE
\e{16}%exer 2.2.16
Consider the circumference to be the interval $[0, 1]$, as in the hint.  Let $A
= 0$.  There are two
cases to consider; $0 < B < 1/2$, and $1/2 < B < 1$. In the first case, $C$ must
lie between 1/2 and $B + 1/2$,
for otherwise there would be a gap of length greater than 1/2, corresponding to
a semicircle containing none of the points.
Similarly, if $1/2 < B < 1$, it can be seen that $C$ must lie between $B - 1/2$
and 1/2.  The probability of one 
of these two cases occurring is 1/4.
\endNE

\vskip 20pt
%||*****||*****||*****||%sec3.1
\vskip 20pt

%\input mydefinitions
\indent\bf SECTION 3.1\rm
\vskip 5pt
\NE
\e {1}%exer 3.1.1 
24
\endNE\NE
\e {2}%exer 3.1.2
$1/12$
\endNE\NE
\e {3}%exer 3.1.3
$2^{32}$
\endNE\NE
\e {4}%exer 3.1.4 
At this writing, 37 Presidents have died.  The probability that no two people
from a group of 37 (all of whom are dead) died on the same day is about .15.
Thus, the probability that at least two died on the same day is .85.  Yes; 
Jefferson, Adams, and Monroe (all signers of the Declaration of Independence) 
died on July 4.
\endNE\NE
\e {5}%exer 3.1.5 
9, \ 6.
\endNE\NE
\e {6}%exer 3.1.6 
Since we do not get a different
situation if we rotate the table we can consider one
person's position as fixed, and then there are $(n-1)!$
possible arrangements for the other $n-1$ people.
\endNE\NE
\e {7}%exer 3.1.7 
$\displaystyle{{5!}\over
{5^5}}.$ \endNE\NE
\e {8}%exer 3.1.8
Each subset $S$ corresponds to a unique $r$-tuple of 0's and 1's, where
a 1 in the $i$'th location means that $i$ is an element of $S$.  Since
each location has two possibilities, there are $2^r$ $r$-tuples, and hence
there are $2^r$ subsets.
\endNE\NE
\e {10}%exer 3.1.10
1/13
\endNE\NE
\e {11}%exer 3.1.11 
$\displaystyle{{{3n-2}\over
{n^3}},\ {7\over 27},\ {28\over 1000}}.$
\endNE\NE
\e {12}%exer 3.1.12
(a)\ \  $30 \cdot 15 \cdot 9 = 4050$
\vskip 3pt
\itemitem{}(b)$\ \ 4050\cdot(3\cdot 2\cdot 1) = 24300$
\vskip 3pt
\itemitem{}(c)$\ \ 148824$
\endNE\NE
\e {13}%exer 3.1.13
(a) \ \ $26^3\times10^3$
\vskip 3pt 
\itemitem {}(b)$\ \ {6\choose 3}\times 26^3\times10^3$
\endNE\NE
\e {14}%exer 3.1.14
(a)$\ \ 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1 = 120$
\vskip 3pt
\itemitem{}(b)\ \ 60
\endNE\NE
\e {15}%exer 3.1.15
$\displaystyle{{{3\choose1}\times(2^n-2)}\over{3^n}}$.
\endNE\NE
\e{16}%exer 3.1.16
The number of possible four-digit
phone numbers is $10^4 = 10,000$, but there are more than
this number of people in Atlanta who have a telephone. 
\endNE\NE
\e {17}%exer 3.1.17
$\displaystyle{1 - {{12\cdot 11 \cdot \ldots \cdot (12 - n + 1)}\over{12^n}}\
,}$
if $n \le 12$, and 1, if $n > 12$.
\endNE\NE
\e {18}%exer 3.1.18 
36
\endNE\NE
\e{20}%exer 3.1.20  
Think of the person
on your right at lunch and  at dinner as determining a
permutation. Do the same for the person on your left at
lunch and at dinner.  We have two examples of the problem
of a random permutation having no fixed point.  The
probability of no match for large n for each random
permutation would be approximately $e^{-1}$ and if they
were independent the probability of no match in either
would be  $e^{-2}$.  They are not quite independent but
for large n they are close enough to being independent to
make this a good estimate.
\endNE\NE
\e{21}%exer 3.1.21
They are the same.
\endNE\NE
\e{22}%exer 3.1.22 
If $x_1,x_2,\dots,x_{15},x_{16}$
are the number observed with a maximum of 56,\ and we
assume there are $N$ counterfeits then
$P(x_1,x_2,\dots,x_{15},x_{16})=({1\over N})^{16}$ for any
$N \ge 56$ and 0 for any $N < 56.$  Thus, this probability
is greatest when  $N = 56.$ Your program should verify
that Watson's guess is much better.
\endNE\NE
\e{23}%exer 3.1.23
(a)\ \ $\displaystyle{{1 \over n},\ {1 \over n}}$
\vskip 3pt
\itemitem{}(b) She will get the best candidate if the second best candidate is
in the 
first half and the best candidate is in the secon half.  The probability that
this 
happens is greater than 1/4.
\endNE

\vskip 20pt
%||*****||*****||*****||%sec3.2
\vskip 20pt

%\input mydefinitions
\indent\bf SECTION 3.2\rm
\vskip 5pt
\NE
\e{1}%exer 3.2.1
(a)\ \  20
\vskip 3pt
\itemitem{}(b)\ \  .0064
\vskip 3pt
\itemitem{}(c)\ \  21
\vskip 3pt
\itemitem{}(d)\ \ 1
\vskip 3pt
\itemitem{}(e)\ \ .0256
\vskip 3pt
\itemitem{}(f)\ \ 15
\vskip 3pt
\itemitem{}(g)\ \ 10
\endNE\NE
\e{2}%exer 3.2.2
$\displaystyle{10\choose 5} =
252$
\endNE\NE
\e{3}%exer 3.2.3
$\displaystyle{9\choose 7} = 36$
\endNE\NE
\e{5}%exer 3.2.5
$\displaystyle{.998,\ .965,\ .729}$
\endNE\NE
\e{6}%exer 3.2.6
If Charles has
the ability, the probability that he wins
is$$
b(10,.75,7)+b(10,.75,8)+b(10,.75,9)+b(10,.75,10) = .776.$$
If Charles is guessing, the probability that Ruth wins is
$$ 1 -
b(10,.5,7)-b(10,.5,8)-b(10,.5,9)-b(10,.5,10) =.828.$$ 
\endNE\NE
\e{7}%exer 3.2.7
\vskip 1pt${\eqalign{\qquad{b(n,p,j)\over
b(n,p,j-1)}= 
{{\displaystyle{n\choose{j}}p^jq^{n-j}}\over{\displaystyle{n\choose{j-1}}p^{j-1}
q^{n-j+1}}}
  &=  {n!\over{j!(n-j)!}}{{(n-j+1)!(j-1)!}\over
{n!}}{p\over{q}}\cr &={(n-j+1)\over{j}}{p\over{q}}}}.$
\vskip 5pt
\itemitem{}But
$\displaystyle{{(n-j+1)\over{j}}{p\over{q}}\ge 1}$ if and
only if $j \le p (n+1)$, and so $j = [p(n+1)]$ gives
\hbox{${b(n,p,\ j)}$} its largest value. If $p(n+1)$ is an
integer there will be two possible values of j,
namely $j = p(n+1)$ and $j = p(n+1) - 1$.
\endNE\NE
\e{8}%exer 3.2.8
$\displaystyle{ b(30,{1\over6},5) =
{30\choose
5}\Bigr({1\over6}\Bigr)^{5}{\Bigr({5\over6}\Bigl)^{25}}
= .1921.}$ The most probable number of times is  5. 
\endNE\NE
\e{9}%exer 3.2.9
$n = 15,\ r = 7$
\endNE\NE
\e{10}%exer 3.2.10
$\displaystyle{{{11}\over{64}} \approx .172}$
\endNE\NE
\e{11}%exer 3.2.11
Eight pieces of each kind of pie.
\endNE\NE
\e{12}%exer 3.2.12
(a)\ \ $4/{52 \choose 5} \approx .0000015$
\vskip 3pt
\itemitem{}(b)\ \ $36/{52\choose 5} \approx .000014$
\vskip 3pt
\itemitem{}(c)\ \ $624/{52\choose 5} \approx .00024$
\vskip 3pt
\itemitem{}(d)\ \ $3744/{52\choose 5} \approx .0014$
\vskip 3pt
\itemitem{}(e)\ \ $5108/{52\choose 5} \approx .0020$
\vskip 3pt
\itemitem{}(f)\ \ $10200/{52\choose 5} \approx .0039$
\endNE\NE
\e{13}%exer 3.2.13
The number of subsets of $2n$
objects of size $j$ is $\displaystyle{{2n}\choose j}$.
$${\displaystyle{2n\choose{i}}\over\displaystyle{{2n\choose{i-1}}}}
= {{2n-i+1}\over{i}} 
\ge 1 \Rightarrow  {i} \le {n + {1\over2}.}$$
\itemitem{} Thus $i = n$ makes
$\displaystyle{{2n}\choose{i}}$ maximum.
\endNE\NE
\e{14}%exer 3.2.14
By Stirling's formula,  $n!\sim
\displaystyle\sqrt{2{{\pi}n}}\displaystyle{(n^n)}e^{-n}.$ 
 Thus,  $$\eqalign{b(2n,{1\over2},n)& =
{\displaystyle{20\choose{n}}{1\over{2^{2n}}}}\cr
&={{2n!}\over{(n!)^2}}{\cdot} {{1\over {2^{2n}}}}
\cr &{\sim} \displaystyle{
{{1\over{2^{2n}}}}{{\displaystyle{\sqrt{2
\pi{2n}}{(2n)^{2n}}{e^{-2n}}}}\over{\displaystyle{2\pi{n}(n^{2n})e^{-2n}}}}
= {{1\over{\sqrt{\pi{n}}}}}}.}$$
\endNE\NE
\e{15}%exer 3.2.15
.3443,\  .441,\ .181,\ .027.
\endNE\NE
\e{16}%exer 3.2.16
There are $8\choose3$ ways of
winning three games.  After winning, there are $5\choose3$
ways of losing three games.  After losing, there is only
one way of tying two games.  Thus the the total number of
ways to win three games, lose three games, and tie two is
${8\choose3}{5\choose3} = 560$.
\endNE\NE
\e{17}%exer 3.2.17
There are $n\choose{a}$ ways of
putting \it a \rm different objects into the 1{st} box,
and then ${n-a}\choose{b}$ ways of putting \it b \rm
different objects into the 2{nd} and then one way to put
the remaining objects into the 3{rd} box. Thus the total
number of ways is $${{{n}\choose{a}}{{n-a}\choose{b}}} =
{{n!}\over{{a!b!(n-a-b)!}}}.$$
\endNE\NE
\e{18}%exer 3.2.18
$P$(no student gets 2 or fewer correct) $= b(340, 7/128, 0) \approx
4.96\cdot 10^{-9}$; $P$(no student gets 0 correct) $= b(340, 1/1024, 0) \approx
.717$.  So Prosser is right to expect at least one student with 2 or fewer
correct, but Crowell is wrong to expect at least one student with none correct.
\endNE\NE
\e{19}%exer 3.2.19
(a)\ \ \ 
${{{\displaystyle{4\choose1}\displaystyle{13\choose10}}\over
\displaystyle{52\choose10}}} = 7.23\times 10^{-8}.$ 
\vskip 7pt
\itemitem{}(b)\ \ \ 
${{\displaystyle{4\choose1}\displaystyle{3\choose2}\displaystyle{13\choose4}\displaystyle
{13\choose3}\displaystyle{13\choose3}}\over
\displaystyle{52\choose10}} = .044.$
\vskip 7pt
\itemitem{}(c)\ \ \ 
${{\displaystyle4!\displaystyle{13\choose4}\displaystyle{13\choose3}
\displaystyle{13\choose2} \displaystyle{13\choose1}}\over
\displaystyle{52\choose13}}{= .315}.$
\endNE\NE
\e{20}%exer 3.2.20
(a)\ \ $\displaystyle{{{13}\choose 6}/{{52}\choose 6} \approx .000084}$
\vskip 3pt
\itemitem{}(b)\ \ $\displaystyle{{4\choose 3}{4\choose 2}{4\choose 1}/
{{52}\choose 6} \approx .0000047}$
\vskip 3pt
\itemitem{}(c)\ \ $\displaystyle{{{{4\choose 2}{{13}\choose 3}
{{13}\choose 3}}/{{{52}\choose 6}}}}$
\endNE\NE
\e{21}%exer 3.2.21
$3({2^5}) - 3 = 93$ (We subtract 3
because the three pure colors are each counted twice.)
\endNE\NE
\e{22}%exer 3.2.22
$\displaystyle{{8\choose 2} = 28}$
\endNE\NE
\e{23}%exer 3.2.23
To make the boxes, you need $n+1$
bars, 2 on the ends and $n-1$ for the divisions.  The
$n-1$ bars and the r objects occupy $n-1+r$ places.  You can
choose any $n-1$ of these $n-1+r$ places for the bars and use
the remaining $r$ places for the objects. Thus the number of
ways this can be done is $${{n-1+r}\choose{n-1}} =
{{n-1+r}\choose{r}}.$$ 
\endNE\NE
\e{24}%exer 3.2.24
$\displaystyle{{{19}\choose{10}}/{{29}\choose{20}} \approx .009}$
\endNE\NE
\e{25}%exer 3.2.25
(a)\ \ $\displaystyle{6!{10\choose 6}/10^6 \approx .1512}$
\vskip 3pt
\itemitem{}(b)\ \ $\displaystyle{{10\choose 6}/{15\choose 6} \approx .042}$
\endNE\NE 
\e{26}%exer 3.2.26
(a) $\displaystyle{pq,\ qp,\ p^2,\ q^2}$
\endNE\NE
\e{27}%exer 3.2.27
Ask John to
make 42 trials and if he gets 27 or more correct accept
his claim.  Then the probability of a type I error is
$$\sum_{k\ge 27}{b(42,.5,k)} = .044,$$ and the probability
of a type II error is $$1 - \sum_{k \ge 27} {b(42,.75,k)}=
.042.$$ 
\endNE\NE
\e{28}%exer 3.2.28
$n = 114,\ m = 81$
\endNE\NE
\e{29}%exer 3.2.29
$\displaystyle{b(n,p,m) = {n\choose{m}}{p^m}{(1-p)^{n-m}}.}$
Taking the derivative with respect to $p$ and setting this
equal to 0 we obtain $m(1-p) = p(n-m)$ and so $p =
m/n$.
\endNE\NE
\e{30}%exer 3.2.30
(a)\ \ $p(.5) = .5,\ p(.6) = .71,\ p(.7) = .87$
\vskip 3pt
\itemitem{}(b)\ \ Mets have a 95.2\% chance of winning in a 7-game series.
\endNE\NE
\e{31}%exer 3.2.31
.999996.
\endNE\NE
\e{32}%exer 3.2.32
If $u = 1$, you only need to be sure to send at least one to each side.  If 
$u = 0$, it doesn't matter what you do.  Let $v = 1 - u$ and $q = 1 - p$.  If
$0 < v < 1$, let $x$ be the nearest integer to
$${n\over 2} - {1\over 2}{{\log(p/q)}\over{\log v}}\ .$$
\endNE\NE
\e{33}%exer 3.2.33
By Stirling's formula,
$${{{\displaystyle{{2n}\choose{n}}^2\over\displaystyle{{4n}\choose{2n}}}={{{{(2n
!)}^2}{{(2n!)}^2}}\over{{n!^4}{(4n)!}}}}}
\sim{{({\sqrt{4\pi{n}}(2n)^{2n}{e^{-2n}}})^4}\over{({\sqrt{2\pi{n}}{(n^n)}{e^{-n
}}})^4\sqrt{2\pi(4n)}(4n)^{4n}{e^{-4n}}}}
=\sqrt{2\over{\pi{n}}}.$$
\endNE\NE
\e{34}%exer 3.2.34
Let $E_i$ be the event that you do
not get the ith player's picture.  Then for any $k$ of
these events
$${P(E_{i1}\cap E_{i2}\cap\dots\cap E_{ik})} =
{\Bigr({{n-k}\over{n}}\Bigl)^m}.$$
You have $\displaystyle{n\choose{k}}$ ways of choosing $k$
different $E_i$'s. Thus the result follows from Theorem 9.
\endNE\NE
\e{35}%exer 3.2.35
Consider an urn with \it n \rm red
balls and \it n \rm blue balls inside.  The left side of
the identity
$$ {2n\choose{n}} =
\sum_{j=0}^n{{n\choose{j}}^2}=\sum_{j=0}^n{n\choose{j}}{n\choose{n-j}}$$
counts the number of ways to choose $n$ balls out of the $2n$
balls in the urn. The right hand counts the same thing
but  breaks the counting into the sum of the cases where
there are exactly  $j$  red balls and $n-j$ blue balls.  
\endNE\NE
\e{36}%exer 3.2.35.5
(a)\ \ $\displaystyle{{n\choose j}}$
\vskip 3pt
\itemitem{}(b)\ \ $\displaystyle{1 - {n-j\choose j}/{n\choose j}}$
\endNE
\e{38}%exer 3.2.37
Consider the Pascal triangle (mod 3) for example.
\hskip 10pt \vskip 20pt
{\settabs\+\hskip 100pt&xxxxxx&\cr 
\+&0& $\underline1$ \cr 
\+&1& 1\ 1\cr
\+&2& 1\ 2\ 1\cr
\+&3&$\underline1$\ 0\ 0\ $\underline1$\cr
\+&4&1\ 1\ 0\ 1\ 1\cr
\+&5& 1\ 2\ 1\ 1\ 2\ 1\cr
\+&6&$\underline1$\ 0\ 0\ $\underline2$\ 0\ 0\ $\underline1$\cr
\+&7& 1\ 1\ 0\ 2\ 2\ 0\ 1\ 1\cr
\+&8&1\ 2\ 1\ 2\ 1\ 2\ 1\ 2\ 1\cr
\+&9& $\underline1$\ 0\ 0\  $\underline0$\ 0\ 0\  $\underline0$\ 0\ 0\
$\underline1$\cr \+&10& 1\ 1\ 0\ 0\ 0\ 0\ 0\ 0\ 0\ 1\ 1\cr
\+&11& 1\ 2\ 1\ 0\ 0\ 0\ 0\ 0\ 0\ 1\ 2\ 1\cr
\+&12& $\underline1$\ 0\ 0\ $\underline1$\ 0\ 0\ $\underline0$\ 0\ 0\
$\underline1$\ 0\ 0\ $\underline1$\cr \+&13& 1\ 1\ 0\ 1\ 1\ 0\ 0\ 0\ 0\ 1\ 1\ 0\
1\ 1\cr \+&14& 1\ 2\ 1\ 1\ 2\ 1\ 0\ 0\ 0\ 1\ 2\ 1\ 1\ 2\ 1\cr
\+&15& $\underline1$\ 0\ 0\ $\underline2$\ 0\ 0\ $\underline1$\ 0\ 0\
$\underline1$\ 0\ 0\ $\underline2$\ 0\ 0\ $\underline1$\cr \+&16& 1\ 1\ 0\ 2\ 2\
0\ 1\ 1\ 0\ 1\ 1\ 0\ 2\ 2\ 0\ 1\ 1\cr
\+&17& 1\ 2\ 1\ 2\ 1\ 2\ 1\ 2\ 1\ 1\ 2\ 1\ 2\ 1\ 2\
1\ 2\ 1\cr 
\+&18& $\underline1$\ 0\ 0\ $\underline0$\ 0\ 0\ $\underline0$\ 0\ 0\
$\underline2$\ 0\ 0\ $\underline0$\ 0\ 0\ $\underline0$\ 0\ 0\
$\underline1$\cr}  \vskip 7pt
\itemitem{}Note first that the entries in the third row
are 0 for $0 <  j <  3$. Lucas notes that this will
be true for any $p$. To see this assume that $0 <  j <
 p.$ Note that $${p\choose{j}} = {{p(p-1)\cdots
{p-j+1}}\over {j(j-1)\cdots 1}}$$ 

\itemitem{} is an integer. Since $p$ is prime and $0 <  j < 
p$, $p$ is not divisible by  any of the terms of $j!$, and  so 
$(p-1)!$ must be divisible by $j!$. Thus for $0 <  j <  p$ we
have $\displaystyle{p\choose{j}} = 0$ mod $p$.
   Let us call the triangle of the first three rows a
\bf basic triangle\rm. The fact that the third row is
\vskip 20pt
 \centerline{1 0 0 1}
\vskip 20pt
\itemitem{} produces two more basic triangles in the next
three rows and an inverted triangle of 0's between these
two basic triangles. This leads to the 6'th row\vskip
10pt
 \centerline{1 0 0 2 0 0 1}. 
\vskip 20pt
\itemitem{}  This produces a basic triangle, a basic
triangle multiplied by 2 (mod 3), and then another basic
triangle in the next three rows. Again these triangles are
separated by inverted 0 triangles.  We can continue this
way to construct the entire Pascal triangle as a bunch of
multiples of basic triangles separated by inverted 0
triangles.  We need only know what the mutiples are.
The multiples in row $np$ occur at positions
$0,p,2p,...,np$.  Looking at the triangle we see that the
multiple at position $(mp,\thinspace jp)$ is the sum of the
multiples at positions $(j-1)p$ and $jp$ in the
$(m-1)p$'{th} row.  Thus these multiples satisfy the same
recursion relation

          $${n\choose{j}} = {{n-1}\choose{j-1}} +
{{n-1}\choose{j}}$$

\itemitem{} that determined the Pascal triangle. Therefore 
the multiple at position $(mp,\thinspace jp)$ in the
triangle is $\displaystyle{m\choose{j}}$.    
Suppose we want to determine the value in the
Pascal triangle mod $p$ at the position $(n,\thinspace j)$.
Let $n = sp + s_0$ and $j = rp + r_0$, where $s_0$ and $r_0$
are $< p$. Then the point $(n,\thinspace j)$ is at position
$(s_0,r_0)$ in a basic triangle multiplied by
$\displaystyle {s\choose{r}}$. 
\itemitem{}  Thus 
 $${n\choose{j}} =
{{s\choose{r}}{{s_0}\choose{r_0}}}\ .$$

\itemitem{}But now we can repeat this process with the pair
$(s,r)$ and continue until $s < p$. This gives us the
result:
 $${n\choose{j}} = {\prod_{i=0}^k
{{s_i}\choose{r_j}}(\hbox{mod}\ p})\ ,$$ 
\itemitem{}where 
$$\eqalign {s &= s_0 + s_1p^1+s_2p^2 + \cdots +s_kp^k\ ,\cr 
j &= r_0+r_1p^1 + r_2p^2 + \cdots +r_kp^k\ .}$$
\itemitem{}If $ r_j >  s_j$ for some $j$ then the
result is 0 since, in this case, the pair $(s_j,r_j)$
lies in one of the inverted 0 triangles.
\vskip 5pt
\itemitem{}  If we consider the row $ p^k - 1$ then for
all $k$, $s_k = p-1$ and $r_k \le p-1$ 
 so the product will be
positive resulting in no zeros in the rows $ p^k - 1$. In
particular for $p =2$ the rows $p^k-1$ will consist of all
1's.  
\endNE\NE 
\e{39}%exer 3.2.38
$$\eqalign {b(2n,{1\over2},n) =
2^{-2n}{{2n!}\over{n!n!}}&= {{2n(2n-1)\cdots
2\cdot1}\over{2{n}\cdot{2(n-1)}\cdots 2\cdot
2{n}\cdot{2(n-1)}\cdots 2}} \cr
&={{(2n-1)(2n-3)\cdots1}\over{2n(2n-2)\cdots2}}\ .}$$
\endNE

\vskip 20pt
%||*****||*****||*****||%sec3.3
\vskip 20pt

%\input mydefinitions
\indent\bf SECTION 3.3\rm
\vskip 5pt
\NE
\e{3}%exer 3.3.1
(a)\ \  $96.99\%$
\vskip 3pt
\itemitem{}(b)\ \ $55.16\%$
\endNE

\vskip 20pt
%||*****||*****||*****||%sec4.1
\vskip 20pt

%\input mydefinitions
\indent\bf SECTION 4.1\rm
\vskip 5pt
\NE
\e{2}%exer 4.1.2
(a)\ \ 1/2
\vskip 3pt
\itemitem{}(b)\ \ 1/4
\vskip 3pt
\itemitem{}(c)\ \ 1/2
\vskip 3pt
\itemitem{}(d)\ \ 0
\vskip 3pt
\itemitem{}(e)\ \ 1/2
\endNE\NE
\e{3}%exer 4.1.3
(a)\ \ 1/2
\vskip 3pt
\itemitem{}(b)\ \ 2/3
\vskip 3pt
\itemitem{}(c)\ \ 0
\vskip 3pt
\itemitem{}(d)\ \ 1/4
\endNE\NE
\e{4}%exer 4.1.4
(a)\ \ 1/2
\vskip 3pt
\itemitem{}(b)\ \ 4/13
\vskip 3pt
\itemitem{}(c)\ \ 1/13
\endNE\NE
\e{5}%exer 4.1.5
(a)\ \ (1) and (2)
\vskip 3pt
\itemitem{}(b)\ \ (1)
\endNE\NE
\e{6}%exer 4.1.6
3/10
\endNE
%FOR exer 4.1.7
$\eqalign{7.\ \ \ \ (a) \quad
&P(A\cap B)=P(A\cap C)=P(B\cap C) = {1\over 4}\ ,\cr
&P(A)P(B) =P(A)P(C) = P(B)P(C) = {1\over 4}\ ,\cr
&P(A\cap B\cap C) = {1\over4} \ne P(A)P(B)P(C) =
{1\over8}\ .\cr}$

$\eqalign{\ \ \ \ \ \ (b)\quad &P(A\cap C) =
P(A)P(C) ={1\over 4}, \hbox{\ \ so $C$ and $A$ are independent},\cr 
&P(C\cap B) = P(B)P(C) = {1\over4}, \hbox{\ \ so $C$
and $B$ are independent},\cr
&P(C\cap (A\cap B)) = {1\over4} \ne P(C)P(A\cap B) =
{1\over8}\ ,\cr 
&\hbox{so $C$ and $A\cap B$ are not independent.}\cr}$

%FOR exer 4.1.8
$\hskip -4pt \eqalign{8.\qquad
&P(A\cap B\cap C) = P(\{a\}) = {1\over 8}\ ,\cr
& P(A) = P(B) = P(C) = {1\over2}\ .\cr
&\hbox{Thus while}\ \ 
P(A\cap B\cap C) = P(A)P(B)P(C)={1\over 8}\ ,\cr
&P(A\cap B) = P(A\cap C) = P(B\cap C) =
{5\over 16}\ ,\cr
& P(A)P(B) = P(A)P(C) =
P(B)P(C)= {1\over 4}\ .\cr
&\hbox{ Therefore no two of these events are
independent.}}$
\NE
\e{9}%exer 4.1.9
(a)\ \ $1/3$ 
\vskip 3pt
\itemitem{}(b)\ \ $1/2$
\endNE\NE
\e{10}%exer 4.1.9.5
It is probably a reasonable estimate.  One might refer to earlier life tables
to see how much this number has changed over the last four or five censuses.
\endNE\NE
\e{12}%exer 4.1.10
.0481
\endNE\NE
\e{13}%exer 4.1.11
1/2
\endNE\NE
\e{14}%exer 4.1.12
1/8
\endNE\NE
\e{15}%exer 4.1.13
(a)
\ \ ${{\displaystyle{
48\choose11}\displaystyle{4\choose2}}\over\displaystyle{{52\choose13}-{48\choose
13}}}{\approx
.307}\ .$ \vskip 20pt
\itemitem{}(b)\ 
${{\displaystyle{48\choose11}\displaystyle{3\choose1}}\over
\displaystyle{51\choose12}}{\approx.328}\ .$
\endNE\NE
\e{16}%exer 4.1.14
$\displaystyle{{P(A)P(B\vert A)P(C\vert A\cap B)} =
P(A)\cdot{{{P(A\cap B)}\over{P(A)}}\cdot{{P(A\cap B\cap
C)}\over{P(A\cap B)}}} ={P(A\cap B\cap C)}}\ .$
\endNE
\vskip .1in
  
%FOR exer 4.1.15
\hskip -10pt$\eqalign{{17.}\hskip 14pt
(a)\ \ \displaystyle P(A\cap
\tilde B) = P(A)-P(A\cap B)& = P(A)-P(A)P(B)\cr &=P(A)(1-P(B))\cr &=
P(A)P(\tilde B)\ .}$ 
\itemitem{}(b)\ \ Use (a), replacing $A$ by $\tilde B$ and
$B$ by $A$.
\NE
\e {18}%exer 4.1.16
$\eqalign{\ \ \ 
&P(D1|+) = 4/9\cr
&P(D2|+) = 1/3\cr
&P(D3|+) = 2/9\cr}$
\endNE\NE
\e{19}%exer 4.1.17
.273.
\endNE\NE
\e{20}%exer 4.1.18
It can be shown that after $n$
draws, $P$($k$ white balls, $n+1-k$ black balls in the urn) =
${1/(n+1)}$ for any ${0 \le k \le n}$. Thus you are
equally likely to have any proportion of white balls after
$n$ draws.  In fact, the fraction of white balls will tend to
a limit but this limit is a random number.
This rather suprising fact is a consequence of the fact
that the Polya Urn model is mathematicallly exactly the
same as the following apparently very different model. 
You have a coin where the probability of heads $p$ is chosen
by {\bf rnd}. Once this random $p$ is chosen, the coin is
tosses $n$ times. If a head turns up you say you have a white
ball, and if a tail turns up you have a black ball.  Then the
probability for any particular sequence of colors for the
balls is exactly the same as the probability of this
sequence occurring in the Polya urn model. In the coin
model it is obvious that the proportion of heads will tend
to a limit which is again a random number since it just depends upon
what kind of a coin was chosen by {\bf rnd}.   This
random coin model will be discussed more in the next
section.
\endNE\NE
\e{21}%exer 4.1.19
No.
\endNE\NE
\e{22}%exer 4.1.20
1/2
\endNE\NE
\e{23}%exer 4.1.21
Put one
white ball in one urn and all the rest in the other urn. 
This gives a probability of nearly 3/4, in particular greater
than 1/2, for obtaining a white ball which is what you would have with an equal
number of balls in each urn.  Thus the best choice must have more
white balls in one urn than the other.  In the urn with
more white balls, the best we can do is to have probability
1 of getting a white ball if this urn is chosen.  In the
 urn with less white balls than black, the best we can do is
to have one less white ball than black and 
then to have as many white balls as possible.  Our
solution is thus best for the urn with more white
balls than black and also for the urn with more black
balls than white. Therefore our solution is the best
we can do.
\endNE\NE 
\e{24}%exer 4.1.22
$P$($A$ head on the $j$th trial and a total of $k$ heads in
$n$ trials) =
$\displaystyle{\Bigl({1\over2}}\Bigr)^n{{n-1}\choose{k-1}}\ .$
\itemitem{}
$P$(Exactly $k$ heads in $n$ trials)$
=\displaystyle{\Bigl({1\over2}\Bigr)^n}{\displaystyle{n}\choose
{k}}\ .$
 \itemitem{}Thus\hfil\break
{$P$({Head on $j$'th trial} $\ \vert$\ {{$k$ heads in $n$
trials}})} $=
{\displaystyle{{n-1}\choose{k-1}}\over\displaystyle{n\choose{k}}}=
{\displaystyle{k\over n}}\ .$ 
\endNE\NE
\e{25}%exer 4.1.23
We must have
$$p{n\choose{j}}{p^k}{q^{n-k}}= 
p{{n-1}\choose{k-1}}p^{k-1}q^{n-k}\ .$$
This will be true if and only if $np = k$.  Thus $p$ must equal $k/n$.
\endNE\NE
\e{26}%exer 4.1.24
$P(A\vert B) = P(B\vert A)$
implies that $P(A) = P(B)\ .$ \vskip 20pt
\indent\indent Thus, since since $P(A\cap B) > 0,$ 
$$
1 = P(A\cup B) = P(A) + P(B) - P(A\cap B) < 2P(A)\ ,
$$
\itemitem{}and $P(A) > {1\over 2}.$
\endNE\NE
\e{27}%exer 4.1.25
\itemitem{\hskip 10pt}(a) $P$(Pickwick has no umbrella, given that it
rains)$ = \displaystyle{2\over9}\ .$
\vskip 20pt
\itemitem{\hskip 10pt}(b) $P$(It does not rain, given that
he brings his umbrella)$\displaystyle= {5\over12}\ .$
\endNE\NE
\e{28}%exer 4.1.26
The most obvious objection is the
assumption that all of the events in question are
independent.  A more subtle objection is that, since Los
Angeles is so large, it is reasonable to ask for the
probability that there is a second couple with the same
discription, given that there is one such couple.\ This
probability is not so small.\ (See Exercise 23 of Section
9.3.)
\endNE\NE
\e{29}%exer 4.1.27
$P$(Accepted by Dartmouth $\vert$
Accepted by Harvard) $= \displaystyle{2\over3}\ .$
\vskip 5pt
\itemitem{}The events `Accepted
by Dartmouth' and `Accepted by Harvard' are not
independent.
\endNE\NE
\e{30}%exer 4.1.28
Neither has a convincing
argument  based upon comparing grouped data only. You
need more information.  For example, suppose that each
defective bulb cost \$10 whether a regular or a softglow bulb.  Then in making
3000 bulbs the loss to $A$ is \$130  and to $B$ is \$110 so B has a smaller loss
than A.  But suppose that a defective regular bulb results in a loss of \$20 and
a defective softglow bulb in a loss of \$10. Now making 3000 bulbs $A$
has a loss of $2\times \$20 + \$11\times \$10 = \$150$ while $B$ has a
loss of $5\times \$20 + 6\times \$10 =  \$160$. Thus $A$ has a smaller loss than
$B$. 
   Paradoxes caused by comparing percentages when data are grouped are
called \bf Simpson paradoxes \rm.  An interesting real life example can be
found in  Parsani, and Purvis,  W.W. Norton l978.  In a study of the
admission to graduate school at the University of
California, Berkeley in 1973 it was found that about 44\%
of the men who applied were admitted,\ but 35\% of the women
who applied were admitted.  Suspecting sex bias, an attempt was made to
locate where it occurred by examining the individual
departments.   But within individual departments there did
not seem to be any bias. Indeed the only department with a
significant difference was one that favored women.  Again
more information is needed. In this case the explantion
lay in the fact that the women applied to majors
that were difficult to get into (lower acceptance rate)
and men applied generally to the majors that were easy to
get in (high acceptance rate).
    In each of these examples you are interested in comparing one trait
in the presence of a second confounding trait. In the first example it was good
or bad bulb confounded by the type of bulb and in the second example it was sex
confounded by the major. Freedman et al. discuss how to control the confounding
trait to make a more valid comparison.

