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ECE 6010 Stochastic Processes
Homework #3
Problems from Grimmet & Stirzaker:

1. Prob 2.7.4

a)

\begin{displaymath}P(\frac{1}{2}<x\leq \frac{3}{2})=F(\frac{3}{2})-F(\frac{1}{2})=\frac{1}{2}.\end{displaymath}

b)

\begin{displaymath}P(1<X<2)=F(2)-F(1)=\frac{1}{2}.\end{displaymath}<2)=F(2)-F(1)=\frac{1}{2}.\end{displaymath}" border="0" height="45" width="282" />

c)

\begin{displaymath}P(Y\leq X)=P(X^{2}\leq X) =P(X\leq 1) =F(1)=\frac{1}{2}. \end{displaymath}

d)

\begin{displaymath}P(X\leq 2Y)=P(X\leq 2X^{2}) =P(X\geq \frac{1}{2})=\frac{3}{4}. \end{displaymath}

e)

\begin{displaymath}P(X+Y\leq \frac{3}{4})=P(X+X^{2}\leq \frac{3}{4})=P(X\leq \frac{1}{2})=\frac{1}{4}\end{displaymath}

f)

\begin{displaymath}P(\sqrt{X}\leq Z)=P(X\leq Z^{2}) =\frac{1}{2}z^{2}\end{displaymath}

if $ 0\leq z\leq \sqrt{2}.$

2.Prob 2.7.7. Let T be the numbers of people on given typical flights of TWA, and B be the numbers of people on given typical flights of BA.

\begin{displaymath}P(T=k)=\left( \begin{array}{c} 10\ k \end{array}\right) ... ...9}{10} \right) ^{k} \left( \frac{1}{10} \right)^{10-k},\end{displaymath}


\begin{displaymath}P(B=k)=\left( \begin{array}{c} 20\ k \end{array}\right) ... ...9}{10} \right)^{k} \left( \frac{1}{10} \right) ^{20-k}.\end{displaymath}

Now P(TWA overbooked)= P(T=10)= $(\frac{9}{10})^{10}$
P(BA overbooked) $=P(B\geq 19) =20(\frac{9}{10})^{19}(\frac{1}{10})+(\frac{9}{10})^{20},$ of which the latter is the larger.

3. Prob 2.7.9. (a)

\begin{displaymath}P(X^{+} \leq x)= \left\{ \begin{array}{ll} 0 & \mbox{if $x<0,$}\\ F(x) & \mbox {$x \geq 0$} \end{array} \right. \end{displaymath}<0,$}\\ F(x) & \mbox {$x \geq 0$} \end{array} \right. \end{displaymath}" border="0" height="54" width="260" />

(b)

\begin{displaymath}P(X^{-} \leq x)= \left\{ \begin{array}{ll} 0 & \mbox{if $x<... ...y\uparrow-x}F(y) & \mbox {if $x \geq 0$} \end{array} \right. \end{displaymath}<... ...y\uparrow-x}F(y) & \mbox {if $x \geq 0$} \end{array} \right. \end{displaymath}" border="0" height="54" width="350" />

(c) $ P(\vert X\vert \leq x) =P(-x \leq X \leq x) $, if $x \geq 0$. Therefore.

\begin{displaymath}P(\vert X\vert \leq x)= \left\{ \begin{array}{ll} 0 & \mbox... ...y\uparrow-x}F(y) & \mbox {if $x \geq 0$} \end{array} \right. \end{displaymath}

(d)

\begin{displaymath}P(-X\leq x) = 1- lim_{y\uparrow-x}F(y).\end{displaymath}


4. Ex 3.3.1. (a) No!
(b) Let X have mass function: $f(-1)=\frac{1}{9},f(\frac{1}{2})=\frac{4}{9},f(2)=\frac{4}{9}.$ Then $E(X) =-\frac{1}{9}+\frac{2}{9}+\frac{8}{9}=1=-\frac{1}{9}+\frac{2}{9}+\frac{8}{9}=E(1/X).$