 \endNE\NE
\e{31}%exer 4.1.29
The probability of a 60 year old
male living to 80 is .41, and for a female it is .62.
\endNE\NE
\e{32}%exer 4.1.30
(a)\ \  $\displaystyle{pq}$
\vskip 3pt
\itemitem{}(b)\ \ $\displaystyle{1 - (1-p)(1-q)}$
\vskip 3pt
\itemitem{}(c)\ \ .958
\endNE\NE
\e{33}%exer 4.1.31
You have to make a lot of
calculations, all of which are like this:

$$\eqalign {P (\tilde {A_1}\cap A_2 \cap A_3) &=
 P(A_2)P(A_3)-
P(A_1)P(A_2)P(A_3)\cr &= P(A_2)P(A_3)(1-P(A_1))\cr& =
P(\tilde A_1)P(A_2)P(A_3).}$$
\endNE\NE
\e{34}%exer 5.1.4
$P_{X_j}= \pmatrix{1&0\cr1\over
4&3\over4\cr}.$
\vskip 5pt
\itemitem{} They are not independent. For example, if
we know that \hfill\break $X_1 = 1, X_2 = 1,
\hbox{and} \ X_3 = 1$, then it must be the case that $X_4 =
1$.
\endNE\NE
\e{35}%exer 5.1.5
The random variables $X_1$ and $X_2$ have the same distributions, and in each
case
the range values are the integers between 1 and 10.  The probability for each
value
is 1/10.  They are independent.  If the first number is not replaced, the two 
distributions are the same as before but the two random variables are not
independent.
\endNE\NE
\e{36}%exer 5.1.6
$p=\pmatrix{1&2&3&4&5&6\cr{11\over
36}&{9\over 36}&7\over 36&5\over 36&3\over 36&1\over 36}.$
\endNE %FOR exer 5.1.7
\vskip .1in
$\eqalign{\hskip -10pt{37.\hskip
15pt}P(\hbox{max}(X,Y)=a)&=P(X=a,Y\le  a)+P(X\le
a,Y=a)-P(X=a,Y=a).\cr P(\hbox{min}(X,Y)=a) &= P(X = a,Y >
a)+P(X>a,Y=a)+P(X=a,Y=a).\cr}$ 
\vskip 5pt 
\itemitem{} Thus
$P(\hbox{max}(X,Y)=a)+P(\hbox{min}(X,Y)=a) = P(X = a) + P(Y
= a)$ 
\itemitem{} and so $u = t+s-r$.
\NE
\e{38}%exer 5.1.8
(a)\quad $p_{_X} =
\pmatrix{0&1&2\cr1\over 4&1\over 2&1\over 4},\quad p_{_Y}
=\pmatrix{0&1\cr1\over 2&1\over 2}.$ \vskip 5pt
\itemitem{}(b) \quad$ p_{_Z} = \pmatrix{0&1&2&3\cr 1\over
8&3\over 8&3\over 8&1\over 8}.$
\vskip 5pt
\itemitem{}(c)\quad $p_{_W} = \pmatrix{-1&0&1&2\cr1\over
8&3\over 8&3\over 8&1\over 8}.$
\endNE\NE
\e{39}%exer 5.1.10
(a)\ \ 1/9
\vskip 3pt
\itemitem{}(b)\ \ 1/4
\vskip 3pt
\itemitem{}(c)\ \ No
\itemitem{}(d)
$\ \ p_{_Z}=\pmatrix{{-2}&{-1}&0&1&2&4\cr 1\over 6&1\over
6&1\over 6&{1\over 6}&1\over 6&{1\over
6}}$
\endNE\NE
\e{40}%exer 5.1.11
$p = 1/2\quad p_X = \pmatrix{{0}&{1}\cr {1/2}&{1/2}}\quad 
p_Y = \pmatrix{{3}&{4}&{5}\cr{1/4}&{3/8}&{3/8}}\quad$ Independent
\vskip 2pt
\itemitem{}$p = 2/3\quad p_X = \pmatrix{{0}&{1}\cr{17/81}&{64/81}}\quad
p_Y = \pmatrix{{3}&{4}&{5}\cr{1/3}&{10/27}&{8/27}}\quad$ Not independent
\endNE\NE
\e{42}%exer 5.1.13
Let $u = N - r$ and $v = N
- s$ be the number of games that $A$ and $B$,
respectively, must win to win the series. Then the
series will surely be over in $u+v - 1$ games, so Fermat
extended the game to assure this many plays.  The player
with the most points in the extended game wins.  Therefore,
           $$P(r,s) = P(u,v) = \sum_{j=
u}^{u+v-1}{{u+v-1}\choose j}p^jq^{u+v-1-j}.$$
    An  alternative formula can be derived without
extending the game.  To win the series in $u+j$ games, $A$
must win  $u-1$ games among the first $u+j-1$
games and then win the  ${(u+j)}{\rm th}$ game with  $j \le
v-1$.\ Thus, $$P(r,s)=P(u,v) =
\sum_{j=0}^{v-1}{{u+j-1}\choose j}p^uq^j.$$  
\endNE\NE
\e{43}%exer 5.1.14
.710.
\endNE\NE
\e{44}%exer 5.1.15
 \exindent (a) \hskip .5em\bf First
convention.\rm \  If you serve $N+1$ times, then your
opponent must serve $N$ times.  The total number of points
played is $2N+1$,  so one
of you must have won at least $N+1$ points.  That is a
contradition, since the game is over when a player has won $N$
points.\hfill\break {\hskip 10pt}\bf Second convention.
\rm\  If you serve $N+1$ times, then except for the first
time, before each time you serve, you have won a point. 
Thus at the $(N+1)${st} time you serve you have
already won $N$ points.  The game should have already
ended. Therefore, you serve at most $N$ times. Before each
serve of your apponent he won the previous point. Thus, as
he serves for the $N${th} time he has already won $N$
points. Therefore, your opponent serves at most $N-1$
times. 
 \vskip 5pt \itemitem{}\exindent
(b)\hskip.5em Since the total number of points for the two
players is $2N-1$, and one player has already got $N$ points
before the game is extended, the other can get at most $N-1$
points in the extended game and hence not change the
winner. 
 \vskip 5pt
\itemitem{}\exindent(c)\hskip 1em For the extended game, 
probabilistically, the two methods are
the same: in one, we have $N$ Bernoulli trials with probability $p$
for success and in the other we have $N-1$ trials with
probability $\bar p$ for success, and in either method you win if
you win the most points. 
\endNE\NE
\e{45}%exer 5.1.15.5
\itemitem{}(a) The probability that the first player wins under
either service convention is equal to the probability that if a
coin has probability $p$ of coming up heads, and the coin is tossed
$2N + 1$ times, then it comes up heads more often than tails.  This
probability is clearly greater than .5 if and only if $p > .5$.
\vskip 5pt
\itemitem{}(b)\hskip 1em If the first team is serving on a given play,
it will win the next point if and only if one of the following sequences
of plays occurs (where `W' means that the team that is serving wins the play,
and 
`L' means that the team that is serving loses the play):
$$W,\ LLW,\ LLLLW,\ \ldots\ .$$
The probability that this happens is equal to
$$p + q^2p + q^4p + \ldots\ ,$$
which equals
$${p\over{1 - q^2}} = {1\over{1+q}}\ .$$
Now, consider the game where a `new play' is defined to be a sequence of
plays that ends with a point being scored.  Then the service convention is
that at the beginning of a new play, the team that won the last new play serves.
This is the same convention as the second convention in the preceding problem.
\itemitem{}\hskip 10pt
>From part a), we know that the first team to serve under the second service
convention will win the game more than half the time if and only if $p > .5$.  
In the present case, we use the new value of $p$, which is $1/(1+q)$.  This is
easily seen to be greater than .5 as long as $q < 1$.  Thus, as long as $p > 0$,
the first team to serve will win the game more than half the time.


\endNE\NE
\e{46}%exer 5.1.19
$\displaystyle{P(X = i) = P(Y = i) = 
{\displaystyle{{4\choose i}{\displaystyle{5-i}\choose 48}}\over
{\displaystyle{52\choose 5}}}}\ ,$
\vskip 3pt
\itemitem{}$\displaystyle{P(X = i,\ Y = j) = 
{\displaystyle{{4\choose i}{4\choose j}{44\choose {5 - i - j}}}
\over\displaystyle{52\choose 5}}}$ if $i \le 4,\ j \le 4,\ $and $i+j \le 5,$
\vskip 3pt
\itemitem{}$\displaystyle{P(X = i,\ Y = j) = 0,\ \hbox{otherwise.}}$
\endNE 

%FOR exer 5.1.24
$\eqalign{ \hskip -8pt 47.\hskip 15pt(a)\quad P(Y_1 = r,
Y_2 = s)& = P(\Phi_1(X_1) = r, \Phi_2(X_2) = s)\cr  
& = \sum_{\Phi_1(a) = r\atop
\Phi_2(b) = s}P(X_1 = a, X_2= b)\ .\cr}$
 \vskip 5pt
\itemitem{}(b) \ If $X_1, X_2$ are independent, then 
$$\eqalign {P(Y_1=r,Y_2 = s)& =\sum_{\Phi_1(a) = r\atop
\Phi_2(b) = s}P(X_1 = a, X_2= b)\cr & =
\sum_{\Phi_1(a) = r\atop
\Phi_2(b) = s}P(X_1 = a)P( X_2= b)\cr & =
\Bigr(\sum_{\Phi_1(a) = r} P(X_1 = a)\Bigl)
\Bigr(\sum_{\Phi_2(b) = s}P(X_2 = b)\Bigl)\cr 
& = P(\Phi_1(X_1) =
r)P(\Phi_2(X_2) = s)\cr &=P(Y_1=r)P(Y_2 = s)\ ,}$$
\itemitem{}\quad  so $Y_1 \ \hbox{and}\  Y_2$ are
independent.
\NE
\e{48}%exer 4.1.32
$$\eqalign{\sum_{\omega\in \Omega} m_{_E}(\omega)& =
{1\over{P(E)}}\sum_{\omega\in\Omega}P(\omega\cap
E)\cr &={1\over{P(E)}}P(E) = 1\ .}$$
\endNE\NE
\e{49}%exer 4.1.33
 $P$(both coins turn up
using (a)$) ={1\over 2}p_1^2+{1\over 2}p_2^2.$
\vskip 5pt
\qquad $P$(both coins turn up heads using (b)) =
$ p_1p_2. $
\itemitem{}Since $(p_1-p_2)^2=p_1^2-2p_1p_2+ p_2^2 > 0,$
we see that $p_1p_2 <{1\over 2}p_1^2+{1\over 2}p_2^2$, and so (a) is 
better.
\endNE\NE
\e{50}%exer 4.1.34
For any sequence $B_1,\dots ,B_n$ with $B_k = A_k$ or
$\tilde A_k$ 
$$P(B_1\cap\cdots\cap B_n) = P(B_1)P(B_2)\cdots P(B_n) >
0\ .$$
\itemitem{}Thus there is at least one sample point $\omega$
in each of the sets $B_1 \cap B_2 \cap \cdots \cap B_n$.  Since
there are $2^n$ such subsets, there must be at least this
many sample points.
\endNE\NE
\e{51}%exer 4.1.35
$$\eqalign{ P(A) & = P(A\vert
C)P(C) + P(A\vert\tilde C)P(\tilde C)\cr&\ge P(B\vert
C)P(C) + P(B \vert \tilde C)P(\tilde C) = P(B)\ .}$$
\endNE\NE
\e{52}%exer 4.1.36
$$\eqalign{P&(\hbox{coin not found in the}\  i{\rm 'th}
\ \hbox{box})\cr&=P(\hbox{coin not in} \  i{\rm 'th}\ 
\hbox{box}) + P(\hbox{coin in} \  i{\rm 'th} \ \hbox{box but
not found)}\cr &=1-p_i+(1-a_i)p_i\cr & = 1 - a_ip_i\ .}$$ 
\itemitem{} Thus $${P(\hbox{coin is in} 
\ \  j{\rm 'th}\ \  \hbox{box} \ \vert\ \hbox{not found in}
 \ \  i{\rm 'th}\ \  \hbox{box}) =
p_j}/(1-a_ip_i) \ \ if \  j \ne i.$$
$$\eqalign{&\hbox{P(coin is in the} \ \ 
i{\rm 'th}\ \  \hbox{box} \ \vert\  \hbox{not found in the}
\ \  i{\rm 'th}\ \  \hbox{box})\cr&= 1- \sum_{j\ne i}
\hbox{P(coin is in}\ \  j{\rm 'th}\ \  \hbox{box} \ \vert\ 
\hbox{not found in}\ \  i{\rm 'th}\  \hbox{ box})\cr &= 1 -
{{1-p_i}\over{(1-a_ip_i)}}\cr & =
{{(1-a_i)p_i.}\over{1-a_ip_i}}.}$$
\endNE
\e{53}%exer 4.1.37
We assume that John and
Mary sign up for two courses.  Their cards are
dropped, one of the cards gets stepped on, and
only one course can be read on this card. 
Call card I the card that was not stepped on and
on which the registrar can read government 35 and
mathematics 23;  call card II the card that was stepped
on and on which he can just read mathematics 23.\ There
are four possibilities for these two cards. They are:
\vskip 20pt {\settabs\+    
\indent\indent\qquad&Mary(other,math)\qquad
&Mary(other,math)\qquad&Prob.\qquad&Cond.Prob\cr
 \+& Card I  & Card II & Prob. & Cond. Prob.\cr
\+& Mary(gov,math) & John(gov, math) &
.0015 & .224\cr
\+&Mary(gov,math) & John(other,math) &
.0025 &.373\cr 
\+&John(gov,math) & Mary(gov,math) & .0015
&.224\cr
\+ &John(gov,math)& Mary(other,math) &.0012
& .179\cr}
\vskip 20pt
\itemitem{}In the third column we have written the
probability that  each case will occur. For
example, for the first one we compute  the probability
that the students will take the appropriate courses:
$.5\times.1\times.3\times.2\ = .0030$ and then we
multiply by 1/2, the probability that it was John's card
that was stepped on. 
Now to get the conditional probabilities we must
renormalize these probabilities so that they add up to one.  In this way
we obtain the results in the last column.  From this we
see that the probability that card I is Mary's is .597
and that card I is John's is .403,  so it is more likely
that that the card on which the registrar sees Mathematics
23 and Government 35 is Mary's. 
\NE 
\e{54}%exer 4.1.38
\IND{(a)} Let 
$$A =
\{(a,\bullet){\bf:} a \in \hbox{a subset $\cal A$  of}
\ \{ \clubsuit,\diamondsuit,\heartsuit,\spadesuit\},
\bullet \in \{2,3,\dots,J,Q,K,A\}\}
$$
\itemitem{}\hskip 10pt \itemitem{} be a suit event and
$$
B = \{(\bullet,b) {\bf:} b \in \hbox{a subset $\cal
B $\ of}\  \rm\{2,3,\dots,J,Q,K,A\}, \bullet \in\ 
\{\clubsuit,\diamondsuit,\heartsuit,\spadesuit\}\}
$$
\itemitem{}\hskip 10pt \itemitem{} be a rank event. Then
\vskip 5pt
$$
\eqalign{P( A)& ={{\hbox{size of} \ {\cal A}}\over{ 4}}\cr
P(B) &= {{\hbox{size of} \ {\cal B}}\over{13}}\cr
P(A\cap B) &= P\{(a,b) {\bf :}  a \in {\cal A}, b \in
{\cal B}\} = {{(\hbox{size of}\ {\cal A})(\hbox{size of}
\ {\cal B})}\over{ 52}} = P(A)P(B)\ .}
$$
\vskip 5pt
\itemitem{}\IND{(b)}  The possible sizes of a rank event are
(4$i$ + 3$j$), where $i$ = 0,\dots,12 and \hfill\break $j$= 0 or 1, and the
possible sizes of a suit event are $(13m + 12n)$, where\hfill\break $m$ =
0,1,2,3, and $n$ = 0,1. By the hint we must have
$${(4i+3j)(13m+12n)}\over 51$$ an integer. This means that either ${4i+3}$
or ${m + 12n}$ must be divisible by 17.  We show that this is not
possible.  Assume for example that ${4i+3j = 17k}$\ where $k$ is 1 or 2.
Since ${17k = 16k + k}$,  when 17$k$ is divided by 4 we
would get a remainder of $k$, that is, 1 or 2.  But when ${4i + 3j}$ is divided
by 4 we get a remainder of 3$j$, which is 0 or 3 depending on the value of
$j$.  Therefore we cannot have ${4i+3j = 17k}$, for $k$= 1 or 2 and $j$ = 0 or
1.
Similarly, assume that ${13m+12n = 17k}$ with $k$ = 1 or 2.  Since ${17k = 13k +
4k}$, when $17k$ is divided by 13 we get a remainder of 4 or 8 depending on
the value of $k$, but when {13$m$ + 12$n$} is divided by 13 we get
a remainder of 0 or 12 depending on the value of\ $n$. Thus
we cannot have ${17k = 13m + 12n}$ for $k$ = 1 or 2 and $n$ = 0 or
1.  

\vskip 5pt
\itemitem{}\IND{(c)} $$\eqalign{A& =
\{(\spadesuit,2),(\spadesuit,3),(\spadesuit,4)\}\cr
B&= \{(\spadesuit,4),\dots,(\spadesuit,7),
(\heartsuit,1),\dots,(\heartsuit,K)\}.}$$
\IN{(d)} Let $a$ be the size of $A$, $b$ the size of $B$, and $c$
the size of $A\cap B.$  Then $A$ and $B$ independent implies
that $${c\over 53} = {a\over 53}\cdot {b\over53}.$$
Thus $ab = 53c.$  But,
since 53 is prime, this means that either $a$ or $b$ must be 53,
which means that either $A$ or\ $B$ must be trivial.
\endNE\NE
\e{55}%exer 4.1.39
$$P(R_1) = {4\over
\displaystyle{52\choose 5}} = {1.54\times 10^{-6}}.$$ 
$$P(R_2 \cap R_1) = {{4\cdot 3 }\over
{\displaystyle{52\choose 5} \displaystyle{47\choose 5}}}\ . 
$$ 
\itemitem{}Thus
$$P(R_2 \ \vert\ R_1) =
\displaystyle{3\over\displaystyle{47\choose 5}} =
1.96\times10^{-6}.$$   \itemitem{} Since $P(R_2\vert R_1) >
P(R_1),$ a royal flush is attractive.
\vskip 20pt
\itemitem{}${P\hbox{(player 2 has a full house)} =
{\displaystyle{13\cdot12\displaystyle{4\choose 3}\displaystyle{4\choose
2}}\over \displaystyle{52\choose 5}}}\ .$
 \itemitem{}$P$(player 1 has a flush
and player 2 has a full house) =
$${{4\cdot8\cdot7\displaystyle{4\choose3}
\displaystyle{4\choose2} +
4\cdot8\cdot5\displaystyle{4\choose3}\cdot
\displaystyle{3\choose2} + 4\cdot
5\cdot8\cdot\displaystyle{3\choose3}{4\choose2} +
4\cdot5\cdot4\displaystyle{3\choose3}\displaystyle{3\choose2}}\over
{\displaystyle{52\choose5}\displaystyle{47\choose5}}}\ .$$
\itemitem{} Taking the ratio of these last two quantities gives:
$$\hbox{P(player 1 has a royal flush} \ \vert \ \hbox{player 2 has
a full house)}
= 1.479\times10^{-6}.$$
 \itemitem{}Since this probability is less
than the probability that player 1 has a royal flush
($1.54\times 10^{-6}$), a full house repels a royal flush.
\endNE\NE
\e{56}%exer 4.1.40
$$\eqalign {P(B\vert A) > P(B) &\Leftrightarrow P(B\cap A)
> P(A)P(B)\cr &\Leftrightarrow P(A\vert B) = {{P(A\cap
B)}\over{P(B)}} > P(A)\ .}$$
\endNE\NE
\e{57}%exer 4.1.41
\vskip -30pt
 $$\eqalign {\cr & P(B\vert A)\le P(B)
\hbox{\ and\ }P(B\vert A) \ge P(A)\cr &\Leftrightarrow
P(B\cap A)\le P(A)P(B)\hbox{\ and\ } P(B\cap A) \ge
P(A)P(B)\cr&\Leftrightarrow  P(A\cap B) = P(A)P(B)\ .}$$
\endNE\NE
\e{58}%exer 4.1.42
$$\eqalign{P(A\vert B) >
P(A)&\Leftrightarrow P(A\cap B) > P(A)P(B) \cr
&\Leftrightarrow P(A\cap B) - P(A)P(A\cap B) > P(A)P(B) -
P(A)P(B\cap A)\cr &\Leftrightarrow P(A\cap B)P(\tilde A)
> P(A)P({B\cap}{\tilde A})\cr &\Leftrightarrow
{{P(A\cap B)}\over{P(A)}} > {{P(B\cap\tilde A)}\over
{P(\tilde A)}}\cr &\Leftrightarrow
{P(B\vert A)} > P(B\vert  \tilde A)\ .}$$
\endNE\NE
\e{59}%exer 4.1.43
Since $A$ attracts $B$,\ $P(B\vert A)
> P(A)$ and $$P(B\cap A) > P(A)P(B)\ ,$$ 
and so 
$$P(A) -
P(B\cap A) < P(A) - P(A)P(B)\ .$$ 
Therefore, 
$$P(\tilde B\cap A) <
P(A)P(\tilde B)\ ,$$  
$$P(\tilde B\vert A) < P(\tilde
B)\ ,$$ 
and  $A$ repels $\tilde B.$  
\endNE\NE
\e{60}%exer 4.1.44
If $A$ attracts $B$ and $C$, and $A$
repels $B\cap C$ then we have
$$P(A\cap B) > P(A)P(B)\ ,$$ 
$$ P(A\cap C) > P(A)P(C)\ ,$$ 
$$P(B\cap C\cap A) < P(B\cap C)P(A)\ .$$
Thus
$$\eqalign{P(A\cap (B\cup C)) &= P((A\cap B)\cup (A\cap
C))\cr & = P(A\cap B) + P(A\cap C) - P(A\cap B\cap C)\cr
&>P(A)P(B) + P(A)P(C) - P(B\cap C)P(A)\cr
&= P(A)(P(B) + P(C) - P(B\cap C))\cr
&= P(A)P(B\cup C)\ .}$$
Therefore $$P(B\cup C \vert A) > P(B\cup C)\ .$$
Here is an example in which $A$ attracts $B$ and $C$ and repels
$B\cup C.$ Let 
$$\Omega = \{a,b,c,d\}\ ,$$ 
$$p(a) = .2,\ p(b) = .25, \ p(c) = .25,\  p(d) = .3\ .$$ 
Let $$A = \{a,d\},
B = \{b,d\}, C = \{c,d\}\ .$$ 
Then $$P(B\vert A) = .6 > P(B)
= .55\ ,$$ 
$$P(C\vert A) = .6 > P(C) = .55\ ,$$ 
and
$$P(B\cup
C\vert A) = .6 < P(B\cup C) = .8\ .$$
\endNE\NE
\e{61}%exer 4.1.45
Assume that $A$ attracts $B_1$, but
$A$ does not repel any of the $B_j$'s. Then
$$P(A\cap B_1) > P(A)P(B_1),$$ and $$P(A\cap B_j) \ge
P(A)P(B_j), \ \  \ \ 1 \le j \le n.$$  Then
$$\eqalign{P(A) &= P(A\cap \Omega)\cr 
&=P(A\cap (B_1\cup\dots\cup B_n))\cr
&=P(A\cap B_1) + \cdots + P(A\cap B_n)\cr
&> P(A)P(B_1)  + \cdots + P(A)P(B_n)\cr
&= P(A)\Bigl(P(B_1)  +\cdots + P(B_n)\Bigr)\cr
&= P(A)\ ,}$$
which is a contradiction.
\endNE

\vskip 20pt
%||*****||*****||*****||%sec4.2
\vskip 20pt

%\input mydefinitions
\vskip 20pt
\indent\bf SECTION 4.2\rm
\NE
\e{1}%exer 4.2.1
(a)\ \ 2/3
\vskip 3pt
\itemitem{}(b)\ \ 1/3
\vskip 3pt
\itemitem{}(c)\ \ 1/2
\vskip 3pt
\itemitem{}(d)\ \ 1/2

\endNE\NE
\e{2}%exer 4.2.2
(a)\ \ $1- e^{-.09}=.086$
\vskip 3pt
\itemitem{}(b)\ \ $1-e^{-.5}= .393$
\vskip 3pt
\itemitem{}(c)\ \ 1
\vskip 3pt
\itemitem{}(d)\ \ $(1-e^{-1})/(1-e^{-2})= .731$
\endNE\NE
\e{3}%exer 4.2.3
(a)\ \ .01
\vskip 3pt
\itemitem{}(b)\ \ $\displaystyle{e^{-.01\,T}}$ where $T$ is the time after
20 hours.
\vskip 3pt
\itemitem{}(c)\ \ $\displaystyle{e^{-.2}\approx .819}$
\vskip 3pt
\itemitem{}(d)\ \ $\displaystyle{1 - e^{-.01} \approx .010}$
\endNE\NE
\e{4}%exer 4.2.4
(a)\ \ $1/2$
\vskip 3pt
\itemitem{}(b)\ \ 1/4
\vskip 3pt
\itemitem{}(c)\ \ 3/4
\vskip 3pt
\itemitem{}(d)\ \ 1/2
\endNE\NE
\e{5}%exer 4.2.5
(a)\ \ 1
\vskip 3pt
\itemitem{}(b)\ \ 1
\vskip 3pt
\itemitem{}(c)\ \ 1/2
\vskip 3pt
\itemitem{}(d)\ \ $\pi/8$
\vskip 3pt
\itemitem{}(e)\ \ 1/2
\endNE\NE
\e{6}%exer 4.2.6
\exindent(a)\quad $f(x) =
\cases{{4/3}, & $\ \hbox{if}\ {1/4} < x < 1$,\cr 0,
&otherwise.\cr}$ \vskip 5pt
\itemitem{} (b) \quad $f(t) =
\cases{{{e^{-t}}/({e^{-1} - e^{-10}})}, &$ \hbox{if}\ 1 < t
< 10,$\cr 0, & otherwise. \cr}$ \vskip 5pt
\itemitem{} (c) \quad $f(x,y) = \cases{{\pi/ 50},& if
$(x,y)$ is in upper half of the target,\cr 0,
&otherwise.\cr}$ \vskip 5pt
\itemitem{} (d) \quad $f(x,y) = \cases{2, &if $x > y, $\cr
0, & otherwise.\cr}$
\endNE\NE
\e{7}%exer 4.2.7
$\displaystyle{P(X > {1\over3}, Y > {2\over3}) =
\int_{1\over3}^1\int_{2\over3}^1{dydx} = {2\over9}\ .}$ 
\vskip 5pt
\itemitem{} But $\displaystyle{P(X > {1\over 3})P(Y > {2\over 3}) =
{2\over 3}\cdot{1\over 3}\ ,}$  so $X$ and $Y$ are
independent.
\endNE\NE
\e{8}%exer 4.2.8
$a$ and
$b$; $c$ and $d$.
\endNE\NE
\e{10}%exer 4.2.9
\exindent (b)\ \ Let $Z$ be a number chosen with uniform density on the
interval $[0,a]$. We find the density for 
max$(Z,a-Z)$. This density is nonzero only on the interval
[${a/ 2}$,$a$]. For $x$ in this interval: \vskip 5pt
\itemitem{}\hskip 20pt$\displaystyle{\eqalign{ P(\hbox{max}(Z,a-Z)) \le x) &= 
P(a-Z \ge Z,a-Z \le x) +P(Z >  a-Z, Z \le x)\cr  &= P(Z \ge
{a-x}, Z \le {a\over 2}) + P(Z > {a\over 2}, Z \le x)\cr
&= {{2x - a}\over a}\ .\cr}}$  
\vskip 5pt
\itemitem{}Taking $a =1,$ we see that the density for the
length of the largest stick from the first cut is
uniform  on the interval $[{1\over 2}, 1]$.  Assume
that the length of this longest piece is $X$. Let $Y$ be
position on the interval $[0,X]$ of the second cut.  Then
we obtain a triangle if\  max$(Y,X - Y) \le {{1\over
2}}$.  But by our first computation with $a = X$ and $x =
{1\over 2}$, we see that $$P({\hbox{max}(Y,X-Y)) \le
{1\over 2}} = {\Bigr({{1-X}\over X}\Bigl)}\ .$$ 
\itemitem{\hskip 1em}Thus  $$P(\hbox{triangle}) =
2\int_{1\over 2}^1\Bigr({{1-x}\over x}\Bigl)dx = {2\log 2}
- 1\ .$$
\endNE\NE 
\e{11}%exer 4.2.10
If you have drawn $n$ times (total number of balls in the urn 
is now $n+2$) and gotten $j$ black balls, (total number of black 
balls is now $j+1$), then the probability of getting a black
ball next time is ${(j+1)}/{(n+2)}$. Thus at each time
 the conditional probability for the next outcome is
the same in the two models.  This means that the models are
determined by the same probability distribution, so either
model can be used in making predictions.  Now in the coin
model, it is clear that the proportion of heads will tend
to the unknown bias $p$ in the long run.  Since the
value of p was assumed to be unformly distributed, this
limiting value has a random value between 0 and 1.  Since
this is true in the coin model, it is also true in the
Polya Urn model for the proportion of black balls.(See
Exercise 20 of Section 4.1.) 
\endNE\NE
\e{12}%exer 4.2.11
A new beta density with $\alpha = 6$ and $\beta = 9$.  It will be 
successful next time with probability .4.
\endNE

\vskip 20pt
%||*****||*****||*****||%sec4.3
\vskip 20pt

%\input mydefinitions
\vskip 20pt
\indent\bf SECTION 4.3\rm
\NE
\e{1}%exer 4.3.1
2/3
\endNE\NE
\e{2. \hskip 10pt}%exer 4.3.2
Let $M$ be the event that the hand has an ace.  Let $N$ be the event that the
hand has at least two aces.  Then
$$P(N|M) = {{P(N \cap M)}\over{P(M)}} = {{P(N)}\over{P(M)}} = 
{\displaystyle{{4\choose 2}{48\choose 11} + {4\choose 3}{48\choose 10} +
{4\choose 4}{48\choose 9}}\over{\displaystyle{52\choose 13}-{48\choose 13}}}
\approx
.3696\ .$$
\itemitem{} Let $S$ be the event that the hand has the ace of hearts.  Then
$$P(N|S) = {{P(S \cap T)}\over{P(S)}} = 
{\displaystyle{{3\choose 1}{48\choose 11} + {3\choose 2}{48\choose 10} + 
{3\choose 3}{48\choose 9}}\over{\displaystyle{52\choose 13} - {51\choose 13}}}
\approx  .5612\ .$$
\endNE\NE
\e{3}%exer 4.3.3
(a)\ \ Consider a tree where the first branching corresponds to the number of
aces held by the player, and the second branching corresponds to whether the
player answers `ace of hearts' or anything else, when asked to name an ace in
his
hand.  Then there are four branches, corresponding to the numbers 1, 2, 3, and
4,
and each of these except the first splits into two branches.  Thus, there are
seven
paths in this tree, four of which correspond to the answer `ace of hearts.'  The
conditional probability that he has a second ace, given that he has answered 
`ace of hearts,' is therefore
$${\displaystyle{\Biggl(\biggl({48\choose 12} + {1\over 2}{3\choose
1}{48\choose 11} + {1\over 3}{3\choose 2}{48\choose 10}
+ {1\over 4}{3\choose 3}{48\choose 9}\biggr)}\Bigl/{{52\choose
13}\Biggr)}\over \displaystyle{\Biggl({51\choose 12}\Bigl/{52\choose
13}\Biggr)}} \approx .6962\ .$$
\vskip 3pt
\itemitem{}(b)\ \ This answer is the same as the second answer in Exercise 2,
namely .5612.
\endNE\NE
\e{5}%exer 4.3.5
Let $x = 2^k$.  It is easy to check that if $k \ge 1$, then
$${p_{x/2}\over{p_{x/2} + p_x}} = {3 \over 4}\ .$$
If $x = 1$, then
$${p_{x/2}\over{p_{x/2} + p_x}} = 0\ .$$
Thus, you should switch if and only if your envelope contains 1.
\endNE

\vskip 20pt
%||*****||*****||*****||%sec5.1
\vskip 20pt

%\input mydefinitions

\indent\bf SECTION 5.1\rm
\NE
\e{1}%exer 5.1.100
(a), (c), (d)
\endNE\NE
\e{2}%exer 5.1.101
Yes.
\endNE\NE
\e{3}%exer 5.1.102
Assume that $X$ is uniformly distributed, and let the countable set of values be
$\displaystyle{\{\omega_1, \omega_2, \ldots\}}$.  Let $p$ be the probability 
assigned to each outcome by the distribution function $f$ of $X$.  If $p > 0$,
then
$$\sum_{i = 1}^\infty f(\omega_i) = \sum_{i = 1}^\infty p\ ,$$
and this last sum does not converge.  If $p = 0$, then 
$$\sum_{i = 1}^\infty f(\omega_i) = 0\ .$$
So, in both cases, we arrive at a contradiction, since for a distribution
function,
we must have
$$\sum_{i = 1}^\infty f(\omega_i) = 1\ .$$
\endNE\NE
\e{4}%exer 5.1.103
One way to help decide whether the given experiment will result in a random
subset
is to ask whether one might reasonably expect any of the possible subsets to
occur
if the experiment is performed.  Since some close friends almost always eat
lunch
together, the first experiment will almost never give a subset which has as a
member exactly one of a pair of close friends.  The second experiment will
always
give the same subset when performed on the same student body.  In addition,
Social
Security numbers are assigned based upon geography, among other things.  Thus,
depending upon the use we are going to make of this subset, this experiment may
or
may not give an adequate subset.  Finally, the third experiment will probably
return each subset with approximately the same probability.  (A tenth-floor
window
would be much better.)
\endNE\NE
\e{5}%exer 5.1.104
(b)\ \ Ask the Registrar to sort by using the sixth, seventh, and ninth digits
in
the Social Security numbers.
\vskip 3pt
\itemitem{}(c)\ \ Shuffle the cards 20 times and then take the top 100 cards. 
(Can
you think of a method of shuffling 3000 cards?
\endNE\NE
\e{6}%exer 5.1.105
The distribution function of $Y$ is given by
$$f(x) = {{(k-x +1)^n - (k - x)^n}\over{k^n}}\ ,$$
for $1 \le x \le k$.  The numerator counts the number of $n$-tuples, all of
whose
entries are at least $x$, and subtracts the number of $n$-tuples, all of whose
entries are at least $x+1$.
\endNE\NE
\e{7}%exer 5.1.16
(a)\quad
$\displaystyle{p_{_j}(n) = {1\over 6}\Bigr({5\over 6}\Bigl)^{n-1}} \hbox{\ for}\ 
j
= 0,1,2,\dots.$ 
\vskip 5pt 
\itemitem{}(b)\quad 
$\displaystyle{{P(T > 3)} = ({5\over 6})^3 = {125\over 216}}\ 
.$ 
\vskip 5pt
\itemitem{}(c)\quad $\displaystyle{ P(T > 6\ \vert \ T > 3) = {({5\over 6})^3} =
{125\over216}}\ .$
\endNE\NE
\e{8}%exer 5.1.18
$\displaystyle{1\over 8}\ .$
\endNE\NE
\e{9}%exer 5.1.106
(a)\ \ 1000
\vskip 3pt
\itemitem{}(b)\ \ $\displaystyle{{{100\choose 10}{N-100\choose
90}}\over{N\choose
100}}$
\vskip 3pt
\itemitem{}(c)\ \ $\displaystyle{N = 999\ {\rm or}\ N = 1000}$
\endNE\NE
\e{10}%exer 5.1.107
(a)\ \ $\displaystyle{{{N \choose k}{N-k \choose n_1 - k}{N - n_1\choose 
n_2 - k}}\over{{N\choose n_1}{N\choose n_2}}}$
\vskip 5pt
\itemitem{}(b)\ \ Let $p_N$ denote the probability that $X = n_{12}$, given that
the population size is $N$.  Then $p_N$ equals the expression in the answer to
part
(a), with $k$ replaced by $n_{12}$.  After some gruesome algebra, one obtains
$${{p_{N+1}}\over{p_N}} = {{N^2 - (n_1 + n_2 - 2)N + (n_1 - 1)(n_2 - 1)}
\over{N^2 - (n_1 + n_2 - n_{12} - 2)N + (n_{12} + 1 - n_1 - n_2}}\ .$$
Let $a_N$ and $b_N$ denote the numerator and denominator of this expression.  We
want the smallest value of $N$ for which the expression is less than or equal to
1,
or equivalently, we want the smallest value of $N$ for which $a_N \le b_N$.  If
we 
solve this inequality for $N$, we obtain
$$N = \Bigl\lceil{{n_1
n_2}\over{n_{12}}}\Bigr\rceil - 1\ .$$
\endNE\NE
\e{13}%exer 5.1.108
.7408, .2222, .0370
\endNE\NE
\e{14}%exer 5.1.109
(a)$\ \ \  e^{-10} \approx 4.54\times 10^{-5}$
\vskip 3pt
\itemitem{}(b)\ \ We need $\displaystyle{e^{-n/1000} = 1/2, \ \hbox{or} \ n =
1000\cdot \hbox{log} 2 \approx 694.}$
\endNE\NE
\e{16}%exer 5.1.111
$P$(miss 0 calls) + $P$(miss 1 call) = .0498+.1494 =
.1992.
\endNE\NE
\e{17}%exer 5.1.112
649741
\endNE\NE
\e{18}%exer 5.1.113
(a)\ $ m = 600\times \D{1\over 500}$ so $P$(no
raisins) =  $e^{-m} = .301.$ \vskip 5pt
\itemitem{} (b)\ $ m = 400\times \D{1\over 500}$,  so
$P$(exactly two chocolate chips) = $\D{m^2e^{-m}\over 2!} =
.144.$ \vskip 5pt
\itemitem{} (c)\  $m = 1000\times\D{1\over 500},$  so$$
P(\hbox{at least two bits}) = 1 - P(\hbox{0 bits}) -
P(\hbox{1 bit}) = 1 - .1353-.2707 = .594.$$
\endNE\NE
\e{19}%exer 5.1.114
The probability of at least one call in a given day
with $n$ hands of bridge can be estimated by $1- e^{-n\cdot
(6.3\times 10^{-12})}.$  To have an average  of one per
year we would want this to be equal to ${1\over 365}.$ 
This would require that $n$  be about 400,000,000 and 
that the players play on the average 8,700
hands a day.  Very unlikely! It's much more likely
 that someone is playing a practical joke.
\endNE\NE
\e{20}%exer 5.1.115
$\displaystyle{e^{-5} \approx .00674}$
\endNE\NE
\e{21}%exer 5.1.116
(a)\ \ $b(32,j,\D{1/ 80}) = \D{32\choose
j}\D\Bigl({1\over 80}\Bigr)^j\D\Bigl({79\over
80}\Bigr)^{32-j}$
\vskip 3pt
\itemitem{}(b)\ \ Use $\lambda = 32/80 = 2/5$.  The approximate probability that 
a given student is called on $j$ times is $\displaystyle{e^{-2/5}(2/5)^j/j!}\,.$ 
Thus, the approximate probability that a given student is called on more than
twice
is
$$1 - e^{-2/5}\biggl({{(2/5)^0}\over{0!}} + {{(2/5)^1}\over{1!}} + 
{{(2/5)^2}\over{2!}}\biggr) \approx .0079\ .$$
\endNE\NE
\e{22}%exer 5.1.29
.0077
\endNE\NE
\e{23}%exer 5.1.117
$$\hbox{$P$(outcome is $j+1$)/P(outcome is
$j$)}\  = \D{{m^{j+1}e^{-m}}\over
{(j+1)!}}\Bigr/{{m^je^{-m}}\over {j!}} = {m\over {j+1}}.$$
Thus when $j+1 \le m$, the probability is increasing, and
when $j+1 \ge m$ it is decreasing. Therefore, $j = m$ is a
maximum value. If $m$ is an integer, then the ratio will
be one for $j = m-1$, and so both $j =m-1$ and $j = m$
will be maximum values. (cf. Exercise 7 of Chapter 3,
Section 2) 
\endNE\NE
\e{24}%exer 5.1.118
The probability that Kemeny
receives no mail on a given weekday can be estimated by
$e^{-10} = 4.54\times 10^{-5}.$  Thus in ten years, the
probability that at least one day brings no mail can be
estimated by $1 - e^{-3000\cdot4.54\times 10^{-5}} =
.127.$  Thus he finds that the probability is .127, which
is inconclusive.
\endNE\NE
\e{25}%exer 5.1.119
Without paying the
meter Prosser pays $$2\cdot {5e^{-5}\over {1!}} + (5\cdot
2){{5^2e^{-5}}\over {2!}} + \cdots (5\cdot n)
{{5^ne^{-5}}\over {n!}} + \cdots = 25 - 15e^{-5}
 = \$24.90.$$ He is better off putting a dime in the
meter each time for a total cost of \$10.
\endNE
\e{26}%exer 9.2.15
{\settabs\+&number\qquad&observed\qquad&expected\qquad&\cr 
\+&\hskip 35pt number&\hskip 35pt observed&
\hskip 30pt expected\cr
\+&&&\cr
\+&\hfill 0&\hfill 229&\hfill 227&\cr
\+&\hfill 1&\hfill 211&\hfill 211&\cr
\+&\hfill 2&\hfill 93&\hfill 99&\cr
\+&\hfill 3&\hfill  35&\hfill 31&\cr
\+&\hfill 4&\hfill 7&\hfill 9&\cr
\+&\hfill 5&\hfill 1&\hfill 1&\cr}
\NE
\e{27}%exer 5.1.120
 $m$ = $100\times (.001)$ = .1.  Thus $P$(at least
one accident) = $1- e^{-.1}$ = .0952.
\endNE\NE
\e{28}%exer 5.1.121
.9084
\endNE\NE
\e{29}%exer 5.1.122
Here $m$ = $500\times (1/ 500)$ = 1, and so
$P$(at least one fake) = $1 - e^{-1}$ = .632.  If the king
tests two coins from each of 250 boxes, then $m$ =$
250\times \D{2\over 500} = 1$, and so the answer is again
.632.
\endNE\NE
\e{30}%exer 5.1.123
$\displaystyle{P({\rm win}\ \ge 3\ {\rm times}) \approx .5071,\ {\rm expected\
winnings}\ \approx\ -2.703}$
\endNE
\e{31}%exer 9.2.20
The expected number of deaths per corps per year is $$1\cdot {91\over
280} + 2\cdot {32\over 280} + 3\cdot {11\over 280} + 4\cdot
{2\over 280} = .70.$$ The expected number of corps with
$x$ deaths would then be $280\cdot
\D{{(.70)^xe^{-(.70)}}\over {x!}}.$ From this we obtain
the following comparison: \vskip 5pt
{\settabs\+&Number of deaths\qquad&corps with $x$
deaths\qquad&Expected number\qquad&\cr 
\+&\hskip 50pt Number of deaths&\hskip 50pt Corps with $x$
deaths&\hskip 50pt Expected number of Corps \cr
\+&&&\cr
\+&\hfill 0&\hfill144&\hfill139.0&\cr
\+&\hfill 1&\hfill 91&\hfill 97.3&\cr
\+&\hfill 2&\hfill 32&\hfill 34.1&\cr
\+&\hfill 3&\hfill 11&\hfill 7.9&\cr
\+&\hfill $\ge 4$&\hfill 2&\hfill 1.6&\cr}
\vskip 5pt
\itemitem{} The fit is quite good.
\NE
\e{32}%exer 9.2.21
$$\eqalign{P(X + Y = j)& = \sum_{k=0}^jP(X = k)P(Y = j-k) =
\sum_{k=0}^j{{{m^k}e^{-m}}\over {k!}}\cdot
{{{\bar m}^{j-k}e^{-\bar m}}\over {(j-k)!}}\cr &=
e^{-(m+\bar m)}\Bigl (\sum_{k=0}^j {{j!m^k\bar
m^{j-k}}\over {k!(j-k)!}}\Bigr ){1\over {j!}} 
\cr 
&= e^{-(m+\bar m)}{1\over {j!}}\Bigl(\sum_{k=0}^j
{j\choose k} m^k\bar m^{j-k}\Bigl) = {{(m+\bar m)^j}\over
{j!}}e^{-{(m+\bar m)}}.}$$
Thus, $X + Y$ has a Poisson density with mean $m+ \bar m.$
\endNE\NE
\e{33}%exer 9.2.22
Poisson with  mean 3.
\endNE\NE
\e{34}%exer 9.2.23
.168
\endNE\NE
\e{35}%exer 5.1.20
\IND{(a)}\hskip .5em In order
to have $d$ defective items in $s$ items, you must choose $d$
items out of $D$ defective ones and the rest from $S-D$ good
ones. The total number of sample points is the number of
ways to choose $s$ out of $S$. \vskip 5pt \itemitem{}(b)
\hskip .5em Since $$\sum_{j=0}^{\hbox{min}(D,s)}P(X = j) = 1,$$
\itemitem{}\quad we get  $$\sum_{j = 0}^{\hbox{min}(D,s)}{D\choose j}
{{s-D}\choose {s-j}} = {S\choose s}\ .$$
\endNE\NE
\e{36}%exer 5.1.21
$D = 20$.  This illustrates the general fact
that the maximum probability is achieved when
$${d\over D} = {s\over S}\ .$$
\endNE\NE
\e{37}%exer 5.1.22
The maximum likelihood principle gives an estimate of 1250 moose.
\endNE\NE
\e{38}%exer 5.1.23
With replacement:  $P(X = 1) \approx .396$
\vskip 2pt
\itemitem{} Without replacement:  $P(X = 1) \approx .440$
\endNE\NE
\e{40}%exer 5.1.24
(a)\quad $
{{\displaystyle{4!\over{2}}\displaystyle{13\choose
4}\displaystyle{13\choose 4}\displaystyle{13\choose
3}\displaystyle{13\choose 2}}\over {\displaystyle{52\choose
13}}} = .2155.$ \vskip 5pt \itemitem{}(b)\quad $
{{\displaystyle{4!\over{2}}\displaystyle{13\choose
5}\displaystyle{13\choose 3}\displaystyle{13\choose
3}\displaystyle{13\choose 2}}\over {\displaystyle{52\choose
13}}} = .1552.$ \vskip 5pt

\endNE\NE
\e{42}%exer 7.1.9
$p_{_{X_1}} = \pmatrix{0&4\cr{1\over 3}&{2\over 3}},\quad
p_{_{X_2}} = \pmatrix{3\cr 1},\quad p_{_{X_3}} = \pmatrix{2&6\cr{2\over
3}&{1\over 3}}, \quad p_{_{X_4}} = \pmatrix{1&5\cr {1\over 2}&{1\over
2}}.$\vskip 5pt\itemitem{} If your friend chooses die 1, choose die 4;
if she chooses die 2, choose die 1; if she chooses die 3, choose die 2;
if she chooses die 4, choose die 3.  Then $$P(X_1< X_4)=P(X_2 < X_1) =
P(X_3< X_2) = P(X_4< X_3) = {2\over 3}.$$  Thus you are assured of
winning with probability 2/3.
\endNE\NE
\e{43}%exer 5.1.25
If the traits were independent, then the probability that we would obtain
a data set that differs from the expected data set by as much as the actual
data set differs is approximately .00151.  Thus, we should reject the hypothesis
that the two traits are independent.
\endNE\NE
\e{44}%exer 5.1.126
The value of $\chi^2$ corresponding to the data is $v = 9931.6$, which is much
greater
than $v_0$, so the hypothesis that the chosen numbers are uniformly distributed
should be rejected.
\endNE

\vskip 20pt
%||*****||*****||*****||%sec5.2
\vskip 20pt

%\input mydefinitions
\indent\bf SECTION 5.2\rm
\NE
\e{1}%exer 5.2.1
(a)\ \ $\displaystyle{f(x) = 1\ {\rm on}\ [2, 3]; F(x) = x - 2\ {\rm on}\ [2,
3]}$.
\vskip 3pt
\itemitem{}(b)\ \ $\displaystyle{f(x) = {1\over 3}x^{-2/3}\ {\rm on}\ [0, 1];\ 
F(x) = x^{1/3}\ {\rm on}\ [0, 1].}$
\endNE 
%FOR EXER 5.2.2
\settabs\+\indent\indent&xxxx&xxxxxxxxxxxxxx&xxxxxxxxxx&xxxx&\cr
{\+\e{2}&(a)& $ F(x) = 2 - {1\over
x}$,&$f(x) = {1\over{x^2}}$&{on} &$[{{1\over 2}, 1}].$\cr
\vskip 5pt 
\+ &(b)& $ F(x) =e^x-1$, & $f(x) = e^x$&on
&$[0, \log 2].$\cr}

%FOR EXER 5.2.4
\settabs\+\indent\indent&xxxx&xxxxxxxxxxxxxx&xxxxxxxxxx&xxxx&\cr
{\+\e{5}&(a)& $F(x) = 2x$,&  $f(x) =2$ &
on&$[0,{1\over 2}].$ \cr  \vskip 5pt
\+&(b)& $F(x) =2\sqrt x$, & $f(x) = {1\over{\sqrt x}}$
& on& $[0,{1\over 4}].$\cr}
\endNE\NE
\e{7}%exer 5.2.6
Using Corollary 5.2, we see that the expression $\sqrt {rnd}$ will simulate the
given
random variable.
\endNE\NE
\e{9}%exer 5.2.8
(a)\quad$F(y)=\cases{{y^2\over 2},
& $0\le y \le 1$;\cr 1-{{(2-y)^2}\over 2},&$1\le y \le
2,$ \cr}$ $f(y) = \cases {y,&$0\le y \le 1$;\cr 2-y &
$1\le y \le 2.$\cr}$ \vskip 5pt
\itemitem{} (b)\quad $F(y) = 2y - y^2, \qquad f(y) =
2-2y,\quad 0 \le y \le 1.$
\endNE\NE
\e{10}%exer 5.2.9
(a)\ \ $F(x) = x^2$ and $f(x) = 2x$ on $[0, 1]$.
\vskip 3pt
\itemitem{}(b)\ \ $F(x) = 2x - x^2$ and $f(x) = 2 - 2x$ on $[0, 1]$.
\endNE\NE
\e{12}%exer 5.2.11
(a)\ \ 1/2
\vskip 3pt
\itemitem{}(b)\ \ 1
\vskip 3pt
\itemitem{}(c)\ \ .2
\endNE
\itemitem{13.}%exer 5.2.12
{\settabs
\+\indent 10pt&\quad xxxxxxx\quad xxxxxxxxxxxxxx &\qquad xxxxxxxxxx
&\cr 
\+\hskip 14pt &(a)\quad $\displaystyle{F(r) =\sqrt r}$\ , & 
$\displaystyle{f(r) ={1\over{2 \sqrt r}}}\ ,$ &on \quad[0,1].\cr
\vskip 5pt 
\+&(b)\quad $\displaystyle{F(s) = 1-\sqrt{1-4s}}$\ , 
&$\displaystyle{f(s)= {2\over{\sqrt{1-4s}}}}$\ , & on \quad $[0,{1\over 4}].$\cr 
\vskip 5pt
\+&(c)\quad $\displaystyle{F(t) = {t\over{1+t}}}\ ,$
&$\displaystyle{f(t) = {1\over {(1+t)}^2}\ ,}$& on\quad $[0,\infty].$\cr}
\NE
\e{14}%exer 5.2.13
(a)\ \ 3/4
\vskip 3pt
\itemitem{}(b)\ \ $\pi/16$
\endNE\NE
\e{15}%exer 5.2.14
$F(d) = 1-(1-2d)^2,\qquad f(d) =
4(1-2d)\qquad \hbox{on}\quad [0,{1\over 2}].$
\endNE\NE
\e{16}%exer 5.2.15
(a)\ \ c = 6
\vskip 3pt
\itemitem{}(b)\ \ $F(x) =3x^2 - 2x^3$
\vskip 3pt
\itemitem{}(c)\ \ .156
\endNE\NE
\e{17}%exer 5.2.16
(a) $f(x) =\cases{{\pi\over 2}\sin ({\pi x}),& $0\le x \le
1$;\cr
 0, & otherwise.\cr}$
\vskip 3pt
\itemitem{}(b) \quad ${{\hbox{sin}^2}({\pi\over 8})} =
.146.$ 
\endNE\NE
\e{18}%exer 5.2.17
$F_{_W}(w)$ =
\vskip 5pt
\itemitem{}$a > 0:\quad F_{_X}({{w-b}\over a});\quad  a =
0:\quad$ $\cases{1,&$ w \ge b$; \cr 0,&otherwise;}$$\quad
a< 0:\quad 1-F_{_X}({{w-b}\over a}).$
\endNE\NE
\e{19}%exer 5.2.18
a $\ne 0:f_{_W}(w) = \
{1\over{\vert {a}\vert}} {f_{_X}({{w-b}\over a})},$ \ \  a
= 0: $f_{_W}(w) = 0 \ \hbox{if}\  w \ne 0. $
\endNE\NE
\e{20}%exer 5.2.19
(a)\ \ $\displaystyle{{1\over {d-c}}}$
\vskip 3pt
\itemitem{}(b)\ \ $\displaystyle{{c\over {c-d}}}$
\endNE\NE
\e{21}%exer 5.2.20
$P(Y \le y) = P(F(X) \le y) = P(X
\le F^{-1}(y)) = F(F^{-1}(y)) = y \quad\hbox{on}\quad
[0,1].$
\endNE\NE 
\e{22}%exer 5.2.2
(a)\ \ $\displaystyle{{{a+b}\over 2}}$
\vskip 3pt
\itemitem{}(b)\ \ $\mu$
\vskip 3pt
\itemitem{}(c)\ \ $\displaystyle{{1\over \lambda} \log 2}$
\endNE\NE
\e{23}%exer 5.2.22
The mean of the uniform density
is ${(a+b)}/ 2$.  The mean of the normal density is $\mu$.  The mean of the
exponential density is ${1/\lambda}$.
\endNE\NE
\e{24}%exer 5.2.23
The mode of the uniform density is any number in $[0, 1]$.  The mode of the
normal
is $\mu$.  The mode of the exponential is 0.
\endNE\NE
\e{25}%exer 5.2.24
(a)\quad .9773,\quad (b)\quad
.159,\quad (c) \quad.0228,\quad (d) \quad.6827.
\endNE\NE
\e{26}%exer 5.2.25
13.4\% are likely to be rejected.  For 1\% rejection rate, let $\sigma = .0012$.
\endNE\NE
\e{27}%exer 5.2.26
A:\quad 15.9\%,\quad B: \quad
34.13\%,\quad C:\quad 34.13\%,\quad D:\quad 13.59\%,\quad
F:\quad 2.28\%.
\endNE\NE
\e{28}%exer 5.2.27
2.4\%
\endNE\NE
\e{29}%exer 5.2.28
$e^{-2},e^{-2}.$
\endNE\NE
\e{30}%exer 5.2.29
The car will last for 4 years with probability $1/e \approx .368$.
\endNE\NE
\e{31}%exer 5.2.30
${1\over 2}.$
\endNE\NE
\e{34}%exer 5.2.33
$$ P(X < Y) =
\int_{x=0}^\infty\int_{y=x}^\infty f(x)g(y)dxdy =
\int_{x=0}^\infty f(x)(1-G(x))dx.$$ \itemitem{} Thus
$$P(X < Y) = \int_0^\infty \lambda e^{-\lambda x}\cdot e^{-\mu
x}dx = {\lambda\over{\lambda + \mu}}.$$
\itemitem{} Therefore, the probability that a 100 watt
bulb will outlast a 60 watt bulb is ${{(1/
200)}\over{{1/200} + {1/ 100}}} = {1/ 3}.$
\endNE\NE 
\e{35}%exer 5.2.34
$P$(size increases) =
$P(X_j<Y_j) = {\lambda/ ({\lambda + \mu}}). $ \vskip 5pt
\itemitem{} $P$(size decreases) = $1-P$(size increases) =
$\mu/ ({\lambda+\mu}).$ 
\endNE\NE
\e{36}%exer 5.2.36
Exponential with parameter
$\lambda/ r$.
\endNE\NE
\e{37}%exer 5.2.37
$\displaystyle{F_Y(y) = {1\over{\sqrt{2\pi y}}}e^{-{{\log^2(y)}\over 2}}}$\ , 
for $y > 0$.
\endNE\NE
\e{38}%exer 5.2.38
$$\eqalign{P(Y_1 \le y_1, Y_2 \le y_2)& = P(X_1
\le \Phi_1^{-1}(y_1),X_2\le \Phi_2^{-1}(y_2))\cr
&= P(X_1 \le \Phi_1^{-1}(y_1))P(X_2\le
\Phi_2^{-1}(y_2))\cr & = P(Y_1\le y_1)P(Y_2 \le
y_2),\cr}$$  \itemitem{}  so $Y_1$ and $Y_2$ are
independent.
\endNE

\vskip 20pt
%||*****||*****||*****||%sec6.1
\vskip 20pt

%Modified on 6/1/96
%\input mydefinitions

\indent\bf SECTION 6.1\rm
\NE
\e{1}%exer 6.1.1
-1/9
\endNE\NE
\e{2}%exer 6.1.2
-1/2
\endNE\NE
\e{3}%exer 6.1.3
$5'\,10.1"$
\endNE\NE
\e{4}%exer 6.1.4
-1/19
\endNE\NE
\e{5}%exer 6.1.5
-1/19
\endNE\NE
\e{6}%exer 6.1.6
Let $U$ and $V$ be independent
identically distributed random variables with the density:

$$p_{_U} = \pmatrix{1&2&3&4&5&6\cr1\over 6&1\over 6&1\over
6&1\over 6&1\over 6&1\over 6}.$$
\itemitem{} Then $$XY = (U+V)(U-V) = U^2 -
V^2,$$\itemitem{} so $$E(XY) = E(U^2) - E(V^2) = 0.$$ 
\itemitem{}Since $$E(Y) = E(U) - E(V) = 0,$$ we have
$$E(XY) = E(X)E(Y) = 0. $$\itemitem{} But $X$ and $Y$ are not
independent.  For example, if we know that $X = 12$, then we
know that $Y = 0$.
\endNE\NE
\e{7}%exer 6.1.7
Since $X$ and
$Y$ each take on only two values, we may choose $a, b, c, d$ so
that$${U = {X+a\over b}},{V = {Y + c\over d}}$$ take only values 0
and 1. If $E(XY) = E(X)E(Y)$ then $E(UV) = E(U)E(V)$.\ If $U$
and $V$ are independent, so are $X$ and $Y$. Thus it is sufficient to
prove independence for $U$ and $V$ taking on values 0 and 1 with $E(UV) =
E(U)E(V)$.Now $$E(UV) = P(U = 1, V = 1) = E(U)E(V) = P(U = 1)P(V = 1),$$
and $$\eqalign{P(U = 1, V = 0)& = P(U = 1) - P(U = 1, V
=1)\cr &= P(U=1)(1-P(V=1)) = P(U=1)P(V=0).}$$
 Similarly,
$$\eqalign{P(U = 0, V = 1)& = P(U=0)P(V = 1)\cr  P(U = 0,V = 0) &= P(U =
0)P(V=0).}$$ \itemitem{} Thus $U$ and $V$ are independent, and hence 
$X$ and $Y$ are also.
\endNE\NE
\e{8}%exer 6.1.8
The expected number of boys and
the expected number of girls are both  $7\over 8$. 
\endNE\NE
\e{9}%exer 6.1.9
The second bet is a fair bet so has expected winning 0.  Thus your expected
winning for the two 
bets is the same as the original bet which was -7/498 = -.0141414...  On the
other hand, you bet 1 dollar with probability  $1/3$ 
and  2 dollars with probability 2/3.  Thus
the expected amount you bet is 1${2\over 3}$ dollars and your expected winning
per dollar bet is
-.0141414/1.666667 = -.0085 which makes this option a better bet in terms of the
amount won per dollar bet.
However, the amount of time to make the second bet is negligible, so 
in terms of the expected winning per time to make one play the answer would
still be -.0141414.
\endNE\NE
\e{11}%exer 6.1.11
The roller has expected winning -.0141; the pass bettor has expected winning
-.0136.
\endNE\NE
\e{12}%exer 6.1.12
0
\endNE\NE
\e{13}%exer 6.1.13
45
\endNE\NE
\e{14}%exer 6.1.14
$\displaystyle{E(X_j) = {1\over N}}\ .$ For $\displaystyle{j \not= k,\ 
E(X_jE_k) = {1\over {N(N-1)}}}$\ . Thus $X_j$ and $X_k$ are not
independent.
\endNE\NE
\e{15}%exer 6.1.16
$E(X) = {1\over 5}$,  so this is
a favorable game.
\endNE\NE
\e{16}%exer 6.1.17
(a)$$ E(X) = {(1+2+3+4+5+6)\over 6} =
{3{1\over 2}}.$$
\itemitem{} \exindent(b) \hskip .5em The large sums are much less likely
to occur than small sums. For example $$P(\hbox{total} = 21) = ({1/6})^6=
2.14\times 10^{-5}$$ {and}  $$P(\hbox{total} = 0) = ({5/6})^6 =.335.$$
\endNE\NE
\e{17}%exer 6.1.18
$\displaystyle{p_k = p(\overbrace{S\cdots S}^{k-1\rm \
times}F) = p^{k-1}(1-p) = p^{k-1}q, \ k = 1,2,3,\dots}\ .$\hfill\break 
\vskip 5pt 
$\displaystyle{\hskip 20pt{\sum_{k=1}^\infty p_k} =
{q\sum_{k=0}^\infty p^k} = {q{1\over {1-p}}} = 1\ .}$\hfill\break
\vskip 5pt 
\hskip 20pt
$\displaystyle{E(X) = q\sum_{k=1}^\infty kp^{k-1} = {q\over {(1-p)^2}} = {1\over
q}\ .}$
(See Example 6.4.)
\endNE\NE
\e{18}%exer 6.1.19
7/2
\endNE\NE
\e{19}%exer 6.1.20
\vskip 5pt
$\displaystyle{\eqalign{\ \ \ \ E(X) = {{{4\choose 4}\over{4\choose 4}}{(3-3)}}
&+
{{{3\choose 2}\over {4\choose 3}}{(3-2)}} + {{{3\choose 3}\over{4\choose
3}}{(0-3)}}+{{{3\choose 1}\over {4\choose 2}}{(3-1)}}\cr
&+{{{3\choose 2}\over{4\choose
2}}{(0-2)}}+{{{3\choose 0}\over{4\choose 1}}{(3-0)}}+{{{3\choose 1}\over
{4\choose 1}}{(0-1)}} = 0}\ .}$
\endNE\NE
\e{20}%exer 6.1.21
(a)\ \ $\displaystyle{E(X) = {1\over 2}\cdot 2 +
{\Bigr({1\over 2}\Bigl)^2 }\cdot 2^2 + \Bigr({1\over 2}\Bigl)^3\cdot 2^3
\cdots  = 1 + 1 + 1 + \cdots =\infty\ ,}$ and so E(X) does not exist. This
means that if we could play the game, it would be favorable now matter
how much we pay to play it.  However, we cannot realize this game, since
it requires arbitrarly large amounts of money.
\vskip 5pt
\itemitem{}(b) $$\eqalign{E(X) & = {1\over 2}\cdot 2 + {\Bigl({1\over
2}\Bigr)^2}\cdot {2^2} + \cdots +\Bigl({1\over 2}\Bigr)^{10} \cdot 2^{10}
+ {2^{10}} {\Bigr({1\over {2^{11}}} + {1\over {2^{12}}}+ \cdots \Bigl)}\cr
&= 10 + {1\over 2} + {1\over 2^2} + \cdots = 11\ .}$$
\vskip 5pt
\itemitem{}(d)\ \ If the utility of $n$ dollars is $\sqrt n$, then the expected
utility of the payment is given by
$$\sum_{i = 1}^\infty {1\over {2^i}} \sqrt{2^i} = {1\over{\sqrt 2 - 1}}\ .$$
\qquad If the uility of $n$ dollars is $\log n$, then the expected utility of
the payment
is given by
$$\sum_{i = 1}^\infty {1\over {2^i}} \log(2^i) = 2\log 2\ .$$
\endNE\NE
\e{22}%exer 6.1.22
The expected number of days in a year with more
than 60 percent boys for the large hospital is
$$ 365\cdot\sum_{k=28}^{k=45}b(45,.5,k) = 24.67\ .$$
For the small hospital it is
$$365\cdot\sum_{k=10}^{k=15}b(15,.5,k) = 55.1\ .$$
\endNE\NE
\e{23}%exer 6.1.23
10
\endNE\NE
\e{25}%exer 6.1.25
\itemitem{}\exindent(b)\hskip .5em Let $S$ be the number of stars and $C$
the number of circles left in the deck.  Guess star if $S > C$
and guess circle if $S < C$.  If $S = C$ toss a coin.
\itemitem{} \exindent(d)\hskip .5em Consider the recursion relation:
$$h(S,C) = {{\hbox{max}(S,C)}\over {S+C}} + {{S\over {S+C}}h(S-1,C)} +
{{C\over {S+C}}h(S,C-1)}$$
 and $h(0,0) = h(-1,0)=h(0,-1) = 0$.  In this equation the
first term represents your expected winning on the current guess and
the next two terms represent your expected total winning on the
remaining guesses. The value of $h(10,10)$ is 12.34.
\endNE\NE
\e{26}%exer 6.1.26
\exindent(a)\hskip 1em Let $L$  be the
horizontal line passing through $S-C$.  If the random walk is below $L$,
then there are more stars than circles in the remaining
deck, and so, using the optimal strategy, you guess star. 
If you are right, the graph goes up.
 If the walk is above $L$, then there are more
circles than stars, and you guess circle.\ If you are right,
the graph goes down. Since $S \ge C$, the graph ends at
$(S+C,S-C)$.  Let $a$ be the number of times  the
graph goes up under $L, b$ the number of times it goes down
under $L, c$ the number of times it goes down above $L$, and
d the number of times it goes up above $L$. Then $a+b+c+d =
S+C, a-b  = S-C, c-d = 0$.  Thus $2a-S+C+2c = S + C$, and
this implies $a+c= S$, i.e., we have $S$ correct guesses.
\vskip 5pt\itemitem{}\exindent(b)\hskip .5em We arrive at $(x,x)$ if
$S-x$ stars turn up and $C-x$ circles turn up in  \hbox{$S + C - 2x$}
guesses. The probability of this happening is $${{{S\choose
{S-x}}{C\choose {C-x}}}\over {{S+C}\choose {S+C - 2x}}} = {{{S\choose
x}{C\choose x}}\over {{S+C}\choose 2x}}\ .$$
\itemitem{}\exindent(c)\hskip .5em The number of correct guesses
equals the number of correct guesses when the graph is
under or above $L$ plus the number of correct guesses when
the graph hits $L$. Thus the expected number of correct
guesses is:  $$S + \sum_{x=1}^C{{S\choose
x}{{C\choose x}}\over {S+C\choose 2x}}\cdot {1\over 2}\ .$$
\endNE\NE
\e{27}%exer 6.1.27
(a)\ \ 4 
\itemitem{} (b)\ \ $\displaystyle{4 + {\sum_{x=1}^4
{{4\choose x}{4\choose x}\over
{8\choose x}}} = 5.79\ .}$
\endNE\NE
\e{28}%exer 6.1.28
\exindent(a)\hskip .5 em Assume that $n = 2^k-1$. Choose the middle
number of the numbers from $1 \ \hbox{to} \ 2^k-1$, and then continue to
choose the middle number until you guess the number correctly. 
If you have not yet succeeded after $k-1$ guesses you will
be down to a single number and will be sure to get it on the $k$th
question. This strategy obviously works just as well if $n < {2^{k-1}}$. 
\vskip 5pt
\itemitem{}\exindent(b)\hskip 1em
Whenever you make a guess and are wrong the search is
narrowed to a new and smaller interval $[a,b]$.  The probability that you
guess correctly on a question when the interval is $[a,b]$ is
$$P\hbox{(correct)} = {(b-a)\over n}\cdot {1\over (b-a)} = {1\over n}\ .$$
Thus the probabililty that you guess the number on the $k$th question is
$a(k)/n$ where $a(k)$ is the number of possible subintervals for the
$k$th question. The probability of guessing the number correctly for a
strategy with at most $k$ guesses is $${{\sum_k a(k)}\over n}\ .$$  Thus any
strategy that makes this sum as large as possible is optimal.  We can at
most double the number of intervals on each question.Thus any strategy
that achieves this is optimal.The resulting probability of guessing the
number in $k$ questions is 
 $${{\sum_{j=0}^{k-1} {2^k}}\over n} = {{2^{k} - 1}\over n}\ .$$
If $n \ge {2^k-1}$ we can achieve this optimal strategy by continuing to
bisect the numbers between 1 and $2^k-1$.
\endNE\NE
\e{29}%exer 6.1.29
If you have no ten-cards and the dealer has
an ace, then in the remaining 49 cards there are 16 ten cards.  Thus
the expected payoff of your insurance bet is:
$$ 2\cdot {16\over 49} - 1\cdot {33\over 49} = {-{1\over 49}}\ .$$
If you are playing two hands and do not have any ten-cards then there
are 16 ten-cards in the remaining 47 cards and your expected payoff on
an insurance bet is:
$$ 2\cdot{16\over 47} - 1\cdot{31\over 47} = {1\over 47}\ .$$
Thus in the first case the insurance bet is unfavorable and in the
second it is favorable.
\endNE\NE 
\e{30}%exer 6.1.30
\IND{(a)} $P(X_k= j) $= $P(j-1$ boxes
have old pictures and the $j$th box has a new picture)$$ =
\Bigr({{k-1}\over n}\Bigr)^j\Bigl({{n-k+1}\over n}\Bigr)\ ,$$ 
and so 
$X_k$ has a geometric distribution with $p = {{(n-k+1)}/ n}$.
\vskip 5pt
 \itemitem{}(c) The expected time for getting the first
half of the players is
$$\eqalign{E(X_1) + \cdots + E(X_n) &= {2n\over{2n-1+1}} +
{2n\over{2n-2+1}} + \cdots + {2n\over{2n-n+1}}\cr & =
2n\Bigr({1\over {2n}} + {1\over {2n-1}} + \cdots {1\over
{n+1}}\Bigl)\ .}$$
The expected time for getting the second half of the
players is: 
 $$\eqalign{E(X_{n+1}) + \cdots +E(X_{2n}) &=
{2n\over {2n-(n+1)-1}} + \cdots + {2n\over {2n-2n+1}}\cr &=
2n\Bigr({1\over n} + {1\over {n-1}}+\cdots + {1\over
1}\Bigl)\ .}$$ 
\itemitem{}(d)$$\eqalign{1+ {1\over 2} +
\cdots + {1\over{n}} &\sim \log n + .5772 + {1\over
{2n}}\ .\cr 
1 + {1\over 2} + \cdots + {1\over n} + {1\over
{n+1}} + \cdots + {1\over{2n}} &\sim \log {2n} + .5772 +
{1\over{4n}}\ .\cr 
{2n}\Bigr({{1\over {2n}} + \cdots +
{1\over{n+1}}}\Bigl) &\sim {2n}\Bigr({\log {2n}  +
{1\over {4n}} - {\log n}  - {1\over {2n}}}\Bigl)\ .\cr
& = 2n\Bigr(\log 2 - {1\over{4n}}\Bigl)\cr & = 2n{\log 2}
- {1\over 2}\cr  2n\Bigr({1 + {1\over 2} +
\cdots + {1\over n}} \Bigl)&\sim {2n\Bigr({\log n} + .5772
+ {1\over {2n}}\Bigl)\ .}}$$
\endNE\NE
\e{31}%exer 6.1.31
(a)\ \ $\displaystyle{1-(1-p)^k\ .}$
\vskip 5pt
\itemitem{}(b)\ \ $\displaystyle{{N\over k}\cdot\Bigr((k+1)(1-(1-p)^k) +
(1-p)^k\Bigl)\ .}$
\vskip 5pt
\itemitem{}\exindent(c)\ \ If $p$ is small, then $(1-p)^k \sim
1-kp$, so the expected number in (b) is  \hfill\break$\sim N[kp + {1\over
k}]$, which will be minimized when $k = {1/{ \sqrt p}}.$
\endNE\NE
\e{32}%exer 6.1.32
Your estimate should
be near $e$ = 2.718...
\endNE\NE
\e{33}%exer 6.1.33
We begin by noting that
$$P(X \ge  j+1) = P((t_1+t_2+\cdots + t_j) \le
n)\ .$$ 
Now consider the $j$ numbers $a_1,a_2,\cdots, a_j$ defined
by
$$\eqalign{a_1 &= t_1\cr a_2 &= t_1 + t_2\cr a_3
&=t_1+t_2+t_3\cr \vdots &\qquad \vdots 
\qquad \vdots \cr a_j&=t_1+t_2+\cdots + t_j\ .}$$ 
The sequence $a_1,a_2,\cdots ,a_j$ is a monotone increasing
sequence with distinct values and with successive
differences between 1 and $n$.  There is a one-to-one
correspondence between the set of all such sequences and
the set of possible sequences $t_1,t_2,\cdots ,t_j$. 
Each such possible sequence occurs with probability
$1/{n^j}.$
  In fact, there are $n$ possible values for $t_1$ and hence for $a_1$. 
For each of these there are $n$ possible values for $a_2$ corresponding to
the $n$ possible values of $t_2.$  Continuing in this way we see that
there are $n^j$ possible values for the sequence $a_1,a_2, \cdots ,a_j.$ 
On the other hand, in order to have $t_1+t_2+\cdots + t_j \le n$ the
values of $a_1,a_2,\cdots ,a_j$ must be distinct numbers lying between 
1 to $n$ and arranged in order. The number of ways that we can do this is
${n\choose j}$.  Thus we have 
$$P(t_1+t_2+\cdots+t_j \le n) = P(X \ge
j+1) = {n\choose j}{1\over{n^j}}\ .$$ 
$$\eqalign {E(X) = P(X = 1) &+ P(X =
2) +P(X=3) \cdots\cr &+P(X = 2) + P(X = 3)\cdots \cr &\hskip 56pt +P(X =
3) \cdots \ .\cr}\ .$$ 
If we sum this by rows we see that 
$$E(X) = \sum_{j=0}^{n-1} P(X \ge j+1)\ .$$
Thus, 
$$E(X) = \sum_{j=1}^n {n\choose j}\Bigr({1\over n}\Bigl)^j = 
{\Bigr(1+{1\over n}\Bigl)^n}\ .$$ The limit of
this last expression as $n \to \infty$ is $e$ = 2.718...\ .\hfill\break 
\itemitem{}There is an interesting connection between this problem and
the exponential density discussed in Section 2.2 (Example 2.17).  Assume that
the
experiment starts at time 1 and the time between occurrences is equally likely
to be any value
between 1 and $n$. You start observing at time $n$.  Let  $T$ be the length of
time
that you wait. This is the amount by which  $t_1+t_2+\cdots + t_j$ is
greater than $n$.   Now imagine a sequence of plays of a game in which you
pay $n/ 2$ dollars for each play and for the $j$'th play you receive
the reward  $t_j.$  You play until the first time your total
reward is greater than $n$. Then $X$ is the number of times you play and
your total reward is $n + T$.  This is a perfectly fair game and your
expected net winning should be 0.  But the expected total reward is  $ n
+ E(T) $. Your expected payment for play is ${n\over 2}E(X).$  Thus by
fairness, we have 
$$n+E(T) = {(n/2)}E(X)\ .$$
Therefore, 
$$E(T) = {n\over 2}E(X) - n\ .$$ 
We have seen that for large $n$, $E(X)\sim
e.$\ Thus for large $n$,
$$E(\hbox{waiting time}) = E(T) \sim n({e\over 2} - 1) = .718n\ .$$ 
Since the average time between occurrences is $n/2$ we have
another example of the paradox where we have to wait on the average
longer than 1/2 the average time time between occurrences.
\endNE\NE
\e{34}%exer 6.1.34
(a) We prove first that for Bernoulli trials
the probability that the $k$th failure precedes the $r$th success is
$$f(k,p,r) = {{r+k-1}\choose k }p^{r-1}q^k\cdot p\ .
$$
To prove this, we note that for the $k$th failure to precede the $r$th
success we must have $r-1$ successes and $k$ failures in the first
$r+k-1$ trials and then have a success.   The probability
that this happens is $f(k,p,r)$.  Now consider a Bernoulli trials process
where success is getting a match from the right pocket. In order to have
$r$ matches in the left pocket when the right pocket has none we must
have $N-r$ failures before the $(N+1)${st} success.  
Thus the probability that there are $r$ matches in the left pocket when
the right pocket has none is 
$$ {f(N-r,{1\over 2},N+1)}\ .$$  
The same argument applies for the probability that there are $r$ matches in the
right pocket when the left pocket has none.  Thus 
$$p_r =2f(N-r,{1\over 2},N+1)  = {{2N-r}\choose N}\Bigr({1\over 2}\Bigl)^{2N-r}\
.$$
\vskip 5pt  
\itemitem{} (c) $$\eqalign{(N-{r\over
2})p_{r+1}& = (N-{r\over 2}){{2N-r-1}\choose N} \Bigr({1\over
2}\Bigl)^{2N-r-1}\cr & = (2N-r){{2N-r-1}\choose N}\Bigr({1\over
2}\Bigl)^{2N-r}\cr & = (N-r){{2N-r}\choose N}\Bigr({1\over
2}\Bigl)^{2N-r}\cr & = (N-r)p_r\ .}$$ 
\vskip 5pt 
\itemitem{} (d)\ \ $\displaystyle{\sum_{r=0}^N p_r = 1\ .}$ 
\vskip 5pt 
\itemitem{} (e) $$\sum_{r=0}^N(N-r)p_r = \sum_{r=0}^N{1\over 2}(2N+1)p_{r+1} -
\sum_{r=0}^N{1\over 2}(r+1)p_{r+1}\ .$$ 
\itemitem{}Thus 
$$ N-E = {1\over
2}(2N+1)(1-p_0)-{1\over 2}E\ ,$$
\itemitem{} and 
$$E = p_0(2N+1)-1\ .$$
\itemitem{}But 
$$p_0 = {2N\choose N}\Bigr({1\over 2}\Bigl)^{2N}
\sim{1\over \sqrt{{\pi}N}}\ ,$$ 
\itemitem{}so 
$$E \sim {2\sqrt{N\over \pi}}\ .$$
\itemitem{} Using this asymptotic expression leads to an estimate of 133
for the number of matches needed in each pocket to make $E =13$.  It
is easy to make an exact calculation with the computer, and this gives 153
matches.
\endNE\NE 
\e{35}%exer 6.1.35 
One can make a conditionally convergent series like the alternating harmonic 
series sum to anything one pleases by properly rearranging the series.  
For example, for the order given we have 
$$\eqalign{E &= \sum_{n = 0}^\infty
(-1)^{n+1}{{2^n}\over n}\cdot {1\over 2^n}\cr & = \sum_{n=0}^\infty
(-1)^{n+1}{1\over n} = \log 2\ .\cr}$$ 
But we can rearrange the terms to add up to a negative value by choosing 
negative terms until they add up to more than the first positive term, 
then choosing this positive term, then more negative terms until they add up 
to more than the second positive term, then choosing this positive term, etc.
\endNE\NE
\e{36}%exer 6.1.37
$\displaystyle{c {k\over {c+d}}}$
\endNE\NE
\e{37}%exer 6.1.36
(a) Under option (a), if red turns up, you win 1 franc, if black turns up, you
lose
1 franc, and if 0 turns up, you lose 1/2 franc.  Thus, the expected winnings are
$$1 \Bigl({{18}\over{37}}\Bigr) + (-1)\Bigl({{18}\over{37}}\Bigr) +
\Bigl({{-1}\over{2}}\Bigr)
\Bigl({{1}\over{37}}\Bigr) \approx -.0135\ .$$
\vskip 5pt
\itemitem{} (b) Under option (b), if red turns up, you win 1 franc, if black
turns up, you
lose 1 franc, and if 0 comes up, followed by black or 0, you lose 1 franc. 
Thus, the expected
winnings are
$$1 \Bigl({{18}\over{37}}\Bigr) + (-1)\Bigl({{18}\over{37}}\Bigr) + (-1)
\Bigl({{1}\over{37}}\Bigr)\Bigl({{19}\over{37}}\Bigr) \approx -.0139\ .$$
\vskip 5pt
\itemitem{} (c) 
\endNE\NE
\e{38}%exer 6.1.38
(Solution by Victor Hern\'{a}ndez)
Let $p_{ij}$ be the probability that book $i$ is above book $j$.  Then the
average depth of
book $j$ is
$$d_j = \sum_{i \ne j} p_{ij}\ ,$$
where the top book is considered to be at depth 0.  Now if book $i$ is above
book $j$, then
the relative order of books $i$ and $j$ is changed after a call if and only if
book $j$ is
consulted.  Hence,
$$\eqalign{p_{ij} &= p_{ij}(1-p_j) + p_{ji}p_i\cr
&= p_{ij}(1 - p_j) + (1-p_{ij})p_i\cr
&= p_i + p_{ij}(1 - p_i - p_j)\ .\cr
}
$$
Thus, we have
$$p_{ij} = {{p_i}\over{p_i + p_j}}\ ,$$
and
$$d_j = \sum_{k \ne j} {{p_k}\over{p_k + p_j}}\ .$$
If $p_i \ge p_j$, then
$${{p_k}\over{p_k + p_i}} \le {{p_k}\over{p_k + p_j}}$$
for $k \ne i,\,j$, and
$${{p_j}\over{p_i + p_j}} \le {{p_i}\over{p_i + p_j}}\ .$$
Since each term in the sum for $d_i$ is less than or equal to the corresponding
term in the
sum for $d_j$, we have $d_i \le d_j$.
\endNE\NE
\e{39}%exer 6.1.39
(Solution by Peter Montgomery) 
The probability that book 1 is in the right place is the probability that the
last phone call
referenced book 1, namely $p_1$.  The probability that book 2 is in the right
place, given
that book 1 is in the right place, is
$$p_2 + p_2p_1 + p_2p_1^2 + \ldots = {{p_2}\over{(1-p_1)}}\ .$$
Continuing, we find that
$$P = p_1 {{p_2}\over{(1 - p_1)}}{{p_3}\over{(1 - p_1 -
p_2)}}\cdots{{p_n}\over{(1 - 
p_1 - p_2 - \ldots - p_{n-1}}}\ .$$
Now let $q$ be a real number between 0 and 1, let
$$p_1 = 1 - q\ ,$$
$$p_2 = q - q^2\ ,$$
and so on, and finally let
$$p_n = q^{n-1}\ .$$
Then
$$P = (1 - q)^{n-1}\ ,$$
so $P$ can be made arbitrarily close to 1.
\endNE\NE
\e{40}%exer 6.1.40
If $a_1,\ a_2,\ \ldots$ is the sequence, then the event $\{a_1 < a_2 < \ldots <
a_k\}$ occurs
with probability $1/k!$, since there are $k!$ different orderings of $k$ real
numbers, and
all of them are equally likely to occur in this experiment.  Therefore,
$$P(X > k) = {1\over{k!}}\ .$$
Now let
$$p_k = P(X = k)\ .$$
Then
$$\eqalign{E(X) &= p_1 + 2p_2 + 3p_3 + \ldots\cr
&= (p_1 + p_2 + \ldots) + (p_2 + p_3 + \ldots) + \ldots\cr
&= P(X > 0) + P(X > 1) + \ldots\cr
&= 1 + {1\over{1!}} + {1\over{2!}} + \ldots\cr
&= e\ .\cr}$$ 
\endNE

\vskip 20pt 
%||*****||*****||*****||%sec6.2
\vskip 20pt

%\input mydefinitions

\indent\bf SECTION 6.2\rm
\NE
\e{1}%exer 6.2.1
$\displaystyle{E(X) = 0,\ V(X) = {2\over 3},\ \  \sigma = D(X) = 
\sqrt{{2\over 3}}}\ .\ $
\endNE\NE
\e{2}%exer 6.2.2
$\displaystyle{E(X) = {4\over 3},\ \ V(X)
= {17\over 9},\ \ \sigma = D(X) = {{\sqrt {17}}\over
3}}\ .$
\endNE\NE
\e{3}%exer 6.2.3
$\displaystyle{E(X) = {-1\over 19},\ \ E(Y) = {-1\over 19},\ \ V(X) = 33.21,\ \ 
V(Y) = .99\ .}$
\endNE\NE 
\e{4}%exer 6.2.4
(a) 10015, \quad (b) 310, \quad (c)
$-$100, \quad (d) 15, \quad (e) $\sqrt {15}$.
\endNE\NE
\e{5}%exer 6.2.5
(a)\ \ $\displaystyle{E(F) = 62,\ \ V(F) = 1.2\ .}$
\vskip 3pt
\itemitem{}(b)\ \ $\displaystyle{E(T) = 0,\ \ V(T) = 1.2\ .}$
\vskip 3pt
\itemitem{}(c)\ \ $\displaystyle{E(C) = {50\over 3}\ ,\ \ V(C) = {10\over 27}\
.}$
\endNE\NE
\e{7}%exer 6.2.7
$\displaystyle{V(X) = {3\over 4}\ ,\ \ D(X) = {\sqrt {3}\over 2}\ .}$
\endNE\NE
\e{8}%exer 6.2.8
$E(S_{2400}) = 960, \quad
V(S_{2400}) = 576, \quad \sigma = D(S_{2400}) = 24.$
\endNE\NE
\e{9}%exer 6.2.9
$\displaystyle{V(X) = {3\over 4}\ ,\ \ D(X) = {2 \sqrt{5}\over 3}\ .}$
\endNE\NE
\e{10}%exer 6.2.10
(a) \quad $V(X+c) = V(X),$  so $D(X+c) = D(X)$.
\vskip 5pt
\itemitem{}(b) \quad $V(cX) = {c^2}V(X), \hbox {\  so\ } D(cX) = {|c|}
X.$
\endNE\NE
\e{11}%exer 6.2.11
$E(X) = (1+2+\cdots + n)/n =
{{(n+1)}/ 2}.$
\vskip 5pt
\itemitem{} $\eqalign{V(X)&= (1^2 + 2^2 + \cdots + n^2)/n -
(E(X))^2\cr &= (n+1)(2n+1)/6 - {(n+1)^2}/ 4 = {(n+1)(n-1)}/12.}$
\endNE\NE
\e{12}%exer 6.2.13
$E\Bigr({{X-\mu}/{\sigma }}\Bigl) =
{(1/ \sigma)}(E(X) - \mu) = 0,$
\itemitem{}${V\Bigr({{X-\mu}/{\sigma}} }\Bigl)^2 = {{(1/
\sigma^2)}}E{(X-\mu)^2} = {\sigma^2/ \sigma^2} = 1.$
\endNE\NE
\e{13}%exer 6.2.14
Let $X_1,\dots , X_n$ be identically
distributed random variables such that $${P(X_i = 1) = P(
X_i=-1)}= {1\over 2}.$$ Then $E(X_i) = 0,$ and $V(X_i) = 1.$ Thus $W_n =
\sum_{j=1}^nX_i.$  Therefore \hfill \break
$E(W_n) = \sum_{i=1}^nE(X_i) = 0,$
and $V(W_n) = \sum_{i=1}^nV(X_i) = n$.
\endNE\NE
\e{14}%exer 6.2.15
Let $X$ be the number of boys and $Y$ be the number of girls.  Then
$$E(X) = E(Y) = {7\over 8}\ ,$$
and
$$V(X) = {7\over 64}\ ,\qquad V(Y) = {71\over 64}\ .$$
\endNE\NE
\e{15}%exer 6.2.16
(a)\ \ $\displaystyle{P_{X_i} = \pmatrix{0&1\cr
{{n-1}\over n}& {1\over{n}}}\ .}$ Therefore, $E(X_i)^2 = {1/ n}$ for $i
\ne j$. 
\vskip 5pt
\itemitem{}(b)\ \ $\displaystyle{P_{X_iX_j} = \pmatrix {0&1\cr 1-{1\over
{n(n-1)}}&{1\over {n(n-1)}}} \ \hbox{for}\  i \ne j\ .}$
\vskip 5pt
\itemitem{}\hskip 20pt Therefore, $\displaystyle{E(X_iX_j) =
{1\over {n(n-1)}}\ .}$ 
\vskip 5pt
\itemitem{}(c)  
$$\eqalign{E(S_n)^2& = \sum_iE(X_i)^2 + \sum_i\sum_
{j\not= i}E(X_iX_j)\cr & = n\cdot {1\over n} + {n(n-1)}\cdot {1\over
{n(n-1)}} = 2\ .}$$
\vskip 5pt
\itemitem{}(d)
$$\eqalign{V(S_n)& = E(S_n)^2 - E(S_n)^2\cr & = 2 -
{(n \cdot {(1/n)})^2} = 1\ .}$$
\endNE
\itemitem{16.}%exer 6.2.17
(a) \quad For $p$ = .5:
{\settabs\+&xxxxxxxxxxxxxxxxxxx&xxxx& .272\quad& .967\quad 
&\quad .994\cr\+&&&&\ \ \  $k$  &\cr\+&& \ &\ \ 1&\ \ \  2&\ \ \ 3\cr
\+&&10&
 .656&\ .979&\ .998\cr\+&\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
\ \ \ \ \ \ \ $N$&30&.638&\ .957&\ .999\cr \+ &&50&.678&\ .967&\ .997\cr}
\vskip 5pt
\hskip 50pt For $p$ = .2:
 {\settabs\+&xxxxxxxxxxxxxxxxxxx&xxxx& .272\quad& .967\quad 
&\quad .994\cr\+&&&&\ \ \ $k$  &\cr\+&& \ &\ \ 1&\ \ \  2&\ \ \ 3\cr
\+&&10&
 .772&\ .967&\ .994\cr\+&\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
\ \ \ \ \ \ \ $N$&30&.749&\ .964&\ .997\cr \+ &&50&.629&\ .951&\ .997\cr}
\vskip 5pt
\itemitem{}\IND {(b)} Use Exercise 12 and the fact that $E(S_n) = np$ and
$V(S_n) = npq$. The two examples in (a) suggests that the probability
that the outcome is within $k$ standard deviations is approximately the
same for different values of $p$. We shall see in Chapter 9 that the
Central Limit Theorem explains why this is true.
\NE
\e{18}%exer 6.2.19
(a)\quad $\D{E(\bar x) = {1\over n}\sum_{i=1}^nE(x_i) = {1\over
n}\cdot n\mu = \mu\ .}$
\vskip 5pt
\itemitem{}(b)\ \ We have
$$E\Bigl((\bar x - \mu)^2\Bigr) = V(\bar x)\ ,$$
which was shown to equal $\sigma^2/n$ in Theorem 6.9.
\vskip 5pt
\itemitem{}\IN {(c)} We have from the hint:
$$\sum_{i=1}^n(x_i-\bar x)^2 = \sum_{i=1}^n(x_i-\mu)^2 - n(\bar
x-\mu)^2\ .$$
Thus, 
$$\eqalign{E(s^2) &= {1\over n}E\left(\sum_{i=1}^n(x_i-\bar
x)^2\right)\cr &= {1\over n}\Bigr(E\left(\sum_{i=1}^n(x_i - \mu)^2\right) -
nE(\bar x - \mu)^2\Bigl )\cr&= {1\over n}(n{\sigma}^2 -
n{{\sigma}^2\over n}) = {{n-1}\over n}\sigma^2\ ,} $$
where we have used the definition of the variance and part (b) to obtain the
penultimate
expression.
\vskip 5pt
\itemitem{}(d)\ \ Since the expectation operator is linear, and the `new' 
$s^2$ is $n/(n-1)$ times the `old' $s^2$, the new $s^2$ has expectation
$${n\over {n-1}}{{n-1}\over n}\sigma^2 = \sigma^2\ .$$
\endNE\NE
\e{19}%exer 6.2.20
Let $ X_1,X_2$ be independent random variables with
$$p_{X_1} = p_{X_2} = \pmatrix{-1&1\cr 1\over 2&1\over 2}\ .$$ 
Then
$$p_{X_1+X_2} = \pmatrix{-2&0&2\cr {1\over 4} & {1\over 2} & {1\over 4}}\ .$$
Then 
$$\bar\sigma_{X_1} = \bar\sigma_{X_2} = 1 ,
\ \bar\sigma_{X_1+X_2} = 1\ .$$
Therefore 
$$V(X_1+X_2) = 1 \not= V(X_1)+ V(X_2) = 2\ ,$$
and
$$\bar\sigma_{X_1+X_2} = 1 \not= \bar\sigma_{X_1} + \bar\sigma_{X_2} =
2\ .$$
\endNE\NE
\e{20}%exer 6.2.21
(a)\ \ $\displaystyle{E(\bar \mu) = \mu}$
\vskip 3pt
\itemitem{}(b)\ \ $\displaystyle{w = {{V(X_2)}\over{V(X_1) + V(X_2)}}}$
\endNE\NE
\e{21}%exer 6.2.22
$$\eqalign{f^\prime (x) &= -\sum_\omega
2(X(\omega)-x)p(\omega)\cr & = -2\sum_\omega X(\omega)p(\omega) +
2x\sum_\omega p(\omega)\cr &= -2\mu + 2x\ .}$$
Thus $x = \mu$ is a critical point.  Since $f^{\prime\prime}(x) \equiv 2,$
we see that $x = \mu$\ is the minimum point. 
\endNE\NE
\e{22}%exer 6.2.23
$X, Y, X+Y$, and $X-Y$ have the same distribution, so they have the same
mean and variance.  Thus $E(X) = E(Y) = E(X) + E(Y)$.  This implies that
$$E(X) = E(Y) = 0\ .$$ 
It also implies that
$$  E(X+Y)^2 = E(X-Y)^2 =
E(X^2) = E(Y^2)\ .$$  
Thus $E(XY) = 0$ and $E(Y^2) = E(X^2) = 0.$ 
Therefore, $P(X = Y = 0) = 1$.
\endNE\NE
\e{23}%exer 6.2.24
If $X$ and $Y$ are independent, then 
$$\hbox{Cov}(X,Y) = E(X-E(X))\cdot E(Y-E(Y)) = 0\ .$$ 
Let $U$ have distribution
$$p_{_U} = \pmatrix{0& {\pi/ 2}& \pi &{3\pi/ 2} \cr {1/4
}&{1/4}&{1/ 4}& {1/4}\cr}\ .$$ Then let $X$ = cos($U$) and $Y$ =
sin($U$).  $X$ and $Y$ have distributions 
$$p_{_X} = \pmatrix{1& 0& -1
&0\cr {1/4}&{1/4}&{1/4}& {1/4}\cr}\ ,$$
$$p_{_Y} = \pmatrix{0& 1& 0 &-1 \cr {1/
4}&{1/4}&{1/4}& {1/4}\cr}\ .$$ 
Thus $E(X) = E(Y) = 0$ and
$E(XY) = 0,$ so Cov$(X,Y) = 0$.  However, since \hfill\break
$\hbox{sin}^2(x) + \hbox{cos}^2(x) = 1$, $X$ and $Y$ are dependent.
\endNE\NE
\e{24}%exer 6.2.25
Consider the variance of $S_n$:
$$V(S_n) = \sum_{i=1}^np_i(1-p_i) =
\sum_{i=1}^np_i - \sum_{i=1}^np_i^2\ ,$$ 
with the constraint
$$\sum_{i=1}^np_i = np\ .$$ 
Assume that we have values of $p_i$ that satisfy
the constraint and that  $m$ of the values of $p_i$ are equal to $x$ and
one is equal to $y$ with $x > y$. By rearranging the terms if necessary we
can assume that the first $m$ are equal to $x$ and the ${(m+1)}{\rm st}$
is equal to $y$. We shall show that we can increase the variance by making
these $m+1$ values equal.  To do this we define 
$${\bar p_i} = p_i -
{\epsilon\over m},\ \  \hbox{for $i$ = 1 to $m$}$$ 
and 
$$\bar p_{m+1} = p_{m+1} + \epsilon\ ,$$ 
where 
$$\epsilon = {m\over{m+1}}(x-y)\ .$$  
Then the new $\bar {p_i}{\rm s}$ satisfy the constraint, and the difference
between
the new variance $\bar V$ and the old variance $V$ is
$${\bar V - V} = {m(x-{\epsilon\over m})^2} + (y+\epsilon)^2 - mx^2 - 
y^2\ .$$  
After simplifying and substituting the value for $\epsilon$
this becomes 
$$\bar V - V = {m\over{m+1}}(x-y)^2\ .$$  
Since this value is positive we have increased the variance by making 
the first $m+1$ values equal. The same argument applies in case $y > x$.
By induction we see that the variance is maximized
by making all the values equal.\ (Note: Students who know about the
technique of Lagrange multipliers will find this easier to prove
using that method.)   
\endNE\NE
\e{25}%exer 6.2.26
(a) The expected value of $X$ is  
$$\mu = E(X) =  \sum_{i = 1}^{5000} iP(X = i)\ .$$  
The probability that a white ball is drawn is
$$P(\hbox{white ball is drawn}) = \sum_{i = 1}^n P(X = i){i\over
{5000}}\ .$$
Thus 
$$P(\hbox{white ball is drawn}) = {\mu\over 5000}\ .$$
\EE (b) To have $P$(white,white) = $P(\hbox{white})^2$ we must have
$$\sum_{i=1}^{5000} ({i\over {5000}})^2P(X = i) =
(\sum_{i=1}^n{i\over 5000}P(X = i))^2\ .$$
But this would mean that\ $E(X^2) = E(X)^2$, or $V(X) = 0$.  
Thus we will have independence only if $X$ takes
on a specific value with probability 1.
\vskip 5pt
\itemitem{}\IND {(c)} From (b) we see that 
$$P(\hbox{white,white}) = {1\over {5000}^2}E(X^2)\ .$$ 
Thus 
$$V(X) = {{(\sigma^2 + \mu^2)}\over {5000^2}}\ .$$
\endNE\NE
\e{26}%exer 6.2.27
(a)\ \ $P(X = k) = pq^{k-1}, k = 1,2,\dots$ Thus by Example 3
we have 
$$E(X_j) = {1\over p},\ V(X_j) = {q\over p^2}\ .$$
\vskip 5pt
\EE (b)\ \ $E(T_n) = {n/ p},\  V(T_n) = {{nq}/ p^2}.$
\vskip 5pt
\EE (c)\ \ $E(T_n) = 2n, \ V(T_n) = 2n.$
\endNE\NE
\e{27}%exer 6.2.28
The number of boxes needed to get the $j$'th picture has a geometric
distribution with 
$$p = {{(2n-k+1)}\over {2n}}\ .$$ 
Thus 
$$V(X_j) =
{{2n(k-1)}\over {(2n-k+1)^2}}\ .$$ 
Therefore, for a team of 26 players the
variance for the number of boxes needed to get the first half of the
pictures would be  
$$\sum_{k = 1}^{13} {{26(k-1)}\over{(26-k+1)^2}} =
7.01\ ,$$ 
and to get the second half would be 
$$\sum_{k = 14}^{26}
{{26(k-1)}\over {(26-k+1)^2}}= 979.23\ .$$  
Note that the variance for the
second half is much larger than that for the first half.
\endNE

\vskip 20pt
%||*****||*****||*****||%sec6.3
\vskip 20pt

%\input mydefinitions

\indent\bf SECTION 6.3\rm
\NE
\e{1}%exer 6.3.1
(a)\ \ $\displaystyle{\mu = 0,\ \sigma^2 = 1/3}$
\vskip 3pt
\itemitem{}(b)\ \ $\displaystyle{\mu = 0,\ \sigma^2 = 1/2}$
\vskip 3pt
\itemitem{}(c)\ \ $\displaystyle{\mu = 0,\ \sigma^2 = 3/5}$
\vskip 3pt
\itemitem{}(d)\ \ $\displaystyle{\mu = 0,\ \sigma^2 = 3/5}$
\vskip 3pt
\endNE\NE
\e 2 %exer 6.3.3
(a)\ \ $\displaystyle{\mu = 0,\ \sigma^2 = 1/5}$
\vskip 3pt
\itemitem{}(b)\ \ $\displaystyle{\mu = 0,\ \sigma^2 = {{\pi^2 -
8}\over{\pi^2}}}$
\vskip 3pt
\itemitem{}(c)\ \ $\displaystyle{\mu = 1/3,\ \sigma^2 = 2/9}$
\vskip 3pt
\itemitem{}(d)\ \ $\displaystyle{\mu = 1/2,\ \sigma^2 = 3/20}$
\endNE\NE
\e 3 %exer 6.3.5
$\displaystyle{\mu = 40,\ \sigma^2 = 800}$
\endNE\NE
\e 4 %exer 6.3.6
(a) $\int_{-1}^1(ax+b)dx = 2b = 1$, so $b = {1\over 2}.$
\vskip 5pt
\ee (b) $ax + {1\over
2 }\ge 0$, so when $x =
1, a \ge -{1\over 2},$
and when $x = -1, a \le
{1\over 2}. $ Thus
$-{1\over 2} \le a \le
{1\over 2}. $ 
\vskip 5pt
\ee (c) $\mu =
\int_{-1}^1 (ax^2 + bx) dx =
{2\over 3}a.$  
\vskip 5pt
\ee (d) $E(X^2) =
\int_{-1}^1(ax^3+bx^2)dx =
{2b\over 3} = 1/3 - (4/9)a^2.$ Thus
$\sigma^2(X) = {2\over 3}b
- {4\over 9}a^2.$
\endNE\NE
\e 5 %exer 6.3.7
(d)\ \ $\displaystyle{a = -3/2,\ b = 0,\ c = 1}$
\vskip 3pt
\itemitem{}(e)\ \ $\displaystyle{a = {45\over 48},\ b = 0,\ c = {3\over 16}}$
\endNE\NE
\e 6 %exer 6.3.8
(a) $\displaystyle{\quad \mu_{_T} = {1\over 3}, \qquad {\sigma_T^2} = {1\over
9}}$.
\vskip 5pt
\ee (b) $\displaystyle{\quad \mu_{_T} = {1\over 3}, \qquad \sigma_T^2 = {2\over
9}}$.
\vskip 5pt
\ee (c) $\displaystyle{\quad \mu_{_T} = {1\over 2}, \qquad \sigma_T^2 = {3\over
4}}$.
\endNE\NE                                              
\e 7 %exer 6.3.9
$\displaystyle{f(a) = E(X-a)^2 = \int(x-a)^2f(x)dx\ .}$
Thus $$\eqalign {f^\prime(a)& = -\int 2(x-a)f(x)dx\cr & = -2\int xf(x)dx
+ 2a\int f(x)dx \cr & = -2\mu(X) + 2a\ .\cr}$$
Since $f^{\prime\prime}(a) = 2$,  $f(a)$ achieves its
minimum when $a = \mu(X)$.
\endNE\NE
\e 8%exer 6.3.10
$\displaystyle{a(\sigma^2 + \mu^2) + b\mu + c\ .}$
\endNE\NE
\e 9%exer 6.3.11
(a)\ \ $\displaystyle{3\mu,\ 3\sigma^2}$
\vskip 3pt
\itemitem{}(b)\ \ $\displaystyle{E(A) = \mu,\ V(A) = {{\sigma^2}\over 3}}$
\vskip 3pt
\itemitem{}(c)\ \ $\displaystyle{E(S^2) = 3\sigma^2 + 9 \mu^2,\ 
E(A^2) = {{\sigma^2}\over 3} + \mu^2}$
\endNE\NE
\e {10}%exer 6.3.12
(a)\ \ $\displaystyle{1\over 3}$
\vskip 3pt
\itemitem{}(b)\ \ $\displaystyle{2\over 3}$
\vskip 3pt
\itemitem{}(c)\ \ $\displaystyle{1\over 3}$
\vskip 3pt
\itemitem{}(d)\ \ $\displaystyle{2\over 3}$
\vskip 3pt
\itemitem{}(e)\ \ $\displaystyle{7\over 6}$
\endNE\NE
\e {11}%exer 6.3.13
In the case that $X$ is uniformly distributed on $[0, 100]$, one finds that
$$E(|X-b|) = {1\over {200}}\Bigl(b^2 + (100 - b)^2\Bigr)\ ,$$
which is minimized when $b = 50$.
\vskip 3pt
\itemitem{} When $f_X(x) = 2x/10{,}000$, one finds that
$$E(|X-b|) = {200\over 3} - b + {{b^3}\over {15000}}\ ,$$
which is minimized when $b = 50\sqrt 2$.
\endNE\NE
\e {12}%exer 6.3.14
$\displaystyle{\int_0^1\int_0^1 x^ydxdy = \log(2) \approx .693\ .}$
\endNE\NE
\e {13}%exer 6.3.15
Integrating by parts, we have
$$\eqalign{E(X) &= \int_0^\infty xdF(x)\cr
& = -x(1-F(x)) \big\vert_0^\infty + \int_0^\infty(1-F(x))dx \cr & =
\int_0^\infty (1-F(x))dx\ .}$$
To justify this argment we have to show that $a(1-F(a))$
approaches 0 as $a$ tends to infinity.  To see this, we note that
$$\eqalign{\int_0^\infty xf(x)dx& = \int_0^a xf(x)dx +
\int_a^\infty xf(x)dx\cr  &\ge \int_0^a xf(x)dx + \int_0^a af(x)dx\cr & =
\int_0^a xf(x)dx + a(1-F(a))\ .\cr}$$
Letting $a$ tend to infinity, we have that
$$E(X) \ge E(X) + \lim_{a\to\infty} a(1-F(a))\ .$$ 
Since both terms are non-negative, the only way this can happen is 
for the inequality to be an equality and the limit to be 0. 
\vskip 5pt
\itemitem{} To illustrate this with the
exponential density, we have
$$\int_0^\infty(1-F(x))dx = \int_0^\infty e^{-\lambda x} dx = {1\over
\lambda} = E(X)\ .$$
\endNE\NE
\e {15}%exer 6.3.16
$\displaystyle{E(Y) = 9.5,\ E(Z) = 10,\ E(|X-Y|) = 1/2,\ E(|X-Z|) = 1/2\ .}$
\itemitem{}$Z$ is better, since it has the same expected value as $X$ and
the variance is only slightly larger.
\endNE\NE
\e {17}%exer 6.3.18
(a) $$\eqalign{\hbox{Cov}(X,Y) &= E(XY) - \mu (X)E(Y) - 
E(X)\mu(Y) + \mu (X)\mu (Y)\cr& = E(XY) - \mu (X)\mu (Y) =
E(XY)-E(X)E(Y)\ .}$$
\vskip 5pt
\ee \IND{(b)} If $X$ and $Y$ are independent, then
$E(XY) = E(X)E(Y),$ and so Cov$(X,Y)$ = 0.
\vskip 5pt
\ee \IND{(c)} \hskip 10pt $$\eqalign{ V(X+Y) &= E(X+Y)^2 -
(E(X+Y))^2\cr& = E(X^2) +2E(XY) + E(Y^2)\cr& -E(X)^2 -2E(X)E(Y) -
E(Y)^2\cr& = V(X) + V(Y) + 2 \hbox{Cov}(X,Y)\ .}$$
\endNE\NE
\e {18}%exer 6.3.19
\IND{(a)} $$\eqalign{0 \le &V\Bigl({X\over \sigma(X)} + {Y\over
\sigma (Y)}\Bigr)\cr &= V\Bigl({X\over \sigma (X)}\Bigr) + V\Bigl({Y\over
\sigma (Y)}\Bigr) + 2\hbox{Cov}\Bigl({X\over \sigma(X))},{Y\over
\sigma(Y)}\Bigr)\cr & = 1 + 1 + 2{\hbox{cov}(X,Y)\over \sqrt{V(X)V(Y)}} =
2(1+ \rho (X,Y))\ .}$$ 
\itemitem{}\IND{(b)}$$\eqalign{0 \le &V\Bigr({X\over
\sigma(X)}- {Y\over \sigma (Y)}\Bigr)\cr & = V\Bigl({X\over \sigma
(X)}) + V({Y\over \sigma (Y)}\Bigr) - 2\hbox{Cov}\Bigl({X\over
\sigma(X))},-{Y\over \sigma(Y)}\Bigr)\cr & = 1 + 1 -
2{\hbox{Cov}(X,Y)\over {\sigma(X)\sigma(Y)}} = 2(1- \rho (X,Y))\ .}$$ 
\itemitem{}\IND{(c)}\quad From (a),
$1 + \rho (X,Y) \ge 0,$ so $\rho (X,Y) \ge -1.$\hfill\break
\itemitem{}\hskip 30pt From (b), $1-\rho(X,Y) \ge 0,$  so $\rho(X,Y)
\le 1.$ Thus $-1 \le \rho(X,Y) \le 1.$ 
\endNE\NE
\e {19}%exer 6.3.20
(a)\ \ 0
\vskip 3pt
\itemitem{}(b)\ \ $\displaystyle{1\over \sqrt 2}$
\vskip 3pt
\itemitem{}(c)\ \ $\displaystyle{-{1\over \sqrt 2}}$
\vskip 3pt
\itemitem{}(d)\ \ 0
\endNE\NE
\e {20}%exer 6.3.21
(a) $$\eqalign{f_X(x) &= {1\over
{2\pi \sqrt{1-\rho^2}}}\int_{-\infty}^{\infty}\exp\Bigl({-{(x^2 - 2\rho
xy + y^2)}\over {2(1-\rho^2)}}\Bigr)dy\cr & = {1\over {2\pi \sqrt
{1-\rho^2}}}
 \int_{-\infty}^\infty \exp({-(y-\rho x)^2})\cdot \hbox{exp}(-{{1\over
2}x^2}) dy\cr &  = {1\over {\sqrt
{2\pi}}}\cdot  \exp({-{1\over 2}x^2})\ .}$$
Thus $X$ has a standard normal density. By symmetry, $Y$ also has a
standard normal density.
\vskip 3pt
\ee (b) $$\eqalign{E(XY)& = {1\over
{2\pi
\sqrt{1-\rho^2}}}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}xy\cdot
\exp\Bigl({-{(x^2 - 2\rho xy + y^2)}\over {2(1-\rho^2)}}\Bigr)dxdy\cr &
= {1\over {2\pi \sqrt {1-\rho^2}}}
 \int_{-\infty}^\infty\Bigl(\int_{-\infty}^\infty y\cdot
\exp\Bigl({{-(y-\rho x)^2}\over {(2(1-\rho^2)}}\Bigr)dy\Bigr) x\cdot
\exp({-{1\over 2}x^2}) dx\cr &  = {1\over {\sqrt
{2\pi}}}\int_{-\infty}^\infty \rho x^2 {\cdot}
 \exp({-{1\over 2}x^2})= \rho\ .}$$
Now 
$$\hbox{Cov}(X,Y) = {{E(XY)-E(X)E(Y)}\over {\sqrt{V(X)V(Y)}}}\ .$$ 
Since $E(X) = E(Y) = 0$ and $V(X) = V(Y) = 1$ ,\ Cov$(X,Y) = E(XY)
= \rho.$
\endNE\NE
\e {21}%exer 6.3.22
We have
$$\eqalign{{f_{_{XY}(x,y)}\over {f_{_Y}(y)}} &= {{{{1\over
{2\pi\sqrt{1-\rho ^2}}}\cdot \exp\Bigl({{-(x^2-2\rho xy + y^2)}\over
{2(1-\rho ^2) }}}}\Bigr)\over {\sqrt{2\pi}\cdot \exp({-{y^2\over 2}})}}\cr
&= {1\over \sqrt{2\pi(1-\rho^2)}}\cdot {\exp\biggl({-{(x-\rho y)^2}\over
{2(1-\rho^2)}}\biggr)}}$$
\vskip 5pt\itemitem{} which is a normal density with
mean $\rho y$ and variance $1-\rho^2.$ Thus, 
$$\eqalign{E(X \vert Y = y)& =
\int_{-\infty}^\infty x{1\over {\sqrt{2\pi(1-\rho^2)}}}\cdot
\exp\Bigl({{-(x-\rho y)^2}\over {2(1-\rho^2)}}\Bigr)\,dx\cr&=\rho y
\int_{-\infty}^\infty {1\over {\sqrt{2\pi(1-\rho^2)}}}\cdot
\exp({-(x-\rho y)^2})\cr &= \cases{\rho y < y,&if $0 < \rho < 1$;\cr
y,& if $\rho = 1.$}}$$
\endNE\NE
\e {22}%exer 6.3.23
We have
$$\eqalign{f_{_X}(x)& = \int_0^1 f_{_{X,Y}}(x,y)dy\cr& = \int_0^1
f_{_{X\vert Y}}(x\vert y)f_{_Y}(y)dy\cr & = \int_x^1{1\over y}dy\cr & =
-\log x\ ,}$$
if $0 < x \le 1$.
\endNE\NE
\e {24}%exer 6.3.25
\exindent(a)\hskip 1em Since the father's height is 72 inches, 
 $Y = (72-68)/2.7 = 1.48$. Therefore the density for $X$ given $Y$  is
normal with mean $.5\cdot 1.48$ = .74 and variance $1-.5^2$ = .75. Thus
the density for the son's height, given that the father's height is 72, is
normal with mean 2.7 $\cdot$ .74 + 68 = 70  and variance $(2.7)^2\cdot
.75 = 5.47$.
\endNE\NE
\e {26}%exer 6.3.27
(a)\ \ Let $\theta$ denote the angle that our path makes with the river bank, 
and assume without loss of generality that $0 \le \theta \le \pi/2$.  Let $X$
denote the distance from $P$ to the river.  Then $X = \sin(\theta)$.  Thus,
the cumulative distribution function of $X$ is given by
$$\eqalign{F_X(x) &= P(X \le x)\cr
&= P\bigl(\sin(\theta) \le x\bigr)\cr
&= P(\theta \le \arcsin x)\cr
&= {2\over \pi}\arcsin x\ .\cr}$$
So,
$$f_X(x) = {2\over \pi}{1\over{\sqrt{1 - x^2}}}\ .$$
Therefore,
$$\eqalign{E(X) &= \int_0^1 {2\over \pi}{x\over {\sqrt{1-x^2}}}\,dx\cr
&= {2\over \pi}\Bigl[(1-x^2)^{1/2}\Bigr]_0^1\cr
&= {2\over \pi}\ .\cr}$$
\vskip 3pt
\itemitem{}(b)\ \ For a fixed $\theta$ between 0 and $\pi/2$, let $A_{\theta}$
denote
the set of angles $\alpha$ that you can choose at $P$ and get back to the river
by 
walking at most 1 mile in the direction $\alpha$.  If $\alpha = 0$ corresponds
to the
direction directly towards the river from $P$, then
$$A_\theta = \Bigl[\theta - {\pi\over 2}, {\pi\over 2} - \theta\Bigr]\ .$$
So the probability that you choose a good angle $\alpha$, given that you are at
$P$, 
is
$${{|A_\theta|}\over{2\pi}} = {{\pi - 2\theta}\over{2\pi}}\ .$$
This must be averaged over all $\theta \in [0, \pi/2]$ to obtain the final
answer:
$$\eqalign{{2\over \pi}\int_0^{\pi/2} {1\over 2} - {1\over \pi}\theta\,d\theta
&=
{2\over \pi}\Bigl[{\theta\over 2} - {1\over {2\pi}}\theta^2\Bigr]_0^{\pi/2}\cr
&= {1\over 4}\ .\cr}$$
\endNE\NE
\e {27}%exer 6.3.28
Let $Z$ represent the payment.  Then
$$\eqalign{P(Z = k | X = x) &= P(Y_1 \le x, Y_2 \le x, \ldots, Y_k \le x, 
Y_{k+1} > x)\cr
&= x^k(1-x)\ .\cr}$$
Therefore,
$$\eqalign{P(Z = k) &= \int_0^1 x^k(1-x)\,dx\cr
&= \biggl[{1\over{k+1}}x^{k+1} - {1\over{k+2}} x^{k+2} \biggr]_0^1\cr
&= {1\over{k+1}} - {1\over{k+2}}\cr
&= {1\over{(k+1)(k+2)}}\ .\cr}$$
Thus,
$$E(Z) = \sum_{k = 0}^\infty k\biggl({1\over{(k+1)(k+2)}}\biggr)\ ,$$
which diverges.  Thus, you should be willing to pay any amount to play this
game.
\endNE


\vskip 20pt
%||*****||*****||*****||%sec7.1
\vskip 20pt

%\input mydefinitions
\vskip 20pt
\indent\bf SECTION 7.1\rm
\NE
\e 1 %exer 7.1.1
(a)\ \ $\displaystyle{.625}$
\vskip 3pt
\itemitem{}(b)\ \ $\displaystyle{.5}$
\endNE\NE
\e 2 %exer 7.1.2
$\displaystyle{\pmatrix{-2&-1&0&1&2&3&4\cr 1\over16 &1\over 4
&5\over{16} &3\over{16} &9\over {64} &1\over {32} &1\over {64}}\ .}$
\endNE\NE
\e 3 %exer 7.1.3
$\displaystyle{\pmatrix{0&1&2&3&4\cr 1\over {64} & 3\over {32} &{17}\over {64} 
&3 \over 8&1\over 4}}$
\endNE\NE
\e 5 %exer 7.1.5
(a)\ \ $\displaystyle{\pmatrix{3 & 4 & 5 & 6\cr 1\over {12} & 4 \over  {12} &
4 \over {12} & 3\over {12}}}$
\vskip 3pt
\itemitem{}(b)\ \ $\displaystyle{\pmatrix{1 & 2 & 3 & 4\cr 1\over {12} & 4 \over
{12} &
4 \over {12}& 3 \over {12}}}$
\endNE\NE
\e 6 %exer 7.1.6
(a) $\displaystyle{P(T_r = k) = {{r+k-1}\choose k} p^rq^k\ .}$
\vskip 5pt
\ee(b) $\displaystyle{P(C_r = k) = b(r,p,k) = {r\choose k}p^kq^{r-k}\ .}$
\vskip 5pt
\ee(c) $\displaystyle{E(C_r) = rp, \quad V(C_r) = rpq\ .}$
\endNE\NE
\e 7 %exer 7.1.7
(a) $P(Y_3 \le j) = P(X_1 \le j,X_2 \le j, X_3 \le j) = P(X_1 \le
j)^3.\hfill\break$ Thus
$$p_{_{Y_3}}=\pmatrix{1&2&3&4&5&6\cr1\over {216}&7\over
{216}&19\over{216}&37\over{216}&61\over{216}&91\over{216}}\ .$$
This distribution is not bell-shaped.
\vskip 5pt
\ee(b)\ \ In general,
$$P(Y_n \le j) = P(X_1 \le j)^3 = \biggl({j\over n}\biggr)^n\ .$$
Therefore,
$$P(Y_n = j) = \biggl({j\over n}\biggr)^n - \biggl({{j-1}\over n}\biggr)^n\ .$$
This distribution is not bell-shaped for large $n$.
\endNE
\NE
\e 8 %exer 7.1.10
(b)\ \ .304
\vskip 3pt
\itemitem{}(c)\ \ .325
\endNE\NE
\e 9 %exer 7.1.11
Let $p_1,\dots, p_6$ be the probabilities for one die and
$q_1,\dots, q_6$ be the probabilities for the other die.  Assume
first that all probabilities are positive.  Then $p_1q_1 > p_1q_6$,
 since there is only one way to get a 2 and several ways to get a 7. 
Thus $q_1 > q_6$.  In the same way $q_6q_6 > q_1p_6$ and so $q_6 >
q_1$.  This is a contradiction. If any of the sides has probability 0,
then we can renumber them so that it is  side 1.  But then the
probability of a 2 is 0 and so all sums would have to have probability
0, which is impossible.
\vskip 5pt 
\itemitem{} Here's a fancy way to
prove it.  Define the polynomials 
$$p(x) = \sum_{k=0}^5 p_{(k+1)}x^k$$
and 
$$q(x) = \sum_{k=0}^5 q_{(k+1)}x^k\ .$$ Then we must have 
$$p(x)q(x) = \sum_{k=0}^{10} {{x^k}\over 11} = {{(1-x^{11})}\over
(1-x)}\ .$$ 
The left side is the product of two fifth degree
polynomials.  A fifth degree polynomial must have a real root 
which will not be 0 if $p_1 > 0.$  Consider
the right side as a polynomial. For $x$ to be a non-zero root of this
polynomial it would have to be a real eleventh root of unity other
than 1, and there are no such roots.  Hence again we have a
contradiction.
\endNE\NE
\e {10} %exer 7.1.12
Let $n = rs$.  Then consider the following
two distributions 
$$p_{_X} = \pmatrix{0&1&2&\ldots&{r-1}\cr{1/ r}&{1/
r}&{1/ r}&\ldots&{1/ r}}\ ,$$ 
$$p_{_Y} =
\pmatrix{0&r&2r&\ldots&(s-1)r\cr {1/s}&{1/ s}&{1/ s}&\ldots&{1/ s}}\ .
$$ 
If $X$ and $Y$ are independent, then $X + Y$ takes on all possible
values from 0 to $n-1$.  Further, there is only one choice of $X$ and
$Y$ that gives $X+Y$ a particular value and the probability for this
choice is ${1/{rs}}$. Thus $X+Y$ has a uniform distribution on the values
from 0 to $n-1$.
\endNE
\vskip 20pt
%||*****||*****||*****||%sec7.2
\vskip 20pt

%\input mydefinitions
\vskip 20pt
\indent\bf SECTION 7.2\rm
\endNE\NE
\e 2 %exer 7.2.1
(a)\ \ $\displaystyle{f_Z(x) = {{x+2}\over 4}\ {\rm on}\ [-2, 0]\ {\rm and}\ 
{{2-x}\over 4}\ {\rm on}\ [0, 2].}$
\endNE\NE
\e 3 %exer 7.2.2
(a) $$f_{{_Z}}(x) = \cases{{x^3/24},& if  $0\le x\le 2$;\cr 
x-{{x^3}/ 24} - {4/ 3},  & if $2\le x \le 4$.\cr}$$

\vskip 5pt
\ee
(b) $$f_{{_Z}}(x) = \cases{(x^3 - 18x^2 + 108x - 216)/24,
& if $6\le x \le 8$; \cr {(-x^3 + 18x^2 - 84x + 40)/24} ,& if $8\le x \le
10$.\cr}$$

\vskip 5pt \ee (c) $$f_{{_Z}}(x)= \cases{
{{x^2}/ 8},& if $0\le x \le 2$;\cr {1/ 2} - {{(x-2)^2}/ 8} ,
& if $2\le x \le 4.$\cr}$$
\endNE\NE
\e 4 %exer 7.2.3
$$f_{_Z}(x) = \cases{x^2/2, & if$\ 0 \le x \le 1;$\cr
(-2x^2 + 6x - 3)/2, & if $\ 1 \le x \le 2;$\cr
(x-3)^2/2, & if $\ 2 \le x \le 3.$\cr}$$
\endNE\NE
\e 5 %exer 7.2.3.5
(a) $$f_{{_Z}}(x) = \cases{{{\lambda \mu}\over {\mu +\lambda}}e^{\lambda
x}, &  $x < 0$;\cr {{\lambda \mu}\over {\mu + \lambda}}e^{-\mu
x}, &  $x \ge 0$.\cr}$$ 
\ee {(b)}$$f_{{_Z}}(x) = \cases{ 1-e^{-\lambda x},&
$0 < x < 1$;\cr (e^\lambda -1)e^{-\lambda x}, & $x \ge 1.$}$$
\endNE\NE
\e 6 %exer 7.2.100
$Z$ is normally distributed with mean $\mu = \mu_1 + \mu_2$ and variance
$\sigma^2 = \sigma_1^2 + \sigma_2^2$.
\endNE\NE
\e 7 %exer 7.2.101
We first find the density for $X^2$ when X has a general normal density 
$$f_{_X}(x) = {1\over {\sigma\sqrt {2\pi}}}  \thinspace
e^{-{{(x-\mu)^2}/{2\sigma^2}}} dx\ .$$ Then (see Theorem 1 of Chapter
5, Section 5.2 and the discussion following) we have  
$$f_{_X}^2(x) = {1\over{\sigma \sqrt {2\pi}}}{1\over {2\sqrt x}}\exp({-{x/
{2\sigma^2}}-{\mu^2/ {2\sigma^2}})}\Bigr ( \exp({{\sqrt x \mu}/{
\sigma^2}})+\exp({{-\sqrt x \mu}/{ \sigma^2}})\Bigl)\ .$$ 
Replacing the last two exponentials by their series representation, we have
$$f_{_X}^2(x) = {e^{-{\mu/{2\sigma^2}}}\sum_{r=0}^\infty
}\Bigr({\mu\over{2\sigma ^2}}\Bigl)^r{1\over {r!}} G({1/
{2\sigma ^2}},r+ {1/ 2}, x)\ ,$$ 
where 
$$G(a,p,x) = {{a^p}\over {\Gamma(p)}}e^{-ax}{x^{p-1}}$$ 
is the gamma density.  We now consider
the original problem with $X_1 \ \hbox{and}\  X_2$ two random
variables with normal density with parameters $\mu_1,\sigma_1$ and
$\mu_2,\sigma_2.$ This is too much generality  for us, and
we shall assume that the variances are equal, and then for
simplicity we shall assume they are 1.  Let 
$$c = \sqrt {\mu_1^2 + \mu_2^2}\ .$$
We introduce the new random variables 
$$Z_1 = {1\over c}(\mu _1X_1 + \mu _2X_2)\ ,$$ 
$$Z_2 = {1\over c}(\mu_2X_1 - \mu_1X_2)\ .$$  
Then $Z_1$ is normal with mean $c$ and variance 1 and
$Z_2$ is normal with mean 0 and variance 1.    Thus,
$$f_{Z_1^2}= 
 e^{-{c^2}/ 2}\sum_{r=0}^\infty \Bigr({{c^2}\over 2}\Bigl)^r{1\over
r!}G({1/ 2},r+{1/ 2},x)\ ,$$ 
and 
$$f_{Z_2^2} =  G({1/2},{1/2},x)\ .$$
Convoluting these two densities and using the fact that the
convolution of a gamma density $G(a,p,x)$ and $G(a,q,x)$ is a gamma
density $G(a,p+q,x)$ we finally obtain 
$$f_{Z_1^2 + Z_2^2} = f_{X_1^2 + X_2^2} = 
e^{-{c^2/ 2}}\sum_{r=0}^\infty\Bigr({c^2\over 2}\Bigl)^r
{1\over r!}G\Bigr({1/ 2},r+1,x\Bigl)\ .$$
(This derivation is adapted from that of C.R. Rao in his book \it
Advanced Statistical Methods in Biometric Research\rm, Wiley, l952.) 
\endNE\NE
\e 8 %exer 7.2.101
$$f_{R^2} = \cases{\pi/4, & if $\ 0 \le x \le 1$;\cr
(1/2)\arcsin\Bigl((2-x)/x\Bigr), & if $\ 1 \le x \le 2$.\cr}$$
$$f_R = \cases{(\pi/2)x, & if $\ 0 \le x \le 1$;\cr
x\arcsin\Bigl((2 - x^2)/x^2\Bigr), & if $\ 1 \le x \le \sqrt 2$.\cr}$$
\endNE\NE 
\e 9 %exer 7.2.102
$P(X_{10} > 22) = .341$ by numerical integration.  This
also could  be estimated by simulation.
\endNE\NE
\e {10} %exer 7.2.9
$P(\hbox{min}(X_1,\dots,X_n > x) = (P(X_1 > x))^n = (e^{-{x/
\mu}})^n = e^{-{(n/\mu}) x}.$\ Thus 
$$f_{\hbox{min}(X_1,\dots,X_n)} = {n\over \mu}e^{-{(n/ \mu)}x}\ .$$ 
This is the exponential density with mean ${\mu/ n}$.
\endNE\NE 
\e {11} %exer 7.2.103
10 hours
\endNE\NE
\e {12} %exer 7.2.104
By Exercise 10 the first claim has the mean of $\mu/ 50$.  If
$\mu $\ is about 30 years, then ${\mu/ 50}$ is about 7 months,
which is practical.  Once we have estimated ${\mu/ 50}$, we have
an estimate for $\mu$.
\endNE\NE
\e {13} %exer 7.2.105
$Y_1 = -\hbox{log} (X_1)$ has an exponential density
$f_{_{Y_1}}(x) = e^{-x}$.  Thus $S_n$ has the gamma density
$$f_{_{S_n}}(x) = {{x^{n-1}e^{-x}}\over{{(n-1)!}}}\ .$$
Therefore
$$ f_{_{Z_n}}(x) = {1\over{(n-1)!}}\Bigr(\hbox{log}{1\over x}\Bigl)^{n-1}\ .$$
\endNE\NE
\e {14} %exer 7.2.106
$X_3 = -X_2$ has density
$$f_{_{-X_2}}(x) = \cases{e^{\lambda x}, &$-\infty < x \le 0$; \cr  
0, & otherwise.}$$
Thus $Z = X_1 + X_3$ has density 
$$\eqalign{f_{_Z}(x) & =
\int_0^\infty e^{\lambda(x-2y)}dy = {1\over {2\lambda}}e^{\lambda
x},  \hskip 40pt x < 0;\cr
&=\int_x^\infty e^{\lambda(x-2y)}dy = {1\over {2\lambda}}e^{\lambda
x}{\Bigr(e^{-2\lambda x}\Bigl) = {1\over {2\lambda}}e^{-{\lambda
x}},\hskip 5pt x \ge 0.}}$$
\endNE\NE
\e {19} %exer 7.2.18
The support of
$X+Y$ is $[a+c,b+d]$.
\endNE\NE
\e {20} %exer 7.2.111
We prove it by induction.  It is true for $n = 1$.  Suppose that
$f_{_{S_k}}$ has support on $[kc,kb]$.  Then $f_{_{S_{k+1}}} =
f_{_{S_k}}*f_{_X}$ has support on [ka + a, kb + b] =
[(k+1)a,(k+1)b] (See Exercise 19 above.)
\endNE\NE
\e{21} %exer 7.2.112
(a) $$f_{_A}(x)
= {1\over {\sqrt {2\pi n}}}\thinspace e^{-{{x^2}/ {(2n)}}}\ .$$
\itemitem{} (b) $${f_{_{A}}}(x) = {{{n^n}{x^n}e^{-nx}}/ {(n-1)!}}\ .$$
\endNE

\vskip 20pt
%||*****||*****||*****||%sec8.1
\vskip 20pt

%\input mydefinitions
\vskip 20pt
\indent\bf SECTION 8.1\rm
\endNE\NE
\e 1 %exer 8.1.1
1/9
\e 3 %exer 8.1.101
We shall see that $S_n - n/2$ tends to
infinity as $n$ tends to infinity. While the difference
will be small compared to $n/2$, it will not tend to 0. 
On the other hand the
difference \hbox{${S_n}/n- 1/2$}\ does tend to 0.
\endNE\NE
\e 4 %exer 8.1.102
You will lose on the average 1.41 percent of the
money that you bet. Thus if you play a long time, you will
lose a lot. The law of large numbers tells you
that the probability that you will be ahead in the long run
tends to 0.
\endNE\NE
\e 5 %exer 8.1.103
$k = 10$
\endNE\NE
\e 6 %exer 8.1.6
$V\left(\displaystyle{S_n\over n} - p\right) = V\left({S_n\over
n}\right) = \D{{p(1-p)}\over n}. $ Thus $P\left(\displaystyle\left |
{{S_n}\over n} - p\right | \ge \epsilon\right) \le 
\displaystyle{{p(1-p)}\over\displaystyle{ n\epsilon^2}}$.
\endNE\NE
\e 7 %exer 8.1.7
$$\eqalign{p(1-p)& = {1\over 4} - \left({1\over 4} - p + p^2\right)\cr &
= {1\over 4} - ({1\over 2} - p)^2 \le {{1\over 4}\ .}}$$  
Thus, $\D\max_{0\le p \le 1}p(1-p) = {1\over 4}.$  From Exercise 6 we have
that 
$$P\left(|{S_n\over n} - p | \ge \epsilon\right) \le {{p(1-p)}\over{
n\epsilon^2}} \le {1\over {4n\epsilon^2}}\ .$$ 
\endNE\NE
\e 8 %exer 8.1.104
No.
\endNE\NE
\e 9 %exer 8.1.105
$$\eqalign{ P(S_n \ge 11)& = P(S_n-E(S_n) \ge 11-E(S_n))\cr& =
P(S_n - E(S_n) \ge 10) \cr& \le
\D{V(S_n)\over {10^2}} = .01.}$$
\endNE\NE
\e {10} %exer 8.1.106
$$\eqalign{ P(X \ge k+1)& = P(X-E(X) \ge k+1 -
E(X))\cr & = P(X - E(X) \ge k) \cr &\le {V(X)\over{k^2}} =
\D{1\over{k^2} }.}$$
\endNE\NE
\e {11} %exer 8.1.107
No, we cannot predict the proportion of heads
that should turn up in the long run, since this will depend
upon which of the two coins we pick.  If you have
observed a large number of trials then, by the Law of
Large Numbers, the proportion of heads should be near the
probability for the coin that you chose.  Thus,
in the long run, you will be able to tell which coin you have from
the proportion of heads in your observations.  To be 95
percent sure, if the proportion of heads is less than .625, 
predict  $p = 1/2$;\ if it is greater than .625, predict 
$p = 3/4$. Then you will get the correct coin if the proportion
of heads does not deviate from the probability of heads by
more than .125.  By Exercise 7, the probability of a
deviation of this much is less than or equal to $ 1/(4n(.125)^2).$ This will
be  less than or equal to $ .05 $\ if $n > 320$.  Thus with
321 tosses we can be 95 percent sure which coin we have.
\endNE\NE
\e {12} %exer 8.1.13
$$\eqalign{P\left(\displaystyle \Bigl |{{S_n} \over n} -{
M_n\over n} \Bigr | > \epsilon\right) & = P\left({1\over n} \Bigr |
\sum_{i = 1}^n (X_i - m_i)\Bigl | > \epsilon\right) \cr &=
P\left(\Bigr | \sum_{i = 1}^n (X_i - m_i)\Bigr | > n\epsilon
\right)\cr &\le \displaystyle{1\over{n^2\epsilon^2}}\sum
_{i = 1}^n \sigma_k^2 < \displaystyle{nR\over
{n^2}\epsilon^2}\cr & =\displaystyle{ R\over
{n\epsilon^2}}\ .}$$
This last expression approaches 0 as $n$ goes to $\infty$.
\endNE\NE
\e {14} %exer 8.1.109
$$\eqalign{E(|X - E(X)|)& = \sum_{\omega}|X(\omega) -
E(X)|\cr & \ge \sum _{\{\omega:|X(\omega) - E(X)|\ge
\epsilon\}}|X(\omega)-E(X)|\cr  & \ge \epsilon P(|X -
E(X)| \ge \epsilon).}$$
\endNE\NE
\e {15} %exer 8.1.16
Take as $\Omega$ the set of all sequences of 0's
and 1's, with 1's indicating heads and 0's indicating tails.
We cannot determine a probability distribution by simply
assigning equal weights to all infinite sequences, since
these weights would have to be 0.  Instead, we assign probabilities
to finite sequences in the usual way, and
then probabilities of events that depend on infinite sequences can be
obtained as limits of these finite sequences. (See Exercise 28 of Chapter 1,
Section 1.2.) 
\endNE\NE
\e {16} %exer 8.1.16.5
The exercise as stated in the text is incorrect.  The following replacement
exercise, sent to us by David Maslen, is correct:  In this exercise, we shall
construct an example of a sequence
of random variables that satisfies the weak law of large numbers, but not the
strong
law. The distribution of $X_i$ will have to depend on $i$, because
otherwise both laws would be satisfied. As a preliminary, we need to prove
a lemma, which is one of the Borel-Cantelli lemmas.  Suppose we have an infinite
sequence of mutually
independent events $A_1, A_2, \dots$. Let $a_i = {\rm Prob}(A_i)$, and let $r$
be a positive integer.
\itemitem{} (a)  Find an expression of the probability that none of the $A_i$
with
$i>r$ occur.
\itemitem{} (b)  Use the fact that $x-1 \leq e^{-x}$ to show that
$$
{\rm Prob}({\rm No}\ A_i\ {\rm with}\ i > r\ {\rm occurs}) \leq
e^{-\sum_{i=r}^{\infty}
a_i}\ .
$$
\itemitem{} (c) Prove that if $\sum_{i=1}^{\infty} a_i$ diverges, then
$$
{\rm Prob}({\rm infinitely\ many\ }A_i\ {\rm occur}) = 1\ .
$$
Now, let $X_i$ be a sequence of mutually independent random variables such
that for each positive integer $i \geq 2$,
$$
{\rm Prob}(X_i = i) = {1\over{2i\log i}}\ , \quad {\rm Prob}(X_i = -i) =
{1\over{2i\log i}}\ , \quad {\rm Prob}(X_i = 0) = 1 - {1\over{i \log i}}\ .
$$
When $i=1$ we let $X_i=0$ with probability $1$. As usual we let $S_n = X_1
+ \cdots + X_n$. Note that the mean of each $X_i$ is $0$.
\itemitem{} (d) Find the variance of $S_n$.
\itemitem{} (e) Show that the sequence $\{ X_i \}$ satisfies the weak
law of large numbers, i.e.\ prove that for any $\epsilon > 0$
$$
{\rm Prob}\Biggl( \biggl|{{S_n}\over{n}}\biggr| \geq \epsilon\Biggr) \rightarrow
0\ ,
$$
as $n$ tends to infinity.  We now show that $\{ X_i \}$ does not satisfy the
strong law of
large numbers. Suppose that $S_n / n \rightarrow 0$. Then because
$$
{{X_n}\over{n}} = {{S_n}\over{n}} - {{n-1}\over n} {{S_{n-1}}\over{n-1}}\ ,
$$
we know that $X_n / n \rightarrow 0$. From the definition of limits, we
conclude that the inequality $|X_i| \geq i/2$ can only be
true for finitely many $i$.
\itemitem{} (f) Let $A_i$ be the event $|X_i| \geq i/2$. Find
${\rm Prob}(A_i)$. Show that 
$$\sum_{i=1}^{\infty} {\rm Prob}(A_i)$$ 
diverges (think Integral Test).
\itemitem{} (g) Prove that $A_i$ occurs for infinitely many $i$.
\itemitem{} (h) Prove that 
$${\rm Prob}\biggl({{S_n}\over n} \rightarrow 0\biggr) = 0\ ,$$
and hence that the Strong Law of Large Numbers fails for the sequence
$\{ X_i \}$.
\endNE\NE
\e {17} %exer 8.1.110
For $x \in [0,1]$, let us toss a biased coin that
comes up heads with probability $x$.  Then
$$\displaystyle E\Bigr({{f(S_n)\over n}}\Bigl) \to
f(x).$$  But $$\displaystyle E\Bigr({{f(S_n)\over
n}}\Bigl) = \sum_ {k = 0}^n f\Bigl({k\over
n}\Bigr){n\choose k} x^k(1-x)^{n-k}.$$ \itemitem{}The
right side is a polynomial, and the left side tends to f(x). Hence  $$\sum_ {k
= 0}^n f\Bigl({k\over n}\Bigr){n\choose k} x^k(1-x)^{n-k} \to
f(x).$$ \itemitem{} This shows that we can
obtain obtain any continuous function $f(x)$ on [0,1] as
a limit of polynomial functions.
\endNE
\vskip 20pt
%||*****||*****||*****||%sec8.2
\vskip 20pt

%\input mydefinitions
\vskip 20pt
\indent\bf SECTION 8.2\rm
\NE
\e 1 %exer 8.2.1
(a)\ \ 1
\vskip 3pt
\itemitem{}(b)\ \ 1
\vskip 3pt
\itemitem{}(c)\ \ 100/243
\vskip 3pt
\itemitem{}(d)\ \ 1/12
\endNE\NE
\e 2 %exer 8.2.2
(a) $E(X) = 10,\  V(X) = {100/ 3}$.
\vskip 5 pt
\itemitem{} (b) $P(|X - 10| \ge 2) = {4/5}, \hskip 10
pt P(|X - 10| \ge 5) = {1/2}.$
\vskip 5 pt
\itemitem{} \hskip 15pt$P(|X - 10| \ge 9) = {1/ 10},
\hskip 10pt P(|X - 10| \ge 20) = 0.$
\endNE\NE
\e 3 %exer 8.2.3
$$f(x) = \cases{1 - x/10, & if $\ 0 \le x \le 10;$\cr
0 & otherwise.\cr}$$
$$g(x) = {{100}\over {3x^2}}\ .$$
\endNE\NE
\e 4 %exer 8.2.100
(a) E($X$) = $1/\lambda$ = 10,  V($X$) = $(1/\lambda)^2 $=
100.
\vskip 5pt
\itemitem{}\IND{(b)} For the first three probabilities
Chebyshev's estimate is greater than 1, and so the best
estimate is 1. For the last one Chebyshev's estimate gives
\hfill\break$P(|X - 10| \ge 20) \le.25. $\vskip 5pt
\itemitem{}(c) Comparing these Chebyshev's estimates with the exact
values,
 we have: $$(1, .852),\ (1,.617), \ (1,.245),
\ (.25,.0498).$$
\endNE\NE
\e 5 %exer 8.2.101
(a)\ \ 1, 1/4, 1/9
\vskip 3pt
\itemitem{}(b)\ \ 1 vs. .3173, .25 vs. .0455, .11 vs. .0027
\endNE\NE
\e 6 %exer 8.2.102
(a) \ \ 1, \ \ (b)\ \  {1/4},\ \ (c)\ \ {1/
9},\ \ (d) \ \ {1/ 16}.
\vskip 5pt
\itemitem{} The exact values are (a) \ \ .3173,\ \  (b)\ \ 
.0455, \ \  (c)\ \  .0027,\ \  (d) \ \ 0.
\endNE\NE
\e 7 %exer 8.2.103
(b)\ \ 1, 1, 100/243, 1/12
\endNE\NE
\e 8 %exer 8.2.104
(a) $$\eqalign{P(|X^*| \ge a)& =
P\Bigl(\Bigl|{{X-\mu}\over \sigma}\Bigr| \ge a\Bigr)\cr& =
P(|X-\mu| \ge a\sigma)\cr &\le {\sigma^2\over
{\sigma^2a^2}} = {1\over {a^2}}.}$$
\vskip 5pt
\itemitem{} \IND{(b)} $P(|X^*| \geq 2) = {1/4}, \hskip 10
pt P(|X^*| \geq 5) = {1/25}.$
\vskip 5 pt
\itemitem{} \hskip 15pt$P(|X^*| \geq 9) = {1/81}.$
\endNE\NE
\e 9 %exer 8.2.105
(a)\ \ 0
\vskip 3pt
\itemitem{}(b)\ \ 7/12
\vskip 3pt
\itemitem{}(c)\ \ 11/12
\endNE\NE
\e {10} %exer 8.2.106
(a) $$\eqalign{P(65\le X \le 75) &= P(65-70 \le X - 70 \le
75-70)\cr & = P(-5 \le X-70\le 5)\cr & = 1 - P(|X - 70| \ge 5)\cr &
\ge 1-25/25 = 0.}$$  \itemitem{}Thus Chebyshev's estimate gives us a
useless lower bound in this case.
\vskip 5pt
\itemitem{} (b)\ $ E(\bar X) = 70, \ V(\bar X) = 25/100
= .25.$  Thus $$\eqalign{P(65 \le \bar X \le 75)& = 1-P(|\bar X - 70|
\ge 5) \cr &\ge 1-{.25\over 25} = .99.}$$ \itemitem{} Therefore,
Chebyshev's estimate gives a lower bound of .99.
\endNE\NE
\e {11} %exer 8.2.107
(a)\ \ 0
\vskip 3pt
\itemitem{}(b)\ \ 7/12
\endNE\NE
\e {12} %exer 8.2.12
(a) $E(Y_2) = 30$, \ \ \ \  $V(Y_2) = {1\over 4}.$ \ \
\ \ \ \ \  Thus\ $ P(25 \le Y_2 \le 35) \ge .99.$
\vskip 5pt
 \itemitem{}(b) $E(Y_{11}) = 30,\ \ \ \ V(Y_{11}) = {10\over 4}.$\ \
\ \ \  Thus\  $P(25\le Y_{11} \le 35) \ge .9.$
\vskip 5pt
\itemitem{} (c) $E(Y_{101}) = 30,\ \ \  V(Y_{101}) = {100\over 4}.$
\ \ Thus\  $P(25\le Y_{101} \le 35) \ge 0.$
\endNE\NE
\e {13} %exer 8.2.108
(a)\ \ 2/3
\vskip 3pt
\itemitem{}(b)\ \ 2/3
\vskip 3pt
\itemitem{}(c)\ \ 2/3
\endNE\NE
\e {17} %exer 8.2.111
$E(X) = \int _{-\infty}^{\infty}xp(x)dx.$  Since $X$ is
non-negative, we have 
$$E(X) \ge \int_{x\ge a}xp(x)dx \ge aP(X \ge a)\ .$$
\endNE\NE
\e {18} %exer 8.2.112
Since $E(X)$ = 20  and $X$ is non-negative, we have:
$$20 = \int_0^\infty xp(x)dx \ge \int_a^\infty xp(x)dx \ge aP(X\ge
a)\ .$$ 
Therefore,
$$ P(X \ge a) \le {20\over a}\ .$$
\itemitem{} This is  interesting only for $a \ge 20$.
\vskip 5pt
\itemitem{}\IND{(b)} Now assume $E(X)$ = 20 and $V(X)$ = 25.  Then
$$E(X^2) = V(X) +E(X)^2 = 425\ .$$  Thus
$$425 = \int_0^\infty x^2p(x)dx \ge \int_a^\infty x^2dx \ge
a^2P(X\ge a)\ .$$ 
Therefore 
$$P(X \ge a) \le {425\over a^2}\ .$$ 
>From part (a) we also have that $P(X \ge a) \le \displaystyle{20\over a}.\ $
Thus our best upper bounds are:
$$P(X \ge a) \le {20\over a}\ \ \ \hbox{if}\   20
\le a \le 21.25\ ,$$
and 
$$P(X \ge a) \le {425\over a^2}\ \ \ \hbox{if}\   a \ge 21.25\ .$$ 
\vskip 5pt
\itemitem{}\IND{(c)} Since $X$ is non-negative and the density is
symmetric with mean 20, we must have $p(x)$ positive only on the
interval [0,40].  Again by symmetry we have 
$$10 = \int_{20}^{40}xp(x)\,dx\ .$$ Thus  for $a \ge 20$,
$$10 = \int_{20}^{40}xp(x)\,dx \ge \int_a^{40}xp(x)\,dx \ge aP(X\ge a)\ .$$ 
Therefore, 
$$P(X \ge a) \le {10\over a}\ .$$ 
Again by symmetry we have
$$\int_{20}^{40}(x-20)^2p(x)dx = 12.5\ .$$  
Then
$$\int_{20}^{40}x^2p(x)dx - 40\int_{20}^{40}xp(x)dx +
400\cdot{1\over 2} = 12.5\ .$$ 
>From this we obtain
$$\int_{20}^{40}x^2p(x)dx = 12.5 + 40\cdot10 - 200 = 212.5\ .$$
Therefore, for $a \ge 20$ we have 
$$212.5 = \int_{20}^{40}x^2p(x)dx
\ge \int_{a}^{40}xp(x)dx \ge a^2P(X \ge a)\ .$$  
Thus for $a \ge 20\ $ we
have 
$$P(X\ge a) \le {212.5\over a^2}\ .$$ 
Combining our two estimates we have: 
$$P(X \ge  a) \le {10\over a} \ \ \ \hbox{if}\ 20\le a \le
21.25\ ,$$ 
and 
$$P(X \ge a) \le {212.5\over a^2} \ \ \ \hbox{if}\  21.25
\le a \le 40\ .$$
\endNE

\vskip 20pt
%||*****||*****||*****||%sec9.1
\vskip 20pt

%\input mydefinitions
\indent\bf SECTION 9.1\rm (The answers to the problems in this chapter do not
use the `1/2 correction 
mentioned in Section 9.1.
\NE
\e 1 %exer 9.1.1
(a)\ \ .158655
\vskip 3pt
\itemitem{}(b)\ \ .6318
\vskip 3pt
\itemitem{}(c)\ \ .0035
\vskip 3pt
\itemitem{}(d)\ \ .9032
\endNE\NE
\e 2 %exer 9.1.2
(a) \ \ .0564
\vskip 3pt
\itemitem{}(b)\ \ .0208
\vskip 3pt
\itemitem{}(c)\ \ $1.033\times10^{-3}$
\endNE\NE
\e 3 %exer 9.1.3
(a)\ \ $P({\rm June\ passes}) \approx .985$
\vskip 3pt
\itemitem{}(b)\ \ $P({\rm April\ passes}) \approx .056$
\endNE\NE
\e 4 %exer 9.1.4
(a) $P(499,500 < S_{1,000,000} < 500,500) \ge 0$
\ by Chebyschev. \vskip 5pt
\itemitem{} (b) $P(499,500 < S_{1,000,000} < 500,500)
\approx .6826$\ by the Central Limit Theorem.
\vskip 5pt
\itemitem{} (a) $P(499,000< S_{1,000,000} < 501,000)\ge
.75$\ by Chebyschev.
\vskip 5pt
\itemitem{} (b) $P(499,000< S_{1,000,000} <
501,000)\approx .9545$\ by the Central Limit Theorem.
\vskip 5pt
\itemitem{} (a) $P(498,500< S_{1,000,000} < 501,500)\ge
.8889$\ by Chebyschev. 
\vskip 5pt
\itemitem{} (b) $P(498,500< S_{1,000,000} <
501,500)\approx .9973$ by the Central Limit Theorem.
\endNE\NE
\e 5 %exer 9.1.5
Since his batting average was .267, he must have had 80 hits.  The
probability that one would obtain 80 or fewer successes in 300 Bernoulli
trials, with individual probability of success .3, is approximately .115.
Thus, the low average is probably not due to bad luck (but a statistician
would not reject the hypothesis that the player has a probability of success
equal to .3).
\endNE\NE
\e 6 %exer 9.1.6
We need to choose $k$ so that $P(S_{1000} \le k) \ge
.99$. This is the same as 
$$P\biggl(S_{1000}^* \le \D{{k-500}\over 15.81}\biggr) \ge .99\ .$$  
Thus we want 
$$\D{{k-500}\over 15.81} = 2.33\ .$$  This will be true if $k$ =  537.
\endNE\NE
\e 7 %exer 9.1.7
.322
\endNE\NE
\e 8 %exer 9.1.8
We want $np + 2\sqrt {npq} = 108$ and $np - 2\sqrt
{npq} = 72.$ Adding and subtracting gives $2np = 180$ and
$4\sqrt {npq}$ = 36. Solving these two equations for $n$
and $p$ gives \hfill\break$p$ = .1 and $n$ = 900. 
\endNE\NE
\e 9 %exer 9.1.9
(a)\ \ 0
\vskip 3pt
\itemitem{}(b)\ \ 1 (Law of Large Numbers)
\vskip 3pt
\itemitem{}(c)\ \ .977 (Central Limit Theorem)
\vskip 3pt
\itemitem{}(d)\ \ 1 (Law of Large Numbers)
\endNE\NE
\e {10} %exer 9.1.10
We want$$ P(S_{10,000} \le 931) = P(S_{10,000}^*
\le \D{{931-1000}\over 30 }) = P(S_{10,000}^*\le -2.3)
\approx .011.$$
\endNE\NE
\e {12} %exer 9.1.12
13
\endNE\NE 
\e {13} %exer 9.1.13
$P\left(S_{1900} \ge
115\right) = P\left(S_{1900}^* \ge \displaystyle{{115-
95}\over \sqrt {1900\cdot .05 \cdot .95}}\right) =
P\left(S_{1900}^* \ge 2.105\right) = .0176. $ 
\endNE\NE
\e {14} %exer 9.1.14
(a)\ \ 64 to 96
\vskip 3pt
\itemitem{}(b)\ \ 6400
\endNE\NE
\e {16} %exer 9.1.16
$\displaystyle{n = 108,\ m = 77}$
\endNE\NE
\e {17} %exer 9.1.17
We want $\displaystyle{{2\sqrt {pq}}\over \sqrt n}$
= .01.  Replacing $\displaystyle\sqrt {pq}$ by its upper
bound ${1\over 2}$, we have $\displaystyle{1\over \sqrt n}
= .01.$  Thus we would need $n$ = 10,000.  Recall that by
Chebyshev's inequality we would need 50,000.
\endNE

\vskip 20pt
%||*****||*****||*****||%sec9.2
\vskip 20pt

%\input mydefinitions
\indent\bf SECTION 9.2\rm
\NE
\e 1 %exer 9.2.100
(a)\ \ .4762
\vskip 3pt
\itemitem{}(b)\ \ .0477
\endNE\NE
\e 2 %exer 9.2.101
.3174
\endNE\NE
\e 3 %exer 9.2.102
(a)\ \ .5
\vskip 3pt
\itemitem{}(b)\ \ .9987
\endNE\NE
\e 5 %exer 9.2.104
(a) $P(S_{210} < 700) \approx .0757.$
\vskip 5pt
\itemitem{} (b) $P(S_{189} \ge 700) \approx  .0528$
\vskip 5pt
\itemitem{}
  $\eqalign {(c)\  P(S_{179} < 700, S_{210} \ge
700) & = P(S_{179} < 700) - P(S_{179} < 700, S_{210}
< 700)\cr & = P(S_{179} < 700) - P(S_{210} < 700)\cr &
\approx .9993- .0757 = .9236\ .}$
\endNE\NE
\e 6 %exer 9.2.105
(a)\ \ The expected loss is .2 cents and the variance
of this loss is .36.
\vskip 5pt
\itemitem{} (b)\ \ .2024
\vskip 5pt
\itemitem{} (c)\ \ .047
\itemitem{} (d)\ \ .9994
\itemitem{} (e)\ \ 54
\endNE\NE
\e 7 %exer 9.2.106
(a)\ \ Expected value = 200, variance = 2
\vskip 3pt
\itemitem{}(b)\ \ .9973
\endNE\NE
\e 8 %exer 9.2.107
$ P(S_{30} = 0) \approx  \D{N(0)\over \sqrt{30\cdot
1.5}} = .0595.$
\endNE\NE
\e 9 %exer 9.2.108
$ P\Bigl(\Bigl\vert \D{S_n\over n} - \mu\Bigr\vert
\ge \epsilon\Bigr) = P\Bigl(\Bigl \vert S_n - n\mu \Bigr
\vert \ge n \epsilon\Bigl)= P\Bigl(\Bigl \vert
\D{{S_n-n\mu}\over \sqrt{n\sigma^2}} \Bigr \vert \ge
\D{{n\epsilon}\over \sqrt{n\sigma^2}}\Bigr) .$ 
\vskip 5pt 
\itemitem{} By the Central Limit Theorem, this probability
is approximated by the area under the normal curve between
$\D{\sqrt n \epsilon\over \sigma}$ and infinity, and this
area approaches 0 as n tends to infinity.
\endNE\NE
\e {10} %exer 9.2.109
\IND {(a)}The law of large numbers
states that the average of Peter's fortune will be close
to 0.  \vskip 5pt
\itemitem{} \IND {(b)} The Central Limit Theorem
states, for example, that with probability .95 Peter will
not have won or lost more than \$2 after the 10,000
plays. 
\endNE\NE
\e {11} %exer 9.2.110
Her expected loss is 60 dollars.  The probability that she lost no
money is about .0013.
\endNE\NE
\e {12} %exer 9.2.111
Betting 1 dollar on red gives $E(X)=  -{1\over 37}$
and Var($X$)= .9787. \hfill\break
\itemitem{} Betting 1 dollar on 17 gives $E(X) = -{1\over 37}$ and
Var($X$) = 34.08.\hfill\break
\itemitem{} Thus, by the Central Limit Theorem, if we bet 1
dollar on red for 100 plays, $$P(S_{100} > 20) =
P(S_{100}^* > 2.295) \approx .011.$$
%\eject
\itemitem{}If we bet 1
dollar on 17 for 100 plays, we have $$P(S_{100} > 20) =
P(S_{100}^* > .389) \approx .3489. $$ \itemitem{}Thus,
betting 1 dollar on 17 for 100 plays gives a higher
probability of winning \$20. (Of course, it also gives a
higher probability of losing \$20.)
\itemitem{} Similar calculations show that if we bet 1 dollar on red
for 100 plays, the probability that we win any money is .3732.  If we
bet 1 dollar on 17 for 100 plays, the probability that we win any money
is .4781.
\endNE\NE
\e {13} %exer 9.2.112
p = .0056
\endNE

\vskip 20pt
%||*****||*****||*****||%sec9.3
\vskip 20pt

%\input mydefinitions
\indent\bf SECTION 9.3\rm
\NE
\e 1 %exer 9.4.1
$$\eqalign{E(X^*)& = {1\over\sigma}(E(X) - \mu) =
{1\over \sigma}(\mu - \mu) = 0\ ,\cr  \sigma^2(X^*)& =
E\Bigl({{X- \mu}\over \sigma}\Bigr)^2
={1\over{\sigma^2}}\sigma^2 = 1\ .}$$
\endNE\NE
\e 2 %exer 9.4.2
$\displaystyle{S_n^* = \D{{S_n-n\mu}\over \sqrt{n}\sigma}= {S_n\over
\sqrt{n}} = {{nA_n}\over \sqrt{n}} = \sqrt{n}A_n\ .}$
\endNE\NE
\e 3 %exer 9.4.3
$\displaystyle{T_n = Y_1 + Y_2 + \cdots +Y_n = \D{{S_n-n\mu}\over
\sigma}\ .}$
 Since each $Y_j$ has mean 0 and variance 1,
$E(T_n) = 0$ and $V(T_n) = n.$  Thus $\displaystyle{T_n^* = \D{T_n\over
\sqrt{n}} = \D{{S_n - n\mu}\over {\sigma\sqrt{n}}} = S_n^*\ .}$
\endNE\NE
\e 4 %exer 9.4.4
For one uniform random variable on [0,20] the mean is
10 and the variance is 400/12.  For the sum of 25 such
random variables the mean is 250 and the standard
deviation is ${\sqrt{25(400/12)}} = 28.87.$  Thus the
normal density used to approximate the sum is: 
$$
f(x) = {1\over 28.87}{1\over \sqrt{2\pi}}\ \exp\left({-{1\over
2}\left({{x-250}\over 28.87}\right)^2}\right)\ .$$ 
\itemitem{} For
the standardized sum $S^*$ the density for the normal
approximation is the density with mean 0 and standard
deviation 1: 
$$f(x) = {1\over \sqrt{2\pi}}\
e^{-{{x^2}/ 2}}\ .$$ 
\itemitem{} The average of 25
numbers $A_{25}$ has expected value 10 and standard
deviation $\sqrt{(400/12)/25} = 1.155.$  Thus the normal
density used to approximate the average is: $$f(x) =
{1\over 1.155}\D{1\over \sqrt{2\pi}}\ exp\left({-{1\over
2}\left(\D{{x-10}\over 1.155}\right)^2}\right).$$ 
\endNE\NE
\e {10}  %exer 9.4.10
By Chebyshev's inequality we would need ${10/ n}
\le .05$ or $n \ge 200$. By the Central Limit Theorem we would
need $2 \sqrt{10/ n} \approx 1$, or $n \approx 40$.  To find the
variance necessary for 10 measurements to suffice using
Chebyshev's inequality, we would need ${\sigma^2/ 10} \approx
.05$, or $\sigma^2 \approx .5.$  Using the Central Limit Theorem we
would need $2\sqrt{\sigma^2/ 10} \approx 1$, or $\sigma^2 \approx
2.5$.(A larger variance is easier to obtain.)
\endNE\NE
\e {11} %exer 9.4.11
(a)\ \ .5
\vskip 3pt
\itemitem{}(b)\ \ .148
\vskip 3pt
\itemitem{}(c)\ \ .018
\endNE\NE
\e {13} %exer 9.4.12.5
.0013
\endNE\NE
\e {14} %exer 9.4.13
(b)\ \ (20.53, 25.87)
\endNE

\vskip 20pt
%||*****||*****||*****||%sec10.1
\vskip 20pt

%\input mydefinitions
\indent\bf SECTION 10.1\rm
\NE
\e 1 %exer 10.1.1
In each case, to get $g(t)$ just replace $z$ by $e^t$ in $h(z)$.
\vskip 3pt
\itemitem{}(a)\ \ $\displaystyle{h(z) = {1\over 2}(1 + z)}$
\vskip 3pt
\itemitem{}(b)\ \ $\displaystyle{h(z) = \sum_{j = 1}^6 z^j}$
\vskip 3pt
\itemitem{}(c)\ \ $\displaystyle{h(z) = z^3}$
\vskip 3pt
\itemitem{}(d)\ \ $\displaystyle{h(z) = {1\over{k+1}}z^n \sum_{j = 1}^k z^j}$
\vskip 3pt
\itemitem{}(e)\ \ $\displaystyle{h(z) = z^n(pz + q)^k}$
\vskip 3pt
\itemitem{}(f)\ \ $\displaystyle{h(z) = {2\over {3 - z}}}$
\endNE\NE
\e 2  %exer 10.1.2
(a) $\ \mu_1 = {1/ 2},\qquad \mu_2 = {1/2},
\qquad h^\prime(1) = {1/ 2} = \mu_1,\qquad
h^{\prime\prime}(1) = 0 = \mu_2-\mu_1.$
\vskip 5pt
\itemitem{}(b)$\  \mu_1 = {7/ 2},\qquad \mu_2 = 91/6,\qquad
h^\prime(1) = 7/2,\qquad h^{\prime\prime}(1) = 70/6 =
\mu_2-\mu_1.$
\vskip 5pt
\itemitem{}(c)$\ \mu_1 = 3,\qquad \mu_2 = 9,\qquad
h^\prime(1) = 3,\qquad h^{\prime\prime}(1) = 6 =
\mu_2-\mu_1.$
\vskip 5pt
\itemitem{} (d)$\ \mu_1 = n + k/2,\qquad \mu_2 = n^2 + nk
+ k(1+2k)/6,\hfill\break$
\itemitem{}$\quad h^\prime(1) = n+ k/2 = \mu_1,\qquad
h^{\prime\prime}(1) = n^2 + (k-1)n + k(k-1)/3 = \mu_2 -
\mu_1.$
\endNE\NE
\e 3 %exer 10.1.3
(a) $\displaystyle{h(z) = {1\over 4} + {1\over 2}z + {1\over
4}z^2\ .}$ 
\vskip 5pt
\itemitem{} (b) $\displaystyle{g(t) = h(e^t) ={1\over 4} + {1\over 2}e^t
+ {1\over 4}e^{2t}\ .}$
\vskip 5pt
\itemitem{}  $\displaystyle{\eqalign{(c)\ g(t) & = {1\over 4} + {1\over
2}\Bigl(\D\Bigl(\sum_{k=0}^\infty {{t^k}\over {k!}}\Bigr) +
{1\over 4}\Bigl(\D\sum_{k=0}^\infty {{2^k}\over
{k!}}t^k\Bigr)\cr & = 1 + \sum_{k=1}^\infty\Bigl({1\over
{2k !}} + {2^{k-2}\over {k!}}\Bigr)t^k = 1 +
\sum_{k=1}^\infty{{\mu_k}\over {k!}}t^k\ .}}$
\vskip 5pt
\itemitem{} \qquad Thus $\mu_0 = 1$, and $\mu_k = {1\over 2} +
2^{k-2} \ \ \hbox{for}  \ k \ge 1.$ \vskip 5pt
\itemitem{} (d) $\displaystyle{p_0 = {1\over 4},\quad p_1 = {1\over 2},\quad p_2
=
{1\over 4}\ .}$
\endNE\NE
\e 4 %exer 10.1.4
$\displaystyle{h(z) = \D\left(1-{3\over 2}\mu_1 + {1\over 2}\mu_2\right) +
(2\mu_1 - \mu_2)z + \D{(\mu_2-\mu_1)\over 2} z^2\ .}$
\I Thus\ $\displaystyle{ p_0 = 1 -\D{3\over 2}\mu_1 + \D{1\over
2}\mu_2, \quad p_1 = 2\mu_1 - \mu_2, \quad p_3 =
\D{{\mu_2-\mu_1}\over 2}\ .}$
\endNE\NE
\e 5 %exer 10.1.5
(a)\ \ $\displaystyle{\mu_1(p) = \mu_1(p') = 3,\ \mu_2(p) = \mu_2(p') = 11}$
\itemitem{}\ \ \hskip 12pt$\displaystyle{\mu_3(p) = 43,\ \mu_3(p') = 47}$
\itemitem{}\ \ \hskip 12pt$\displaystyle{\mu_4(p) = 171,\ \mu_4(p') = 219}$
\endNE\NE
\e 6 %exer 10.1.6
(a) $\displaystyle{p_2 =
\pmatrix{0&1&2&3&4\cr0&0&{1/4}&4/9&4/9}\ .}$ 
\vskip 5pt
\itemitem{}(b) $h(z) = \D{z\over 3}(1+2z), \qquad h_2(z) =
\D{z^2\over 9}(1+4z+4z^2) = \bigl(h(z)\bigr)^2.$

\I (c) $h_n(z) = \Bigl(\D{z\over 3}(1+ 2z)\Bigr)^n.$
\I \IND {(d)} $\mu_1 = \D{5\over 3}n,$\qquad
$\mu_2 = \D{{25}\over 9}n^2 + \D{2\over 9}n.$\hfill\break
\I Thus the mean of $p_n = {5\over 3}n$, and the variance
of $p_n = {2\over 9}n.$ Since $p$ has mean ${5\over 3}$, we
see that the mean of $p_n$ is $n$ times the mean of $p$. \I
(e)\quad $p_n(j) > 0$ for $j = n,\cdots ,2n.$
\endNE\NE
\e 7 %exer 10.1.7
(a)\ \ $\displaystyle{g_{-X}(t) = g(-t)}$
\vskip 3pt
\itemitem{}(b)\ \ $\displaystyle{g_{X + 1}(t) = e^tg(t)}$
\vskip 3pt
\itemitem{}(c)\ \ $\displaystyle{g_{3X}(t) = g(3t)}$
\vskip 3pt
\itemitem{}(d)\ \ $\displaystyle{g_{aX+b} = e^{bt}g(at)}$
\endNE\NE
\e 8 %exer 10.1.8
$$\eqalign{
(a)\hskip 5pt \quad h(z) &= {1\over 3}(z + 2)\ ,\cr 
(b)\quad h_2(z)& = \Bigl({1\over 3}(z+2)\Bigr)^2\ ,\cr
(c) \quad h_n(z) & = \Bigl({1\over 3}(z+2)\Bigr)^n\ ,\cr
(d)\quad  h_n(z) &= \Bigl({1\over 3}(z^{1/n}+2)\Bigr)^n\ ,\cr
(e) \quad h_n(z)& = {1\over {z^{\sqrt n \mu/\sigma}}}\Bigl({1\over
3}(z^{1/\sqrt n \sigma} + 2)\Bigr)^n\ , }
\eqalign{ \quad g(t)
&= {1\over 3}(e^t+2)\ .\cr
\quad g_2(t) & = \Bigr({1\over3}(e^t+2)\Bigr)^2\ .\cr
\quad g_n(t) &= \Bigl({1\over3}(e^t+2)\Bigr)^n\ .\cr
\quad g_n(t)& = \Bigl({1\over3}(e^{t/n}+2)\Bigr)^n\ .\cr
  \quad g_n(t) &= e^{-t\sqrt n\mu/\sigma}\Bigl(\D{1\over 3} (e^{t/\sqrt n\sigma}
+
2)\Bigr)^n\ .}$$
%
\itemitem{}\qquad Note:\ $ \mu = 1/3\  \hbox{and}\  \sigma
= \sqrt{2}/3.$
\endNE\NE
\e 9 %exer 10.1.9
(a)$\quad \D h_{_X}(z) = \D\sum_{j=1}^6 a_jz^j,\qquad
\D h_{_Y}(z) = \D\sum_{j=1}^6 b_jz^j.$
\I (b) $\quad h_{_Z}(z) =  \Bigl(\D\sum_{j=1}^6 a_jz^j\Bigr)
\Bigl(\D\sum_{j=1}^6 b_jz^j\Bigr)\ .$
\I (c) \quad Assume that\quad $ h_{_Z}(z)  = {{(z^2 +
\cdots +                         z^{12})}/ 11\ .}$ 
\I Then $\Bigl(\D\sum_{j=1}^6 a_jz^{j-1}\Bigr)
\Bigl(\D\sum_{j=1}^6 b_jz^{j-1}\Bigr) = {{1+z+\cdots z^{10}}\over
{11}} = {z^{11}-1\over {11(z-1)}}\ .$
\itemitem{} Either $\D\sum_{j=1}^6 a_jz^{j-1} \hbox{or}
\D\sum_{j=1}^6 b_jz^{j-1}$ is a polynomial of degree 5 (i.e., either
$a_6 \ne 0$ or $b_6 \ne 0)$. Suppose that $\D\sum_{j=1}^6
a_jz^{j-1}$ is a polynomial of degree 5.  Then it must have a real
root, which is a real root of ${{(z^{11}-1)}/{(z-1)}}$. 
However  ${{(z^{11}-1)}/ {(z-1)}}$ has no real roots. 
This is because the only real root of $z^{11} - 1$ is 1,
which cannot be a real root of ${{(z^{11}-1)}/
{(z-1)}}$.  Thus, we have a contradiction.  This means that
you cannot load two dice in such a way that the
probabilities for any sum from 2 to 12 are the same. (cf.
Exercise 11 of Section 7.1).
\endNE\NE 
\e {10} %exer 10.1.10
$$\eqalign{h(1)& =
\D{{1-\sqrt{1-4pq}}\over {2q}} =
\D{{1-\sqrt{1-4p+4p^2}}\over {2q}} =
{{1-\vert2p-1\vert}\over {2q}}\cr & = \cases{{q/ p}, & if
$p \le q$,\cr 1,&if $p \ge q$.\cr}}$$  
$$h^\prime(z) =
\cases{\D{{4pqz}\over {2qz\sqrt{1-4pqz^2}}} - \D{1\over
{2qz^2}}(1-\sqrt{1-4pqz})\ ,&if $p > q, $\cr -\D{1\over
{z^2}}\bigl(1-\sqrt{1-z^2}\bigr) +  \D{1\over
{\sqrt{1-z^2}}}\ .& if $p = q$.\cr}$$ 
\I Therefore,
$$h^\prime(1) = \cases {\D{1/(p-q)}, &if $ p > q$,\cr
\infty, & if $p = q$.}$$
\endNE\NE
\e {11} %exer 10.1.11
Let $p_n$ = probability that the gambler is ruined at play
$n$. \I Then  
$$\eqalign{p_n &= 0, \qquad \hbox{if $n$
is even,} \cr p_1 &= q,\cr p_n& = p(p_1p_{n-2} + p_3p_{n-4} +
\cdots + p_{n-2}p_1), \qquad \hbox{if $n > 1$ is odd}.}$$
\I 
Thus
$$h(z) = qz + pz\Bigl(h(x)\Bigr)^2\ ,$$  
so
$$h(z) = {{1-\sqrt{1-4pqz^2}}\over {2pz}}\ .$$
\I 
By Exercise 10 we have $$h(1) = \cases{\D{q/p},& if $q
\le p$,\cr 1,& if $q \ge p,$\cr}$$
$$ h^\prime(1) = \cases{\D{1/{(q-p)},}& if $q > p$,\cr
\infty, & if $q = p$.}$$
\I This says that when $q > p$, the gambler must be
ruined, and the expected number of plays before ruin is
$1/(q-p)$.  When $p > q$, the gambler has a probability
$q/p$ of being ruined.  When $p = q$, the gambler must be
ruined eventually, but the expected number of plays until ruin is
not finite.
\endNE\NE
\e {12} %exer 10.1.12
$${{d^n}\over {{dx}^n}}\Bigl((1-x)^{1/2}\Bigr)\Big|_{x =
0} = (-1)^n{1\over 2}\Bigl({1\over 2}-1\Bigr)\cdots \Bigl({1\over 2}
- n + 1\Bigr)\ .$$
\I 
Thus, 
$$\eqalign {(1-x)^{1/2}& =
\D\sum_{n=0}^\infty (-1)^n{{1\over 2}\Bigl({1\over 2}-1\Bigr)\cdots
\Bigl({1\over 2} - n + 1\Bigr)\over {n!}}(-x)^n\cr &=
\D\sum_{n=0}^\infty{{1\over 2}\Bigl({1\over 2}-1\Bigr)\cdots
\Bigl({1\over 2} - n + 1\Bigr)\over {n!}}x^n\cr & = \D
\sum_{n=0}^{\infty} {1/2\choose n}x^n.}$$
\I 
Therefore,
$$\eqalign{h(z)& = \D{{1-\sqrt{1-4pqz^2}}\over
{2pz}} = {1-\D\sum_{n=0}^\infty{{1/2}\choose
n}(-4pqz^2)^n\over {2pz}}\cr & = \sum_{n=1}^\infty {1\over
{2p}}(-1)^{n}{{1/2}\choose n} (4pq)^nz^{2n-1}\ ,}$$
\I
and
$$ p_{_T}(n) = \cases {\D{1\over
{2p}}(-1)^{k}{{1/2}\choose k} (4pq)^k,& if $n = 2k-1$ =
odd,\cr 0,&if $n$ = even.}$$
\endNE\NE
\e {13} %exer 10.1.13
(a) From the hint:$$ h_k(z) = h_{_U{_{_1}}}(z) \cdots
h_{_U{_{_k}}}(z) = \Bigl(h(z)\Bigr)^k\ .$$
\I (b) $$h_k(1) = \Bigl(h(1)\Bigr)^k = \cases {(q/p)^k &if
$q \le p$,\cr 1&if $q\ge p.$}$$
\I $$ h^\prime(1) = \cases{k/(q-p) &if $q > p$,\cr \infty
&if  $q = p$.}$$ \I Thus the gambler must be ruined if $q
\ge p$. The expected number of plays in this case is
$k/(q-p)$ if $q > p$ and $\infty$ if $q = p$.  When $q < p$
he is ruined with probability $(q/p)^k$.
\endNE\NE
\e {14} %exer 10.1.14
$$\D g_{_{X^*}}(t) = E(e^{X^*t}) = \D E(e^{{{X-\mu}\over
\sigma}t}) =  \D e^{-{{\mu }\over \sigma} t}E(e^{{X\over\sigma }t}) =
\D e^{-{{\mu }\over \sigma}t}g_{_X}\Bigl({t\over
\sigma}\Bigr)\ .$$
\endNE

\vskip 20pt
%||*****||*****||*****||%sec10.2
\vskip 20pt

%\input mydefinitions
\indent\bf SECTION 10.2\rm
\NE
\e 1 %exer 10.2.1
(a)\ \ $d = 1$
\vskip 3pt
\itemitem{}(b)\ \ $d = 1$
\vskip 3pt
\itemitem{}(c)\ \ $d = 1$
\vskip 3pt
\itemitem{}(d)\ \ $d = 1$
\vskip 3pt
\itemitem{}(e)\ \ $d =1/2$
\vskip 3pt
\itemitem{}(f)\ \ $d \approx .203$
\endNE\NE
\e 2 %exer 10.2.2
(a) \quad .618, \quad (b)\quad .414, \quad (c)\quad 0
if $t$ = 0,\quad $\D{{-1 + \sqrt{5}}\over 2}$ =.618 \ if
$t$ $\ne$ 0
\endNE\NE
\e 3 %exer 10.2.3
(a)\ \ 0
\vskip 3pt
\itemitem{}(b)\ \ 276.26
\endNE\NE
\e 4 %exer 10.2.4
$$\eqalign{h(z) = E\Bigl(Z^{S_n}\Bigr)& = \sum_kE(Z^{S_k}\vert
N = k)P(N=k) \cr &= \sum_k\Bigl(E(Z^{X_1})\Bigr)^kP(N=k)\cr &=
\sum_k(f(z))^kP(N = k)\cr & = g(f(z))\ .}$$
\endNE\NE
\e 5 %exer 10.2.5
Let $Z$ be the number of offspring of a single
parent.  Then the number of offspring after two
generations is 
$$S_N = X_1 + \cdots + X_N\ ,$$ 
where $N = Z$ and $X_i$ are independent with generating function $f$.
Thus by Exercise 4, the generating function after two
generations is $h(z) = f(f(z))$.
\endNE\NE
\e 6 %exer 10.2.6
(a) $\ f(z) = p + qz,\  g(z) = \D{{rz}\over
{1-(1-r)z}}\ .$
\vskip 3pt
\itemitem{}(c)\ \ If $p < r$, then the expected length of the busy period
is finite.  (If $p = r$, then the expected length of the busy period is 
infinite, but with probability 1, the busy period will be of finite length.)
\endNE\NE
\e 7 %exer 10.2.7
Let $N$ be the time she needs to be served.  Then the number of 
customers arriving during this time is \hbox{$X_1 + \cdots +X_N$},
where $X_i$ are identically distributed independent of $N$.
$ P(X_0 = 0) = p, \ P(X_i = 1) = q$.  Thus by Exercise 4,
${h(z) = g(f(z))}$.
\endNE\NE
\e 8 %exer 10.2.8
The server ultimately has a time when he is not busy if
the branching procees dies out.  For this we need $m \le
1$ or $h^\prime(1)$ = 1.  But ${h(z) = g(f(z)),}$
so we need 
$$h^\prime(1) = g^\prime(1) f^\prime(1) =\ 
\hbox{mean arrival rate} \cdot\  \hbox{mean service
time}\  \le 1\ .$$ 
This means that we need  $  q/r \le 1.$ 
\endNE\NE 
\e 9 %exer 10.2.9
If there are $k$ offspring in the first
generation, then the expected total number of offspring
will be $kN$, where $N$ is the expected total numer for a
single offspring.  Thus we can compute the expected total
number by counting the first offspring and then the
expected number after the first generation.  This gives
the formula 
$$ N = 1 + \Bigl(\sum_k kp_k \Bigr) = 1 + mN\ .
$$  
>From this it follows that $N$ is finite if and only if
$m < 1$, in which case \hfill\break $N = 1/(1-m)$.
\endNE\NE
\e {10} %exer 10.2.10
You can work this by passing to the limit in the
expressions given in Example 4, but it is easier to do it
directly as follows: The generating function $h_1(z) =
h(z)$ for the population after one generation is 
$$h(z) =
\sum_{k=0}^{\infty}(1/2)^{j+1}z^j =  {1/2\over {1 -
{(1/2)}z}}={1\over{(2-z)}}\ .$$ 
Then we can get the
generating functions for future generations by using the
relation {$h_{n+1}(z) = h_n(h(z))$}. For
example, 
$$h_2(z) = {1\over ({2-{1\over (2-z)})}} =
{{2-x}\over {3-2x}}\ .$$  
Continuing in this way, we get
$$h_3(z) = {{3-2x}\over {4-3x}}\ .$$ 
These results suggest that the general case is 
$$h_n(z) = {{n-x(n-1)}\over
{(n+1)-nx}}\ .$$  
It is easy to check that this satisfies the
equation $h_{n+1} = h_n(h(z))$, so by induction we see
that our guess for $h_n(z)$ is correct. Then
$$\eqalign{h_n(z) &= {{n-x(n-1)}\over {n+1}}\Bigl({1\over
{1-{n\over {n+1}}}x}\Bigr)\cr& = {{n-z(n-1)}\over
{n+1}}\Bigl(\sum_{j=0}^\infty
\Bigl({n\over{n+1}}\Bigr)^jz^j\ .}$$ 
The constant term is $p_n(0) = n/(n+1).$ Collecting the 
coefficients of $z^j$ and simplifying gives $p^{(n)}(j) = \D{1\over
{n(n+1)}}\Bigl({n\over {n+1}}\Bigr)^j.$ 
\vskip 5pt
\IN {(b)} The probability that the population
dies out at the $n$th generation is equal to the
difference between the probability that it has died out by
the $n$th generation and the probability that it has died
out by the $(n-1)$st generation.  This is:
$$p^{n}_0 - p^{n-1}_0 = {n\over {n+1}} - {{n-1}\over n} = {1\over
{n(n+1)}}\ .$$ 
\vskip 5pt \itemitem{}(c) The expected
lifetime is 
$$\eqalign{\sum_{n}n\cdot \hbox{P(population
dies out on the $n$th generation)} & = \sum_{n=1}^\infty
{n\over {n(n+1)}}\cr & = \sum_{n=1}^\infty {1\over {n+1}}
= \infty\ .}$$
\endNE

\vskip 20pt
%||*****||*****||*****||%sec10.3
\vskip 20pt

%\input mydefinitions
\indent\bf SECTION 10.3\rm
\NE
\e 1 %exer 10.3.1
(a)\ \ $\D{g(t) = {1\over {2t}}(e^{2t} - 1)}$
\vskip 3pt
\itemitem{}(b)\ \ $\D{g(t) = {{e^{2t}(2t - 1) + 1}\over{2t^2}}}$
\vskip 3pt
\itemitem{}(c)\ \ $\D{g(t) = {{e^{2t} - 2t - 1}\over {2t^2}}}$
\vskip 3pt
\itemitem{}(d)\ \ $\D{g(t) = {{e^{2t}(ty-1) + 2e^t - t - 1}\over{t^2}}}$
\vskip 3pt
\itemitem{}(e)\ \ $\D{(3/8)\biggl({{e^{2t}(4t^2 - 4t + 2) - 2}\over{t^3}}
\biggr)}$
\endNE\NE
\e 2 %exer 10.3.2
(a)\quad $\D{\mu_1 = 1 = g^\prime(0),\qquad \mu_2 =
{4\over 3} = g^{\prime\prime}(0)\ .}$
\vskip 5pt
\itemitem{}(b)\quad $\D{\mu_1 = {4\over 3} = g^\prime(0),
\qquad \mu_2 = 2 = g^{\prime\prime}(0)\ .}$
\vskip 5pt
\itemitem{}(c)\quad $\D{\mu_1 = {2\over 3} = g^\prime(0),
\qquad \mu_2 = {2\over 3} = g^{\prime\prime}(0)\ .}$
\vskip 5pt
 \itemitem{}(d)\quad $\D{\mu_1 = 1 = g^\prime(0), \qquad \mu_2
= {3\over 2} = g^{\prime\prime}(0)\ .}$
\vskip 5pt
 \itemitem{}(e)\quad $\D{\mu_1 = {3\over 2} =
g^\prime(0),\qquad \mu_2 = {12\over 5} =
g^{\prime\prime}(0)\ .}$
\endNE\NE
\e 3 %exer 10.3.3
(a)\ \ $\D{g(t) = {2\over {2-t}}}$
\vskip 3pt
\itemitem{}(b)\ \ $\D{g(t) = {{4 - 3t}\over{2(1-t)(2-t)}}}$
\vskip 3pt
\itemitem{}(c)\ \ $\D{g(t) = {4\over{(2-t)^2}}}$
(d)\ \ $\D{g(t) = \Bigl({\lambda
\over {\lambda + t}}\Bigr), \quad t < \lambda\ .}$
\endNE\NE
\e 4 %exer 10.3.4
(a) \quad $\D{\mu_1 = {1\over 2} = g^\prime(0), \qquad \mu_2 =
{1\over 2} = g^{\prime\prime}(0)\ .}$
\vskip 5pt
\itemitem{}(b)\quad $\D{\mu_1 = {3\over 4} = g^\prime(0), \qquad \mu_2 =
{5\over 4} = g^{\prime\prime}(0)\ .}$
\vskip 5pt
\itemitem{}(c)\quad $\D{\mu_1 = 1 = g^\prime(0), \qquad \mu_2 =
{3\over 2} = g^{\prime\prime}(0)\ .}$
\vskip 5pt
\itemitem{}(d)\quad  $\D{\mu_1 = {n\over \lambda} = g^\prime(0), \qquad
\mu_2 = {{n(n+1)}\over {\lambda^2}} = g^{\prime\prime}(0)\ .}$
\endNE\NE
\e 5 %exer 10.3.5
(a)\ \ $\D{k(\tau) = {1\over{2i\tau}}(e^{2i\tau} - 1)}$
\vskip 3pt
\itemitem{}(b)\ \ $\D{k(\tau) = {{e^{2i\tau}(2i\tau - 1) + 1}\over{-2\tau^2}}}$
\vskip 3pt
\itemitem{}(c)\ \ $\D{k(\tau) = {{e^{2i\tau} - 2i\tau - 1}\over{-2\tau^2}}}$
\vskip 3pt
\itemitem{}(d)\ \ $\D{k(\tau) = {{e^{2i\tau}(i\tau -1) + 2e^{i\tau} - i\tau - 1}
\over{-\tau^2}}}$
\vskip 3pt
\itemitem{}(e)\ \ $\D{k(\tau) = (3/8)\biggl({{e^{2i\tau}(-4\tau^2 - 4i\tau + 2}
\over{-i\tau^3}}\biggr)}$
\endNE\NE
\e 6 %exer 10.3.6
$$\eqalign{f_X(x)& = {1\over {2\pi}}\int_{-\infty}^\infty
e^{-i\tau }e^{|\tau|}d\tau = {1\over {2\pi}}\int_{0}^\infty e^{-i\tau
}e^{-\tau}d\tau + {1\over {2\pi}}\int_{-\infty}^0 e^{-i\tau
}e^{\tau}d\tau\cr & = {1\over {2\pi}}\int_0^\infty\Bigl(e^{-i\tau
x}+e^{i\tau x}\Bigr)e^{-\tau}d\tau = {1\over
\pi}\int_0^{\infty}\hbox{cos}(\tau x)e^{-\tau}d\tau\ .}$$
\itemitem{} Now to calculate this last integral:

$$\eqalign{{1\over
\pi}\int_0^{\infty}\hbox{cos}(\tau x)e^{-\tau}d\tau& = {1\over
\pi}\Bigl[-e^{-\tau}\hbox{cos}{(\tau x)}\Big|_0^\infty - x\int_0^\infty
e^{-\tau}\hbox{sin}(\tau x) d\tau\Bigr]\cr
&= {1\over \pi}\Bigl[ 1 - x(-e^{-\tau}\hbox{sin}(\tau x))
\Big|_0^\infty + x^2\int_0^\infty e^{-\tau}\hbox{cos}(\tau x)\Bigr] 
\cr & = {1\over \pi}\Bigl[1 - x^2\int_0^\infty
e^{-\tau}\hbox{cos}(\tau x)d \tau \Bigr]\ .}$$
\itemitem{} Solving this for the integral we obtain:
$${1\over\pi}\int_0^{\infty}\hbox{cos}(\tau x)e^{-\tau}d\tau = {1\over
\pi(1+x^2)}\ .$$ 
\itemitem{}Thus,
 $$ f_X(x) = {1\over
\pi}\int_0^{\infty}\hbox{cos}(\tau x)e^{-\tau}d\tau = {1\over
\pi(1+x^2)}\ .$$
\endNE\NE
\e 7 %exer 10.3.7
(a)\ \ $\D{g(-t) = {{1 - e^{-t}}\over{t}}}$
\vskip 3pt
\itemitem{}(b)\ \ $\D{e^tg(t) = {{e^{2t} - e^t}\over{t}}}$
\vskip 3pt
\itemitem{}(c)\ \ $\D{g(et) = {{e^{3t} - 1}\over{3t}}}$
\vskip 3pt
\itemitem{}(d)\ \ $\D{e^bg(at) = {{e^b(e^{at} - 1)}\over{at}}}$
\endNE\NE
\e 8 %exer 10.3.8
(a) $\D{\quad g(t) = \D{{e^{at}-e^{bt}}\over {t(b-a)}}\ .}$
\vskip 5pt
\itemitem{} (b) $\D{\quad g(t) = \D\Bigl({{e^{at}-e^{bt}}\over
{t(b-a)}}\Bigr)^2\ .}$
\vskip 5pt
\itemitem{} (c) $\D{\quad g(t) = \D\Bigl({{e^{at}-e^{bt}}\over
{t(b-a)}}\Bigr)^n\ .}$
\vskip 5pt
\itemitem{} (d) $\D{\quad g(t) = \D\Bigl({{e^{at/n}-e^{bt/n}}\over
{t(b-a)}}\Bigr)^n\ .}$
\vskip 5pt
\itemitem{} (e) $\D{\quad g(t) =
e^{-\sqrt n\mu/\sigma}\D\Bigl({{e^{at/\sqrt n
\sigma}-e^{bt/\sqrt n\sigma}}\over {t(b-a)}}\Bigr)^n\ .}$
\endNE\NE
\e 9 %exer 10.3.9
(a)\ \ $\D{g(t) = e^{t^2 + t}}$
\vskip 3pt
\itemitem{}(b)\ \ $\D{\Bigl(g(t)\Bigr)^2}$
\vskip 3pt
\itemitem{}(c)\ \ $\D{\Bigl(g(t)\Bigr)^n}$
\vskip 3pt
\itemitem{}(d)\ \ $\D{\Bigl(g(t/n)\Bigr)^n}$
\vskip 3pt
\itemitem{}(e)\ \ $\D{e^{t^2/2}}$
\endNE\NE
\e {10} %exer 10.3.10
(a)\quad $\D{m = 0, \qquad \sigma^2 = 2\ .}$
\vskip 5pt
\itemitem{} (b) \quad $\D{g{_{_{X_1}}}(t) = {1\over {1-t^2}}, \qquad
g{_{_{S_n}}}(t) = \Bigl({1\over {1-t^2}}\Bigr)^n, \quad t < 1\ ,}$
\vskip 5pt
\itemitem{}\qquad\ \ $\D{g{_{_{A_n}}}(t) = \Bigl({1\over {1-({t\over
n})^2}}\Bigr)^n \qquad g{_{_{S_n^*}}}(t) = \Bigl({1\over {1-{t^2\over
{2n}}}}\Bigr)^n\ .}$ 
\itemitem{}(c) \quad $\D{g{_{_{S_n^*}}}(t) \to
e^{-{{t^2}\over 2}}\  \hbox{as}\  n \to \infty\ .}$
\vskip 5 pt
\itemitem{}(d)\quad $\D{g{_{_{A_n}}}(t) \to 1\  \hbox{as}\  n \to \infty\ .}$
\endNE
\vskip 20pt
%||*****||*****||*****||%sec11.1
\vskip 20pt

%\input mydefinitions
\indent\bf SECTION 11.1\rm
\NE
\e 1 %exer 11.1.1
$\D{{\bf w}(1) = (.5,\ .25,\ .25)}$
\itemitem{}$\D{{\bf w}(2) = (.4375,\ .1875,\ .375)}$
\itemitem{}$\D{{\bf w}(3) = (.40625,\ .203125,\ .390625)}$
\endNE\NE
\e 2 %exer 11.1.2
${\\P} = \pmatrix {1&0\cr {1\over 2}&{1\over 2}},\quad
{\\P}^2 = \pmatrix{1&0\cr {3\over 4}&{1\over 4}},  \quad
{\\P}^3 = \pmatrix{1&0\cr {7\over 8}&{1\over 8}}.$
\vskip 5pt
\itemitem{}$ {\\P}^n = \pmatrix{1&0\cr {2^n-1}\over
2^n&{1\over 2^n}}\to \pmatrix{1&0\cr 1&0}.$
\vskip 5pt
\itemitem{} Whatever the President's decision, in
the long run each person will be told that he or she is
going to run.
\endNE\NE
\e 3 %exer 11.1.3
$\D{{\bf P}^n = {\bf P}}$ for all $n$.
\endNE\NE
\e 4 %exer 11.1.4
.7.
\endNE\NE
\e 5 %exer 11.1.5
1
\endNE\NE
\e 6 %exer 11.1.6
${\\w}^{(1)} = {\\w}^{(2)} = {\\w}^{(3)} =
{\\w}^{(n)} = (.25,.5,.25).$
\endNE\NE
\e 7 %exer 11.1.7
(a)\ \ $\D{{\bf P}^n = {\bf P}}$
\vskip 3pt
\itemitem{}(b)\ \ $\D{{\bf P}^n = \cases{{\bf P}, & if $n$ is odd,\cr
{\bf I}, & if $n$ is even.\cr}}$
\endNE\NE
\e 8 %exer 11.1.8
$ {\\P} = \bordermatrix{&0&1\cr 0&1-p&p\cr 1&p&1-p}.$
\endNE\NE
\e 9 %exer 11.1.9
$\D{p^2 + q^2,\ q^2, \bordermatrix{&0&1\cr 0&p&q\cr 1&q&p\cr}}$
\endNE\NE
\e {11} %exer 11.1.11
.375
\endNE\NE
\e {12} %exer 11.1.12
(a)\quad $ {\\P} = \bordermatrix{&P&SL&UL&NS \cr
P&.64&.08&.08&.2\cr SL&.16&.48&.16&.2 \cr
UL&.2&.2&.4&.2\cr NS&0&0&0&1}.$
\vskip 5pt
\itemitem{} (b)\quad .24.
\endNE\NE
\e {19} %exer 11.1.19
(a)\quad $5/6.$
\vskip 5pt
\itemitem{} (b)\quad The `transition matrix' is
$${\\P} = \bordermatrix{&H&T\cr
H&5/6&1/6\cr
T&1/2&1/2}\ .$$
\vskip 5pt
\itemitem{} (c)\quad 9/10.
\vskip 5pt
\itemitem{} (d)\quad No.  If it were a Markov chain, then the answer to (c)
would be the same as the answer to (a).
\endNE

\vskip 20pt
%||*****||*****||*****||%sec11.2
\vskip 20pt

%\input mydefinitions
\indent\bf SECTION 11.2\rm
\NE
\e 1 %exer 11.2.1
$\D{a = 0\ {\rm or\ } b = 0}$
\endNE\NE
\e 2 %exer 11.2.2
H is the absorbing state.  Y and D are transient
states. It is possible to go from each of these states to
the absorbing state, in fact in one step.
\endNE\NE
\e 3 %exer 11.2.3
Examples 11.10 and 11.11
\endNE\NE
\e 4 %exer 11.2.4
$\\{N} = \bordermatrix{&Gg&gg\cr GG&2&0\cr gg&2&1\cr}.$
\endNE\NE
\e 5 %exer 11.2.5
The transition matrix in canonical form is 
$$\\{P} = \bordermatrix{&GG,Gg&GG,gg&Gg,Gg&Gg,gg&GG,GG&gg,gg\cr
GG,Gg&1/2&0&1/4&0&1/ 4&0\cr
GG,gg&0&0&1&0&0&0\cr Gg,Gg&1/ 4&1/ 8&1/
4&1/4&1/16&1/16\cr Gg,gg&0&0&1/
4&1/2&0&1/ 4\cr GG,GG&0&0&0&0&1&0\cr
gg,gg&0&0&0&0&0&1\cr}.$$  Thus $$\\{Q} =
\bordermatrix{&GG,Gg&GG,gg&Gg,Gg&Gg,gg\cr GG,Gg&1/2&0&1/4&0\cr
GG,gg&0&0&1&0\cr Gg,Gg&1/ 4&1/ 8&1/
4&1/4\cr Gg,gg&0&0&1/
4&1/2\cr. 
},$$ and $$\\{N} = (I - \\{Q})^{-1}
=\bordermatrix{&GG,Gg&GG,gg&Gg,Gg&Gg,gg\cr
GG,Gg&8/3&1/6&4/3&2/3\cr GG,gg&4/3&4/3&8/3&4/3\cr
Gg,Gg&4/3&1/3&8/3 &4/3&\cr Gg,gg&2/3&1/6&4/3
&8/3\cr}.$$ From this we obtain $$\\{t} = \\{Nc} = \bordermatrix{&\cr
GG,Gg&29/6\cr GG,gg&20/3\cr Gg,Gg&17/3\cr Gg,gg&29/6\cr},$$ and
$${\\B} = {\\ NR} = \bordermatrix{&GG,GG&gg,gg\cr GG,Gg&3/4&1/4\cr
GG,gg&1/2&1/2 \cr Gg,Gg&1/2&1/2\cr Gg,gg&1/4&3/4\cr}.$$
\endNE\NE
\e 6 %exer 11.2.6
The canonical form of the transition matrix is
$${\\P} = \bordermatrix{&N&S&R\cr
N&0&1/2&1/2\cr S&1/4&1/2&1/4\cr R&0&0&1\cr},$$
$${\\N} = \bordermatrix{&N&S\cr
N&4/3&4/3\cr S&2/3&8/3\cr },$$

 $${\\t} = {\\Nc} =\bordermatrix{&\cr N&8/3\cr S&10/3\cr},$$
$${\\B} = {\\NR} = \bordermatrix{&\cr N&1\cr S&1\cr}.$$
\itemitem{} Here is a typical interpretation for an entry of {\\N}.  If
it is snowing today, the expected number of nice days before the first
rainy day is 2/3. The entries of {\\t} give the expected number of days
until the next rainy day. Starting with a nice day this is 8/3, and
starting with a snowy day it is 10/3.\ The entries of {\\B} reflect the
fact that we are certain to reach the absorbing state (rainy day)
starting in either state N or state S.
\endNE\NE
\e 7 %exer 11.2.7
\ \ $\D{{\bf N} = \pmatrix{2.5 & 3 & 1.5\cr
2 & 4 & 2\cr
1.5 & 3 & 2.5\cr}}$
\vskip 3pt
\itemitem{}\ \ $\D{{\bf Nc} = \pmatrix{7\cr 8\cr 7\cr}}$
\vskip 3pt
\itemitem{}\ \ $\D{{\bf B} = \pmatrix{5/8 & 3/8\cr 1/2 & 1/2\cr 3/8 & 5/8\cr}}$
\endNE\NE
\e 8 %exer 11.2.8
The transition matrix in canonical form
is$${\\P}=\bordermatrix{&1&2&3&0&4\cr
1&0&2/3&0&1/3&0\cr2&1/3&0&2/3&0&0\cr3&0&1/3&0&0&2/3\cr0&0&0&0&1&0\cr4&0&0&0&0&1\
cr},$$
$${\\N} = \bordermatrix{&1&2&3\cr
1&7/5&6/5&4/5\cr2&3/5&9/5&6/5\cr3&1/5&3/5&7/5\cr},$$ $$ {\\B} = {\\NR} =
\bordermatrix{&0&4\cr 1&7/15&8/15\cr 2&3/15&12/15\cr 3&1/15&14/15\cr},$$
$$ {\\t} = {\\NC} = \bordermatrix{&\cr 1&17/5\cr2&18/5\cr3&11/5\cr}.$$
\endNE\NE
\e 9 %exer 11.2.9
2.08
\e {12} %exer 11.2.11
$${\bf P} = \bordermatrix{&ABC&AC&BC&A&B&C&none\cr
ABC&5/18 & 5/18 & 4/18 & 0&0& 4/18 & 0\cr
AC & 0 & 5/12 & 0 & 5/2 & 0 & 1/12 & 1/12\cr
BC & 0 & 0 & 10/18 & 0 & 5/18 & 2/18 & 1/18\cr
A & 0 & 0 & 0 & 1 & 0 & 0 & 0 \cr
B & 0 & 0 & 0 & 0 & 1 & 0 & 0\cr
C & 0 & 0 & 0 & 0 & 0 & 1 & 0\cr
none & 0 & 0 & 0 & 0 & 0 & 0 & 1\cr
}$$
\vskip 3pt
\itemitem{}$\D{{\bf N} = \pmatrix{1.385 & .659 & .692\cr 0 & 1.714 & 0\cr 
0 & 0 & 2.25\cr}}$
\vskip 3pt
\itemitem{}$\D{{\bf Nc} = \pmatrix{2.736\cr 1.714 \cr 2.25 \cr}}$
\vskip 3pt
\itemitem{}$\D{{\bf B} = \bordermatrix{& A & B & C & none\cr
ABC & .275 & .192 & .440 & .093 \cr
AC & .714 & 0 & .143 & .143\cr
BC & 0 & .625 & .25 & .125 \cr}}$
\endNE\NE
\e {13} %exer 11.2.12
Using timid play, Smith's fortune is a Markov chain with
transition matrix
$${\\P} = \bordermatrix{&1&2&3&4&5&6&7&0&8\cr1&0&.4&0&0&0&0&0&.6&0\cr
2&.6&0&.4&0&0&0&0&0&0\cr3&0&.6&0&.4&0&0&0&0&0\cr4&0&0&.6&0&.4&0&0&0&0\cr5&0&0&0&
.6&0&.4&0&0&0\cr
6&0&0&0&0&.6&0&.4&0&0\cr7&0&0&0&0&0&.6&0&.4&0\cr0&0&0&0&0&0&0&0&1&0\cr
8&0&0&0&0&0&0&0&0&1\cr}.$$ For this matrix we have $${\\B} =
\bordermatrix{&0&8\cr1&.98&.02\cr2&.95&.05\cr3&.9&.1\cr4&.84&.16\cr5&.73&.27\cr6
&.58&.42\cr7&.35&.65\cr}.$$
\itemitem{}For bold strategy, Smith's fortune is governed instead by the
transition matrix $${\\P} = \bordermatrix{&1&2&4&0&8\cr1&0&.4&0&.6&0\cr
2&0&0&.4&.6&0\cr4&0&0&0&.6&.4\cr 0&0&0&0&1&0\cr8&0&0&0&0&1\cr},$$
with $${\\B} =
\bordermatrix{&0&8\cr1&.936&.064\cr2&.84&.16\cr4&.6&.4\cr}.$$
\itemitem{} From this we see that the bold strategy gives him a
probability .064 of getting out of jail while the timid strategy
gives him a smaller probability .02. Be bold!
\endNE\NE
\e {14} %exer 11.2.13
It is the same.
\endNE\NE
\e {15} %exer 11.2.14
(a) $${\\P} =\bordermatrix{&3&4&5&1&2\cr
3&0&2/3&0&1/3&0\cr 4&1/3&0&2/3&0&0\cr 5&0&2/3&0&0&1/3\cr
1&0&0&0&1&0\cr 2&0&0&0&0&1\cr}.$$
\itemitem{}(b) $${\\N} =
\bordermatrix{&3&4&5\cr 3&5/3&2&4/3\cr 4&1&3&2\cr 5&2/3&2&7/3},$$
$${\\t}= \bordermatrix{&\cr 3&5\cr 4&6\cr 5&5\cr},$$ $${\\B}
=\bordermatrix{&1&2\cr 3&5/9&4/9\cr4&1/3&2/3\cr 5&2/9&7/9\cr}.$$
\IN{(c)} Thus when the score is deuce (state 4), the expected
number of points to be played is 6, and the probability that B wins (ends
in state 2) is 2/3.
\endNE\NE
\e {16} %exer 11.2.15
$$\bordermatrix{& g,GG & G,Gg & g,Gg & G,gg & G,GG & g,gg\cr
g,GG & 0 & 1 & 0 & 0 & 0 & 0 \cr
G,Gg & .25 & .25 & .25 & 0 & .25 & 0\cr
g,Gg & 0  & .25 & .25 & .25 & 0 & .25 \cr
G,gg & 0 & 0 & 1 & 0 & 0 & 0 \cr
G,GG & 0 & 0 & 0 & 0 & 1 & 0 \cr
g,gg & 0 & 0 & 0 & 0 & 0 & 1\cr}\ ,$$
\vskip 3pt
$${\bf N} = \pmatrix{1.667 & 2.667 & 1.333 & .333\cr
.667 & 2.667 & 1.333 & .333\cr
.333 & 1.333 & 2.667 & .667 \cr
.333 & 1.333 & 2.667 & 1.667 \cr}\ ,$$
\vskip 3pt
$${\bf Nc} = \pmatrix{6 \cr 5 \cr 5 \cr 6 \cr\ ,}$$
\vskip 3pt
$${\bf B} = \pmatrix{.667 & .333\cr
.667 & .333\cr
.333 & .667\cr
.333 & .667\cr}\ .$$
\endNE\NE
\e {17} %exer 11.2.16
For the color-blindness example, we have $${\\B} =
\bordermatrix{&G,GG&g,gg\cr g,GG&2/3&1/3\cr G,Gg&2/3&1/3\cr
g,Gg&1/3&2/3\cr G,gg&1/3&2/3\cr},$$ \itemitem{} and for Example 9
 of Section 11.1, we have $${\\B} = \bordermatrix{&GG,GG&gg,gg\cr
GG,Gg&3/4&1/4\cr GG,gg&1/2&1/2\cr Gg,Gg&1/2&1/2\cr
Gg,gg&1/4&3/4\cr}.$$\itemitem{}In each case the probability of ending
up in a state with all G's is proportional to the number of G's in
the starting state.  The transition matrix for Example 9 is $${\\P} =
\bordermatrix{&GG,GG&GG,Gg&GG,gg&Gg,Gg&Gg,gg&gg,gg\cr
GG,GG&1&0&0&0&0&0\cr GG,Gg&1/4&1/2&0&1/4&0&0\cr GG,gg&0&0&0&1&0&0\cr
Gg,Gg&1/16&1/4&1/8&1/4&1/4&1/16\cr Gg,gg&0&0&0&1/4&1/2&1/4\cr
gg,gg&0&0&0&0&0&1\cr}.$$\itemitem{}Imagine a game in which your
fortune is the number of G's in the state that you are in.  This is
a fair game.  For example, when you are in state Gg,gg your fortune
is 1.  On the next step it becomes 2 with probability 1/4, 1 with
probability 1/2, and 0 with probability 1/4.  Thus, your expected
fortune after the next step is equal to 1, which is equal to your current
fortune. You can check that the same is true no matter what state
you are in.  Thus if you start in state Gg,gg, your expected final
fortune will be 1.  But this means that your final fortune must
also have expected value 1. Since your final fortune is either 4 if
you end in $GG,GG$ or 0 if you end in $gg,gg$, we see that the
probability of your ending in $GG,GG$ must be 1/4.
\endNE\NE
\e {18} %exer 11.2.17
(a)\ \ $\D{\bordermatrix{& 1 & 2 & 3 & F & G\cr
1 & r & p & 0 & q & 0\cr
2 & 0 & r & p & q & 0\cr
3 & 0 & 0 & r & q & p\cr
F & 0 & 0 & 0 & 1 & 0 \cr
G & 0 & 0 & 0 & 0 & 1\cr}}$
\vskip 3pt
\itemitem{}(b)\ \ Expected time in second year = 1.09.
\itemitem{}\ \ \ \ \ \ Expected time in med school = 3.3 years.
\vskip 3pt
\itemitem{}(c)\ \ Probability of an incoming student graduating = .67.
\endNE\NE
\e {19} %exer 11.2.18
(a) $${\\P} = \bordermatrix{&1&2&0&3\cr
1&0&2/3&1/3&0\cr2&2/3&0&0&1/3\cr0&0&0&1&0\cr3&0&0&0&1\cr}.$$
\e {} (b) $${\\N} = \bordermatrix{&1&2\cr 1&9/5&6/5\cr 2&6/5&9/5},$$
$${\\B} =\bordermatrix{&0&3\cr1&3/5&2/5\cr 2&2/5&3/5},$$ $${\\t} =
\bordermatrix{&\cr 1&3\cr 2&3\cr }.$$\itemitem{} (c) The game will last
on the average 3 moves.  \vskip 5pt \itemitem{} (d) If Mary deals, the
probability that John wins the game is 3/5.
\endNE\NE
\e {20} %exer 11.2.19
Consider the Markov chain with state $i$ (for $1 \le i < k$) the
length of the current run, and $k$ an absorbing state.  Then when in
state $i < k$, the chain  goes to $i+1$ with probability $1/m$ or to 1
with probability $(m-1)/m$.  Thus, starting in state 1, in order to get
to  state $j+1$ the chain must be in state $j$ and then move to $j+1$. 
This means that $$N_{1,j+1} = N_{1,j}(1/m)\ ,$$ or $$N_{1,j} =
mN_{1,j+1}\ .$$  This will be true also for $j+1 = k$ if we interpret
$N_{1,k}$ as the number of times that the chain enters the state $k$,
namely, 1.  Thus, starting with $N_{1,k} = 1$ and working backwards, we
see that $N_{1,j} = m^{k-j}$ for $j = 1,\cdots ,k.$\ Therefore, the
expected number of experiments until a run of $k$ occurs is$$ 1+m + m^2
+\cdots + m^{k-1} = {{m^k-1}\over {m-1}}\ .$$ (The initial 1 is to start
the process off.) 
   Putting $m = 10$ and\ $k = 9$ we see that the expected number of
digits in the decimal expansion of $\pi$ until the first run of length 7
would be about 111 million if the expansion were random. Thus we should
not be surprised to find such a run in the first 100,000,000 digits of
$\pi$ and indeed there are runs of length 9  among
these digits.
\endNE\NE 
\e {21} %exer 11.2.20
The problem should assume that a fraction $$q_i = 1-
\sum_{j}q_{ij} > 0$$ of the pollution goes into the atmosphere and
escapes.  \IN{(a)} We note that {\\u} gives the amount of pollution in
each city from today's emission, {\\uQ} the amount that comes from
yesterday's emission, ${\\uQ}^2$ from two days ago, etc. Thus $${\\w}^n =
{\\u} + {\\uQ} + \cdots {\\uQ}^{n-1}\ .$$ 
\IN{(b)}    Form a Markov chain with  {\\Q}-matrix {\\Q} and
with one absorbing state to which the process moves with probability
$q_i$ when in state $i$. Then $$ {\\I} + {\\Q} +{\\Q}^2 + \cdots +
{\\Q}^{ n-1} \to {\\N}\ ,$$  so $${\\w}^{(n)} \to {\\w} = {\\uN}\ .$$ 
\IN{(c)} If we are given {\\w} as a goal, then we can achieve this by
solving {\\w} = {\\Nu} for {\\u}, obtaining $${\\u} =
{\\w}({\\I}-{\\Q})\ .$$
\endNE\NE
\e {22} %exer 11.2.21
\IND{(a)}  The total amount of goods
that the $i$th industry needs to produce \$1 worth of goods is $$
x_1q_{1i} + x_2q_{2i} + \cdots + x_nq_{ni}\ . $$ This is the $i$'th
component of the vector {\\xQ}. \vskip 5pt
\IN{(b)} By part (a) the amounts the industries need to meet
their internal demands is {\\xQ}.  Thus to meet both internal and
external demands, the companies must produce amounts given by a vector
{\\x} satifying the equation $${\\x} = {\\xQ} + {\\d}\ .$$ 
 \vskip 5pt \IN{(c)} From
Markov chain theory we can always solve the equation $${\\x} = {\\xQ} +
{\\d} $$ by writing it as $${\\x}({\\I}-{\\Q}) = {\\d}$$ and then using
the fact that $({\\I}-{\\Q}){\\N} = {\\I}$ to obtain $$ {\\x} =
{\\dN}\ .$$ \vskip 5pt \IN {(d)} If the row sums of {\\Q} are all less
than 1, this means that every industry makes a profit.  A company can
rely directly or indirectly on a profit-making company.  If for any
value of $n, q_{ij}^n > 0$, then $i$ depends at least indirectly on $j$.
Thus \it depending upon\rm\  is equivalent in the Markov chain
interpretation to \it being able to reach. \rm \  Thus the demands can
be met if every company is either profit-making or depends upon a
profit-making industry. \vskip 5pt \itemitem{} (e) Since $  {\\x}=  {\\d
N}$, we see that $$  {\\xc} =  {\\d N c} = {\\dt}\ . $$
\endNE\NE
\e {24} %exer 11.2.23
When the walk is in state $i$, it goes to $i+1$
with probability $p$ and $i-1$ with probability $q$.  Condition (a) just
equates the probability of winning in terms of the current state to the
probability after the next step. Clearly, if our fortune is 0, then the
probability of winning is 0, and if it is $T$, then the probability is
1.   Here is an instructive way (not the simplest way) to see that the
values of {\\w} are uniquely determined by (a), (b), and (c).  Let {\\P}
be the transition matrix for our absorbing chain.  Then these equations
state that $${\\Pw} = {\\w}\ .$$ That is, the column vector {\\w} is a
fixed vector for {\\P}.  Consider the transition matrix for an arbitrary
Markov chain in canonical form and assume that we have a vector {\\w}
such that {\\w} = {\\Pw}.  Multiplying through by {\\P}, we see that
${\\P}^2{\\w} = {\\w}$, and in general ${\\P}^n{\\w} ={\\w}$.  But
$${\\P}^n \to \pmatrix{{\\0}&{\\B}\cr {\\0}&{\\I}\cr}\ .$$ Thus $$ {\\w} =
\pmatrix{{\\0}&{\\B}\cr {\\0}&{\\I}\cr}{\\w}\ .$$ If we write $${\\w} =
\pmatrix{\\{w}_T\cr \\{w}_A}\ ,$$ where $T$ is the set of transient states
and $A$ the set of absorbing states, then by the argument above we have 
$${\\w} = \pmatrix{{\\w}_T\cr \\{w}_A} = \pmatrix{\\{Bw}_A\cr \\{w}_A}\ .$$
\itemitem{} Thus for an absorbing Markov chain, a fixed column vector
{\\w} is determined by its values on the absorbing states.  Since in our
example we know  these values are (0,1), we know that {\\w} is completely
determined.
  The solutions given clearly satisfy (b) and (c), and a direct
calculation shows that they also satisfy (a).
\endNE\NE 
\e {26} %exer 11.2.25
Again, it is easy to check that the proposed solution $f(x)
= x(n-x)$ satisfies conditions (a) and (b).  The hard part is to prove
that these equations have a unique solution.   As in the case
of Exercise 23, it is most instructive to consider this problem more
generally.  We have a special case of the following situation. 
Consider an absorbing Markov chain with transition matrix {\\P} in
canonical form and with transient states $T$ and absorbing states $A$. 
Let {\\f} and {\\g} be column vectors that satisfy the following system
of equations $$ \pmatrix{{\\Q}&{\\R}\cr
{\\0}&{\\I}}\pmatrix{{\\f}_{_A}\cr {\\0}} + \pmatrix{{\\g}_{_A}\cr
{\\0}} = \pmatrix{{\\f}_{_A}\cr {\\0}}\ ,$$  where ${\\g}_{_A}$ is given
and it is desired to determine ${\\f}_{_A}$.  In our example,
${\\g}_{_A}$ has all components equal to 1.  To solve for ${\\f}_{_A}$
we note that these equations are equivalent to $${\\Q} {\\f}_{_A} +
{\\g}_{_A} = {\\f}_{_A}\ ,$$ or $$({\\I}-{\\Q}){\\f}_{_A} = {\\g}_{_A}\ .$$
Solving for ${\\f}_{_A}$, we obtain $${\\f}_{_A} = {\\N}{\\g}_{_A}\ .$$ 
Thus ${\\f}_{_A}$ is uniquely determined by ${\\g}_{_A}$.
\endNE\NE
\e {27} %exer 11.2.29
Use the solution to Exercise 24 with {\\w} = {\\f}.
\endNE\NE
\e {28} %exer 11.2.26
Using the program \bf Absorbing Chain \rm for the transition
matrix corresponding to the pattern HTH, we find that $$ {\\t} =
\bordermatrix{&\cr HT&6\cr H&8\cr \emptyset&10}\ .$$ Thus $E(T)$ = 10. 
For the pattern HHH the transition matrix is
$${\\P}= \bordermatrix{&HHH&HH&H&\emptyset\cr HHH&1&0&0&0\cr
HH&.5&0&0&.5\cr H&0&.5&0&.5\cr\emptyset&0&0&.5&.5}\ .$$ Solving for {\\t}
for this matrix gives $${\\t} = \bordermatrix{&\cr HH&8\cr H&12\cr
\emptyset&14\cr}\ .$$  Thus for this pattern E(T) = 14.
\endNE\NE
\e {29} %exer 11.2.27
For the chain with pattern HTH we have already verified that
the conjecture is correct starting in HT.  Assume that we start in H. 
Then the first player will win 8 with probability 1/4, so his expected
winning is 2.  Thus $E(T|H)$ = $10-2$ = 8, which is correct according to
the results given in the solution to Exercise 28.  The conjecture can be
 verified similarly for the chain HHH by comparing the results
given by the conjecture with those given by the solution to Exercise 28.
\endNE\NE
\e {30} %exer 11.2.28
$T$ must be at least 3. Thus when you sum the terms $$P(T >
n) = 2P(T = n+1) + 8P(T = n+3),$$ the coefficients of the 2 and the 8
just add up to 1 since they are all possible probabilies for $T$.
Let $T$ be an integer-valued random variable.  We write $$\eqalign{
E(T) = P(T = 1) &+P(T = 2) + P(T = 3)+\cdots\cr& +P(T = 2) + P(T =
3)+\cdots\cr&\hskip .75in  + P(T = 3)+\cdots\cr}$$ If we
add the terms by columns, we get the usual definition of expected
value; if we add them by rows, we get the result that $$E(T) =
\sum_{n=0}^\infty P(T > n)\ .$$ That the order does not matter
follows from the fact that all the terms in the sum are positive.
\endNE\NE
\e {31} %exer 11.2.30
You can easily check that the proportion of $G$'s in the state
provides  a harmonic function. Then by Exercise 27 the proportion at
the starting state is equal to the expected value of the proportion
in the final aborbing state.  But the proportion of 1s in the
absorbing state $GG,GG$ is 1. In the other absorbing state $gg,gg$ it is
0.  Thus the expected final proportion is just the probability of ending
up in state $GG,GG$. Therefore, the probability of ending up in $GG,GG$
is the proportion of $G$ genes in the starting state.(See Exercise 17.)
\endNE\NE
\e {32} %exer 11.2.31
The states with all squares the same color are absorbing
states.  From any non-absorbing state it is possible to reach any
absorbing state corresponding to a color still represented in the
state. To see that the game is fair, consider the following argument. 
In order to decrease your fortune by 1 you must choose a red square and
then choose a neighbor that is not red.  With the same probability you
could have chosen the neighbor and then the red square and your fortune
would have been increased by 1. Since it is a fair game, if at any time
a proportion $p$ of the squares are red, for example, then $p$ is also
the probability that we end up with all red squares.
\endNE\NE
\e {33} %exer 11.2.32
In each
case Exercise 27 shows that $$f(i) = b_{iN}f(N) + (1-b_{iN})f(0)\ .$$ Thus
$$b_{iN} = {{f(i) - f(0)}\over {f(N) -f(0)}}\ .$$  Substituting the values
of \ $f$ \ in the two cases gives the desired results.
\endNE

\vskip 20pt 
%||*****||*****||*****||%sec11.3
\vskip 20pt

%\input mydefinitions
\indent\bf SECTION 11.3\rm
\NE
\e 1 %exer 11.3.1
(a), (f)
\endNE\NE
\e 2 %exer 11.3.2
(a) ${\\P}^3 = \pmatrix{.5&.333&.167\cr
.562&.250&.187\cr .375&.500&.125}$. Since ${\\P}^3$ has no
zero entries,  ${\\P}$ is regular.  
\itemitem{} (b) 1/6.
\vskip 5pt
\itemitem{} (c) {\\w} = (1/2,\ 1/3,\ 1/6).
\endNE\NE
\e 3 %exer 11.3.3
(a)\ \ $a = 0$ or $b = 0$
\vskip 3pt
\itemitem{}(b)\ \ $a = b = 1$
\vskip 3pt
\itemitem{}(c)\ \ ($0 < a < 1$ and $0 < b < 1$) or ($a = 1$ and $0 < b < 1$) or
($0 < a < 1$ and $b = 1$).
\endNE\NE
\e 4 %exer 11.3.4
${\\w} = (b/ (b+a),\ a/ (b+a)).$
\endNE\NE
\e 5 %exer 11.3.5
(a)\ \ $\D{(2/3, 1/3)}$
\vskip 3pt
\itemitem{}(b)\ \ $\D{(1/2, 1/2)}$
\vskip 3pt
\itemitem{}(c)\ \ $\D{(2/7, 3/7, 2/7)}$
\endNE\NE
\e 6 %exer 11.3.6
Let {\\P} = $\pmatrix{0&1\cr1&0}$. Then
${\\P}^{2n+1} = {\\P}$ and ${\\P}^{2n} = {\\I}$.  Thus
{\\P} is not regular.  However, the average\ ${\\A}_n =
{1\over n}\left({\\P}+{\\P}^2+\cdots + {\\P}^n\right)$  of
these matrices  converges to $$\pmatrix{1/2&1/2\cr
1/2&1/2}.$$ The vector {\\w} = (1/2,\ 1/2) is a fixed
vector for {\\P}. Its components represent the average
number of times the process will be in each state in the
long run.
\endNE\NE
\e 7 %exer 11.3.7
The fixed vector is $(1, 0)$ and the entries of this vector are not strictly
positive, as required for the fixed vector of an ergodic chain.
\endNE\NE
\e 8 %exer 11.3.8
The vectors (1,0,0) and (0,0,1) are fixed vectors,
and so is any vector of the form $a$(1,0,0) +
$(1-a)$(0,0,1) for $0< a< 1$.$${\\P}^n \to
\pmatrix{1&0&0\cr1/2&0&1/2\cr 0&0&1}.$$
\endNE\NE
\e 9 %exer 11.3.9
Let $${\\P} = \pmatrix{p_{11}&p_{12}&p_{13}\cr
p_{21}&p_{22}&p_{23}\cr p_{31}&p_{32}&p_{33}},$$ with
column sums equal to 1.  Then
$$\eqalign{(1/3,\ 1/3,\ 1/3){\\P}& =
(1/3\sum_{j=1}^3p_{j1},\ 1/3\sum_{j=1}^3p_{j2},\ 1/3\sum_{j=1}^3p_{j3})\cr
& = (1/3,1/3,1/3)\ .}$$ The same argument shows that if
{\\P} is an $n\times n$ transition matrix with columns that add
to 1 then $${\\w} = ({1/ n},\cdots, {1/ n})$$ is
a fixed probability vector.  For an ergodic chain this
means the the average number of times in each state is
$1/n$.
\endNE\NE
\e {10} %exer 11.3.10
In Example 11.10 of Section 11.1,  the state
$GG$ is an absorbing state, and it is impossible to go from
this state to either of the other two states.
\endNE\NE
\e {11} %exer 11.3.10.5
In Example 11.11 of Section 11.1, the state ($GG,GG$) is absorbing, and the
same reasoning as in the immediately preceding answer
applies to show that this chain is not
ergodic. 
\endNE\NE
\e {12} %exer 11.3.11
The chain is ergodic but not regular.  Note that it is impossible to reach
states
\ 1\  and 3 from state \ 0\  in an even number of steps, and it is
impossible to reach states 0,\ 2,\  and\  4\  from state \ 0 \ in an
odd number of steps. 
\endNE\NE
\e {13} %exer 11.3.12
The fixed vector is\ $ {\\w} = (a/(b+a),b/(b+a)).$
 Thus in the long run a proportion
${b/ (b+a)}$ of the people will be told that the
President will run.  The fact that this is independent of
the starting state means it is independent of the
decision that the President actually makes.  (See Exercise
2 of Section 11.1)
\endNE\NE
\e {14} %exer 11.3.13
The fixed vector is the common row of ${\bf P}$.
\itemitem{}The chain is regular if and only if the entries of this vector are
strictly
positive.
\endNE\NE
\e {15} %exer 11.3.14
It is clearly possible to go between any two
states, so the chain is ergodic.  From 0 it is
possible to go to states 0,\ 2,\ and 4 only in an even number
of steps,  so the chain is not regular.  For the general
Erhrenfest Urn model the fixed vector must statisfy the
following equations:

$${1\over n}w_1=w_0\ ,$$
$$w_{j+1}{{j+1}\over n} + w_{j-1}{{n-j+1}\over n} = w_j, 
\qquad  \hbox{if}\  0 < j < n ,$$  $${1\over n}w_{n-1} =
w_n\ .$$ \itemitem{} It is easy to check that the binomial
coefficients satisfy these conditions.
\endNE\NE
\e {16} %exer 11.3.15
${\bf P}^2$ is strictly positive.  The fixed vector is ${\bf w} = (1/4, 1/2,
1/4)$.
\endNE\NE
\e {17} %exer 11.3.16
Consider the Markov chain whose state is the
value of $S_n$ mod(7),  that is, the remainder when $S_n$
is divided by 7.  Then the transition matrix for this chain
is \vskip 5pt  $${\\P} =
\bordermatrix{&0&1&2&3&4&5&6\cr0&0&1/6&1/6&1/6&1/6&1/6&1/6\cr
1&1/6&0&1/6&1/6&1/6&1/6&1/6\cr2&
1/6&1/6&0&1/6&1/6&1/6&1/6\cr 3&
1/6&1/6&1/6&0&1/6&1/6&1/6\cr 4&
1/6&1/6&1/6&1/6&0&1/6&1/6\cr 5&
1/6&1/6&1/6&1/6&1/6&0&1/6\cr 6&
1/6&1/6&1/6&1/6&1/6&1/6&0\cr}.$$\itemitem{} Since the
column sums of this matrix are 1, the fixed vector is
$${\\w} = (1/7,1/7,1/7,1/7,1/7,1/7,1/7)\ .$$
\endNE\NE
\e {18} %exer 11.3.17
$\D{2^r + 1}$ (by this power there must have been a repetition of the pattern of 
positive numbers in the matrix so nothing new can occur).  $N(3) = 5$.  See
Exercise
19.
\endNE\NE
\e {19} %exer 11.3.18
\IN{(a)} For the general chain it is
possible to go from any state $i$ to any other state $j$ in
$r^2 -2r + 2$ steps.  We show how this can be done
starting in state 1. To return to 1, circle
$(1,2,..,r-1,1)\  r-2$ times ($r^2-3r+2$ steps) and
$(1,...,r,1)$ once (r steps). For $k = 1,...,r-1$ to reach
state $k+1$, circle $(1,2,\dots,r,1) \ r-k$ times ($r^2 -
rk$ steps)  then $(1,2,\dots,r-1,1)\  k-2$ times $(rk -2r
-k + 2$ steps) and then move to $k+1$ in $k$ steps.You have
taken $r^2-2r+2$ steps in all.   The argument is the same
for any other starting state with everything translated the
appropriate amount. 
\vskip 5pt
\IN{(b)} $${\\P}=\pmatrix{0&*&0\cr
*&0&*\cr *&0&0\cr},\ {\\P}^2=\pmatrix{*&0&*\cr *&*&0\cr
0&*&0\cr},\  {\\P}^3 = \pmatrix{*&*&0\cr *&*&*\cr
*&0&*},$$ $${\\P}^4 = \pmatrix{*&*&*\cr *&*&*\cr
*&*&0\cr}, \ {\\P}^5 = \pmatrix{*&*&*\cr *&*&*\cr
*&*&*\cr}\ .$$
\endNE\NE
\e {20} %EXER 11.3.19
The transition matrix is
$$\bordermatrix{&0&1&2&\ldots&n-1&n\cr
0&1-p&p&0&\ldots&0&0\cr
1&r(1-p)&pr+(1-p)(1-r)&p(1-r)&\ldots&0&0\cr2&0&r(1-p)&pr+(1-p)(1-r)&\ldots&0&0\cr
\vdots&\vdots&\vdots&\ddots&\vdots\cr
n&0&0&\ldots&\ldots&r&1-r\cr}\ .$$ This transition matrix has
a property called \it reversibility \rm which will be
discussed in Section 11.5.  For such a chain the fixed
vector {\\w} satisfies the condition $$w_ip_{ij} =
w_jp_{ji}\ .$$ When this condition is satisfied, it is
easy to determine the fixed vector.  For this
example,\ reversibility in this chain means that
$$w{_i}p(1-r) = w_{i+1}r(1-p)\ , \quad 0 < i < n\ ,$$ or
$${w_{i+1}\over w_{i}} = {{p(1-r)}\over{r(1-p)}}\ ,\quad 0 <
i < n\ .$$ Thus $$w_i = c\rho^i,\quad \quad 0 < i < n,$$
where $$\rho = {{p(1-r)}\over {r(1-p)}}\ .$$ \itemitem{} The
values for $w_0$ and $w_n$ are obtained from $w_0p =
w_1r(1-p)$ and $w_nr = w_{n-1}p(1-r)$.  The constant $c$ is
then chosen 
 to make the probabilities add to 1.  If
the traffic intensity $s = p/r$ is greater than
1  then $\rho$ is greater than  1, and if it
is less than 1 then  $\rho$ is less than one.  Thus when the traffic
intensity is greater than 1 the fixed vector
is large for large values of $i$, and when it is less than
 1 the fixed vector is small for large values
of $n$.  This means that the queue size will build up 
when the traffic intensity is greater than or equal to 1.
The case $\rho$ eaual to 1  is a border-line case, and in this
case the equilibrium vector is  constant for $0 < i < n.$
\endNE\NE
\e {22} %exer 11.3.21
(a) $$\eqalign{P(S = j) &=
P(\hbox{customer does not finish in $j-1$ sec.,but does in $j$th
sec.}\ )\cr & = (1-r)^{j-1}r\ ,}$$ $$\eqalign{P(T = j)& =
P(\hbox{no arrival in $j-1$ seconds,but  arrival in $j$th
second}\ )\cr & = (1-p)^{j-1}p\ .}$$  \I {(b)} $S$ and $T$
have geometric distributions, and so E($S$) = ${1/ r}$ and
E($T$) = ${1/ p}$. \vskip 5pt \IN{(c)} The traffic
intensity  \ s \ greater than 1 means $p$ is greater than  $r$, or $E(S)$ is
greater
than  $E(T)$. This means that the arrival rate is faster than the
service rate,  so the queue size builds up; $s$ equal to 1 means
$p$ is equal to $r$ or that $E(S)$ is equal to $E(T)$;
 $s$ less than 1  means that $p$ is less than  $r$, or $E(S)$ is less than
$E(T)$
and the queue size does not build up.
\endNE\NE
\e {24} %exer 11.3.23
Fixed probability vector is (1/5,\ 4/5).  Thus {\\wP} =
{\\w} implies $${1\over 5}\times .5 + {4\over 5}p = {1\over
5}\ ,$$ so $p = .125$.
\endNE\NE
\e {25} %exer 11.3.24
To each Markov chain we can
associate a {\it directed graph}, whose vertices are the
states $i$ of the chain, and whose edges are determined by
the transition matrix: the graph has an edge from $i$ to $j$
if and only if $p_{ij} > 0$.  Then to say
that {\\P} is
 ergodic means that from any state $i$ you can find
a path following the arrows until you reach any state $j$. 
If you cut out all the loops in this path you will
then have a path that never interesects itself, but
still leads from $i$ to $j$. This path will have at most
$r-1$ edges, since each edge leads to a different state
and none leads to $i$.  Following this path requires at
most $r-1$ steps.
\endNE\NE
\e {26} %exer 11.3.25
Assume that {\\P} is ergodic.
Let $M$ be the maximum of the steps required to go between
two states.  Then it is possible to go from any state $i$
to any state $j$ in $M$ steps.  To see this, assume that
it is possible to go from $i$ to $j$ in $m$ steps with $m<
M$.  Then just stay in $i$ for $M-m$ steps before starting
on your route to $M$. Thus {\\P} is regular.
\endNE\NE
\e {27} %exer 11.3.26
If {\\P} is ergodic it is possible to go between
any two states.  The same will be true for the chain
with transition matrix ${1\over 2}$({\bf I}+{\bf P}).  But
for this chain it is possible to remain in any state; 
therefore, by Exercise 26, this chain is regular. 


\endNE\NE
\e {28} %exer 11.3.27
Assume that {\\wP} = {\\w}.  Then$$
{\\w}{{({\bf I}+{\bf P})}/ 2} = {{({\\w} +
{\\w})}/ 2} ={\\w}\ .$$ Conversely, if $${\\w}
({\bf I}+{\bf P})/2 
 = {\\w} $$ then {\\wP}/2 = {\\w}/2,
and {\\wP} = {\\w}.   
\endNE\NE
\e {29} %exer 11.3.28
\IN {(b)} Since {\\P} has rational transition
probabilities, when you solve for the fixed vector you
will get a vector {\\a} with rational components.  We can
multiply through by a sufficiently large integer to obtain
a fixed vector {\bf u} with integer components  such
that each component of  {\bf u} is an integer multiple of
the corresponding component of {\bf a}.  Let ${\bf 
a}^{(n)}$ be the vector resulting from the $n$th
iteration. Let ${\\b}^{(n)} = {\\a}^{(n)}{\\P}$. Then
${\\a}^{(n+1)}$ is obtained by adding chips to
${\\b}^{(n+1)}$.  We want to prove that ${\\a}^{(n+1)} \ge
{\\a}^{(n)}$.  This is true for $n = 0$ by construction.
Assume that it is true for $n$.  Then multiplying the
inequality by {\\P} gives that ${\\b}^{(n+1)} \ge
{\\b}^{(n)}$.  Consider the component ${a}^{(n+1)}_j$. 
This is obtained by adding chips to ${b}^{(n+1)}_j$
until we get a multiple of ${a}_j$.  Since
${b}^{(n)}_j \le {b}^{(n+1)}_j$, any multiple of
${a}_j$ that could be obtained in this manner to define
${a}^{(n+1)}_j$ could also have been obtained to define
${a}^{(n)}_j$ by  adding more chips if necessary.
Since we take the smallest possible multiple ${a}_j$, 
 we must have ${a}^{(n)}_j \le
{a}^{n+1}_j$. Thus the results after each iteration are
monotone increasing.  On the other hand, they are always
less than or equal to ${\bf u}$. Since there are only a
finite number of integers between components of {\bf a}
and {\bf u}, the iteration will have to stop after a finite
number of steps.
\endNE\NE
\e {30} %exer 11.3.29
Assume that the tape can hold at most $n$ units.Then the
transition matrix is $${\\P} =
\bordermatrix{&0&1&2&3&\ldots &n-1&n\cr 0&0&1&0&0&\ldots
&0&0\cr 1&0&p&q&0&\ldots&0&0\cr 2&0&p^2& {2\choose
1}pq&q^2&\ldots&0&0\cr
\vdots&\vdots&\vdots&\vdots&\vdots&\ddots&\vdots&\vdots\cr
n&p^n& {n\choose 1}p^{n-1}q&{n\choose 2} p^{n-2}
q^2&{n\choose 3}p^{n-3}q^3& \ldots &{n\choose
{n-1}}pq^{n-1}&q^n\cr}.$$ 
   (Note that we assume that the request has no effect
when the tape is full). When the chain is in state $i$ the
expected value of the next position is $1-ip$. No matter
how small $p$ is, for large enough $i$ this will be
negative.   Thus, no matter how small\ $p$ is, we see that
if the tape is big enough, there  will be a tendancy to free up space 
when a large number of spaces
are occupied.
\endNE\NE
\e {31} %exer 11.3.30
If the maximum of a set of numbers is an average
of other elements of the set, then each of the elements
with positive weight in this average must also be
maximum. By assumption, {\\P}{\\x} = {\\x}. This implies
${\\P}^n {\\x} = {\\x}$ for all $n$. Assume that $x_i =
M$, where $M$ is the maximum value for the $x{_k}$'s, and
let $j$ be any other state. Then there is an $n$ such that
$p^n_{ij} > 0$.  The $i$th row of the equation
${\\P}^n{\\x} = {\\x}$ presents $x_i$ as an average of
values of $x_k$ with positive weight,one of which is $x_j$.   Thus
$x_j = M$, and {\\x} is constant.
\endNE\NE
\e {32} %exer 11.3.31
If 0 is the
average of  non-negative numbers, then any element of this
average occurring with positive weight must be 0. Assume
that {\\w} is a fixed probability vector for an ergodic
chain {\\P}.  Then {\\wP} = {\\w} and ${\\wP}^n = {\\w}$. 
Assume that $w_i = 0$.  For any $j$ there is an $n$ such
that $p_{ij}^n > 0$. Thus the $i$th column of ${\\wP}^n$
presents $w_i$ as an average of other values of\ $ w_k$,
with $w_j$ occurring with positive weight.  Hence $w_j =
0$. 
\endNE

\vskip 20pt 
%||*****||*****||*****||%sec11.4
\vskip 20pt

%\input mydefinitions
\indent\bf SECTION 11.4\rm
\NE
\e 1 %exer 11.4.1
$\D{\pmatrix{1/3\cr 1/3\cr}}$
\endNE\NE
\e 2 %exer 11.4.2
${\bf P}^n \to {\bf cw}.$  Thus ${\bf P}^n{\bf y} \to
{\bf cwy} = {\bf wyc}$, since {\bf wy} is a number.
\endNE\NE
\e 3 %exer 11.4.3
For regular chains, only the constant vectors are fixed column vectors.
\endNE\NE
\e 4 %exer 11.4.4
All vectors of the form $a${\bf y} + $b${\bf z} for
$a$ and $b$ constants are fixed vectors for the matrix
{\\P} of Exercise 3.  There are no other fixed vectors. 
\endNE\NE
\e 6 %exer 11.4.6
Let ${\\y}^{(n)} = {\bf P}^n{\bf y}.$  Then
${y}^{(n+1)}_i = \sum_j{p}_{ij}y_j^{(n)}.$ Thus
${y}^{(n+1)}_i \le \sum_jp_{ij}M_n = M_n.$ This
means that $M_n$ is a upper bound for ${\\y}^{(n+1)}_i$. 
Therefore, $M_{n+1} \le M_n$. A similar argument shows
that $m_{n+1} \ge m_n$.  Hence $M_{n+1} - m_{n+1} \le
M_n-m_n$ for any $n\ge 1$. Thus if we can show that  a
subsequence of differences $M_n - m_n$ tends to 0, then the
same will be true for the entire sequence of differences,
since this sequence is monotone decreasing.
\endNE\NE
\e 8 %exer 11.4.8
This game has the flavor of Doeblin's coupling idea.  Once you and your 
friend happen to be looking at the same number, from that time on you
will continue together.  If this happens you will end up
together and you can successfully state that she stopped
the same place you did.  To see how successful you will
be, you will have to estimate the probability that the
coupling takes place.  It is easy to write a program
using random permutations of 52 objects to simulate
this process.  If you do, you will find that about 75
percent of the time the coupling is successful.
\endNE

\vskip 20pt
%||*****||*****||*****||%sec11.5
\vskip 20pt

%\input mydefinitions
\indent\bf SECTION 11.5\rm
\NE
\e 1 %exer 11.5.1
$${\bf Z} = \pmatrix{11/9 & -2/9\cr -1/9 & 10/9\cr}\ .$$
and
$${\bf M} = \pmatrix{0 & 2\cr 4 & 0\cr}\ .$$
\endNE\NE
\e 2 %exer 11.5.2
$ {\\P} = \bordermatrix{&S&A&W\cr S&1/2&1/2&0\cr
A&1/4&1/2&1/4\cr W&0&1/3&2/3\cr}.$
\vskip 5pt
${\\Z} = \bordermatrix{&S&A&W\cr
S&1.333&0&-1\cr
A&-.222&.889&-.333\cr W&-.889&-.444&1.667\cr}$,
$ {\\w} = (2/9,\ 4/9,\ 3/9).$
\vskip 5pt
\itemitem{} (a) Mean recurrence time for $S = 1/w_1 =
4.5.$
\vskip 5pt
\itemitem{} (b) $m_{31} = 10.$
\endNE\NE
\e 3 %exer 11.5.3
2
\endNE\NE
\e 4 %exer 11.5.4
Mean recurrence time for Yes is $1 + a/b$, and for No
it is $1 + b/a.$  For $a$ = .5 and $b$ = .75 this gives a mean
recurrence time of 5/3 for Yes and 5/2 for No.
\endNE\NE
\e 5 %exer 11.5.5
The fixed vector is {\\w} =
(1/6,1/6,1/6,1/6,1/6,1/6),  so the mean recurrence time
is 6 for each state.
\endNE\NE
\e 6 %exer 11.5.6
${\\P} = \bordermatrix{&R&N&S\cr R&1&0&0\cr
N&1/2&0&1/2\cr S&1/4&1/4&1/2\cr},$
\vskip 5pt
\itemitem{}${\\N} = \bordermatrix{&N&S\cr N&4/3&4/3\cr
S&2/3&8/3\cr},$ \vskip 5pt\itemitem{} ${\\t} = {\\Nc} =
\bordermatrix{&\cr N&8/3\cr S&10/3\cr}.$
\vskip 5pt
\itemitem{}This tells us that if it is nice today,\ then the
expected number of days until rain is 8/3.  If it is snowy
today, then the expected number of days until rain is 10/3.
\endNE\NE
\e 7 %exer 11.5.7
(a)$$\bordermatrix{&1&2&3&4&5&6\cr
1&0&0&1&0&0&0\cr
2&0&0&1&0&0&0\cr
3&1/4&1/4&0&1/4&1/4&0\cr
4&0&0&1/2&0&0&1/2\cr
5&0&0&1/2&0&0&1/2\cr
6&0&0&0&1/2&1/2&0\cr}$$
\vskip 5pt
\itemitem{}(b)\ \ The rat alternates between the sets $\{1, 2, 4, 5\}$ and 
$\{3, 6\}$.
\vskip 3pt
\itemitem{}(c)\ \ ${\bf w} = (1/12, 1/12, 4/12, 2/12, 2/12, 2/12)$.
\vskip 3pt
\itemitem{}(d)\ \ $m_{1,5} = 7$
\endNE\NE
\e 8 %exer 11.5.8
The mean recurrence time for state\ \  0\ \  is
the average time between times that the server is busy.
\endNE\NE
\e 9 %exer 11.5.9
(a) if $n$ is odd, {\\P} is regular.  If $n$ is even, {\\P}
is ergodic but not regular.
\vskip 5pt
\itemitem{} (b) ${\\w} = (1/n,\cdots, 1/n).$
\itemitem{} (c) From the program \bf{ Ergodic}\rm \ we obtain\vskip
5pt $${\\M} = \bordermatrix{&0&1&2&3&4\cr 0&0&4&6&6&4\cr
1&4&0&4&6&6\cr 2&6&4&0&4&6\cr 3&6&6&4&0&4\cr
4&4&6&6&4&0\cr}.$$
\itemitem{} This is consistent with the conjecture that $m_{ij}
= d(n-d)$, where $d$ is the clockwise distance from $i$ to $j$.
\endNE\NE
\e {10} %exer 11.5.10
The transition matrix is: $${\\P} =
\bordermatrix{&0&1&2&3&4&5\cr 0&0&1&0&0&0&0\cr1&1/2&0&1/2&0&0&0\cr
2&0&1/2&0&1/2&0&0\cr 3&0&0&1/2&0&1/2&0\cr 4&0&0&0&1/2&0&1/2\cr
5&0&0&0&0&1&0\cr },$$
$${\\w} = (.25,.125,.125,.125,.125,.25),$$ and $$ {\\M} =
\bordermatrix{&0&1&2&3&4&5\cr0&0&1&4&9&16&25\cr1&9&0&3&8&15&24\cr2&16&7&0&5&12&2
1\cr
3&21&12&5&0&7&16\cr
4&24&15&8&3&0&9\cr5&25&16&9&4&1&0\cr}.$$ \itemitem{} Note
that the entries of the first passage matrix are all
integers.  They also form arithmetic progressions going
down diagonals. General formuals for basic quantities for
random walks can be found in \it Finite Markov Chains\rm\
by John G. Kemeny and J. Laurie Snell (New York:
Springer-Verlag, l976). 
\endNE\NE
\e {11} %exer 11.5.11
Yes, the reverse transition matrix is the same matrix.
\endNE\NE
\e {12} %exer 11.5.12
${\\P}=\pmatrix{1/2&0&1/2\cr 2/3&1/3&0\cr 0&2/3&1/3
\cr},$
\vskip 5pt
\itemitem{}$ {\\w} = (4/10,3/10,3/10), \qquad w_1p_{12} = 0
\ne w_2p_{21}.$
\endNE\NE
\e {13} %exer 11.5.13
Assume that {\\w} is a fixed vector for {\\P}.  Then
\vskip 5pt
$$\sum_i w_ip_{ij}^* = \sum_i {{w_iw_jp_{ji}}\over w_i} =
\sum_i {w_jp_{ji}}= w_j,$$ \vskip 5pt  \itemitem{}so {\\w} is a
fixed vector for {\\P}*. Thus if {\\w}* is the unique fixed vector
for {\\P}* we must have {\\w} = {\\w}*.
\endNE\NE
\e {14} %exer 11.5.14
No. In the Land of Oz example we
found the mean first-passage matrix to be
$${\\M}=\bordermatrix{&R&N&S\cr R&0&4&10/3\cr
N&8/3&0&8/3\cr S&10/3&4&0\cr}.$$  Note, for example, that
$m_{NR} = 8/3 \ne m_{RN} = 4.$
\endNE\NE
\e {15} %exer 11.5.15
If $p_{ij} = p_{ji}$ then {\\P} has column sums
1.  We have seen (Exercise 9 of Section 11.3) that in this
case the fixed vector is a constant vector.  Thus for any two states $s_i$ and
$s_j$, $w_i = w_j$ and $p_{ij} = p_{ji}.$ 
Thus $w_{i}p_{ij} = w_{j}p_{ji}$, and
the chain is reversible.

\endNE\NE
\e {16} %exer 11.5.18
We first show that
$$( {\bf I} +  {\bf P} +\cdots+  {\bf P}^{n - 1})( {\bf I} -  {\bf P} +  {\bf
W}) =  {\bf I} -
 {\bf P}^n + n {\bf W}\ .
$$
Recall that $ \bf P \bf W = \bf W$ so also $\bf P^k\bf W = \bf W$.  Thus, just
multiplying out the left side 
 gives this equality.  Now if we divide through by $n$  and pass to the limit we
have
$${{ {\bf I} +  {\bf P} +\ldots+  {\bf P}^{n - 1}}\over n} (\bf I- \bf P+ \bf W)
\to  \bf W $$
Multiplying both sides on the right by $\bf Z$ and recalling that $\bf W \bf Z =
\bf W$  we see that
$${{ {\bf I} +  {\bf P} +\ldots+  {\bf P}^{n - 1}}\over n}\to  \bf W $$

\endNE\NE
\e {17} %exer 11.5.19
We know that {\bf w\bf Z} = {\bf w}. We also know
that $m_{ki} = (z_{ii} - z_{ki})/ w_i$ and $w_i = {1/
r_i}.$  Putting these in the relation $$\bar m_i = \sum_k
w_km_{ki} + w_ir_i\ ,$$ we see that $$\eqalign{\bar m_i &= \sum_k
w_k{{z_{ii}-z_{ki}}\over w_i} + 1\cr & = {z_{ii}\over
w_i}\sum_k w_k - {1\over w_i}\sum_k w_kz_{ki} + 1\cr & =
{z_{ii}\over w_i} - 1 + 1 = {z_{ii}\over w_i}\ .}$$

\e {18} %exer 11.5.20
Form a Markov chain whose states are the possible
outcomes of a roll. After 100 rolls we may assume that the
chain is in equilibrium. We want to find the mean time
in equilibrium to obtain snake eyes for the first time.
For this chain $m_{ki}$ is the same as $r_i$, since the
starting state does not effect the time to reach another
state for the first time. The fixed vector has all entries
equal to 1/36,  so $r_i = 36$. Using this fact, we obtain $$\bar
m_i = \sum_k w_k m_{ki} + w_ir_i =35 + 1 = 36.$$ We see
that the expected time to obtain snake eyes is 36, so the
second argument is correct. 

\endNE\NE
\e {19} %exer 11.5.22
Recall that
$$m_{ij} = \sum_j{{z_{jj}-z_{ij}}\over w_j}.$$
Multiplying through by $w_j$ summing on $j$ and, using the fact that $\bf Z$ has
row sums 1,  we obtain
 $$ m_{ij} = \sum_jz_{jj} -
\sum_jz_{ij} = \sum_jz_{jj} - 1= K,$$ which is independent
of $i$.

\endNE\NE
\e {20} %exer 11.5.23
 Assume that you start in state $a$.  Then the expected amount you
win on the nth step is $\sum_j P^n_{a,j}f_j.$  From this it follows that
your expected winning on the first $n$ steps can be represented by the
column vector~${\bf g}^{(n)}$,  with
$${\bf g}^{(n)} = ( {\bf I} +  {\bf P} +  {\bf P}^2 +\cdots+  {\bf P}^n)
 {\bf f}.$$  But
since  $\bf w \bf f = 0$ also $\bf W \bf f = 0.$  Thus we  have
$${\bf g}^{(n)} = ( {\bf I} +  ({\bf P} - {\bf W}) +  ({\bf P}^2 -  {\bf W})
+\cdots+  ({\bf P}^n - {\bf W}))
 {\bf f}$$
 Letting $n \to \infty $ we obtain,
$${\bf g}^{(n)} \to  \bf Z {\bf f}.$$
We used here that if  $\bf P$ is the transition matrix for a 
regular chain then  $Z$ is equal to the infinite series:
$$ \bf Z =  {\bf I} +  ({\bf P}- {\bf W}) +  ({\bf P}^2-  {\bf W}) + ({\bf P}^3- 
{\bf W})\cdots.$$

\endNE\NE
\e {21} %exer 11.5.24
The transition matrix is $${\\P}=\bordermatrix{&GO&A&B&C\cr
GO&1/6&1/3&1/3&1/6\cr A&1/6&1/6&1/3&1/3\cr
B&1/3&1/6&1/6&1/3\cr C&1/3&1/3&1/6&1/6\cr }.$$
Since the column sums are 1, the fixed vector is
$${\\w} = (1/4,\ 1/4,\ 1/4,\ 1/4)\ .$$ From this we see that
{\\wf} = {\\0}.  From the result of Exercise 20 we see that your
expected winning starting in GO is the first component of the vector 
 {\bf Z\bf f} where 
$$ {\bf f} = \pmatrix{ 15 \cr -30 \cr -5 \cr 20 }\ .$$
Using the program {\bf ergodic} we find that the long run expected winning
starting in GO is 10.4.
\endNE\NE
\e {22} %exer 11.5.27
{\bf P\bf W = \bf W}  follows from the fact the columns of $\bf W$  are constant
and the row sums of {\bf P} are 1.
Similarly  ${\bf W\bf W  = \bf W}$  follows from the fact that ${\bf W}$ has row
sums 1 and constant columns.
Thus ${\bf W^2} = {\bf W}$.  Multiplying both sides by {\bf W} gives ${\bf W^3}
= {\bf W^2} = {\bf W}$.  
Continuing in this
way we obtain ${\bf W^k} = {\bf W.}$

\endNE\NE
\e {23} %exer 11.5.26
Assume that the chain is started in state $s_i$.  Let $X_j^{(n)}$ equal 1 if
the chain is in state $s_i$ on the nth step and 0 otherwise. 
Then $$S_j^{(n)} = X_j^{(0)} + X_j^{(1)} + X_j^{(2)} + \dots X_j^{(n)}.$$  and
$$E(X_j^{(n)}) = P_{ij}^n.$$
Thus
$$E(S_j^{(n)}) = \sum_{h = 0}^n p^{(n)}_{ij}.$$
If now follows then from Exercise 16 that
$$\lim_{n\to \infty}{{E(S_j^{(n)})}\over n} = w_j.$$

\endNE\NE
\e {24} %exer 11.5.29
He got it!
\vskip 15pt
\endNE

\bye




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