5. Ex 3.4.1. Let $I_{j}$ be the indicator function of the event that the outcome of the (i+1)th toss is different from the outcome of the jth toss. The number R of distinct runs is given by $R=1+\sum_{j=1}^{n-1}I_{j}$. Hence

\begin{displaymath}E[R] =1+(n-1)E[I_{1}]=1+(n-1)2pq,\end{displaymath}

where $q=1-p$. Now remark that $I_{j}$ and $I_{k}$ are independent if $\vert j-k\vert > 1$ 1$" align="middle" border="0" height="37" width="92" />, so that

\begin{displaymath}E\{(R-1)^{2}\}=E\left\{ (\sum_{j=1}^{n-1}I_{j})^{2} \right\} =(n-1)E[I_{1}] +2(n-2)E[I_{1}I_{2}] \end{displaymath}


\begin{displaymath}+\{(n-1)^{2}-(n-1)-2(n-2)\}E[I_{1}]^{2}. \end{displaymath}

Now $E[I_{1}^{2}]=E[I_{1}]=2pq$ and $E[I_{1}I_{2}]=p^{2}q+pq^{2}=pq$, and therefore

\begin{displaymath}var(R) =var(R-1) =(n-1)E[I_{1}]+2(n-2)E[I_{1}I_{2}] -\{(n-1)+2(n-2)\}E[I_{1}]^{2}\end{displaymath}


\begin{displaymath}=2pq(2n-3-2pq(3n-5)).\end{displaymath}


6. Ex 3.4.2. The required total is $T=\sum_{i=1}^{k}X_{i}$, where $X_{i}$ is the number shown on the ith ball. Hence $E[T]=kE[X_{1}]=\frac{1}{2}k(n+1)$

\begin{displaymath}E[T^{2}]=E\left\{(\sum_{i=1}^{k}X_{i})^{2} \right\}=kE[X_{1}^{2}]+k(k-1)E[X_{1}X_{2}] \end{displaymath}


\begin{displaymath}=\frac{k}{n}\sum_{1}^{n}j^{2}+\frac{k(k-1)}{n(n-1)}\sum_{i\neq j}ij \end{displaymath}


\begin{displaymath}=\frac{1}{6}k(n+1)(2n+1)+\frac{1}{12}k(k-1)(3n+2)(n+1).\end{displaymath}

Hence

\begin{displaymath}Var(T)= E[T^{2}]-E[T]^{2}=\frac{1}{12}(n+1)k(n-k)\end{displaymath}


7. Ex 3.5.2. The total number H of heads satisfies

\begin{displaymath}P(H=X)=\sum_{n=x}^{\infty}P(H=x\vert N=n)P(N=n) \end{displaymath}


\begin{displaymath}=\sum_{n=x}^{\infty}\left( \begin{array}{c} n\ x \end{array} \right) p^{x}(1-p)^{n-x}\frac{\lambda^{n}e^{-\lambda}}{n!} \end{displaymath}


\begin{displaymath}=\frac{(\lambda p)^{x}e^{-\lambda p}}{x!}\sum_{n=x}^{\infty}\frac{\{\lambda (1-p)\}^{n-x} e^{-\lambda (1-p)}}{(n-x)!}.\end{displaymath}

The last summation equals 1, since it is the sum of the values of the Poission mass function with parameter $\lambda(1-p).$

8. Ex 3.6.5. (a) $log y\leq y-1$ with equality if and only if $y=1$. Therefore,

\begin{displaymath}E \left[ log \frac{f_{Y}(X)}{f_{X}(X)}\right] \leq E\left[ \frac{f_{Y}(X)}{f_{X}(X)}-1 \right] =0,\end{displaymath}

with equality if an only if $f_{Y}=f_{X}.$

(b)

\begin{displaymath}E\left[ log\frac{f(x,y)}{f_{X}(x)f_{Y}(y)} \right] \leq E\left[ \frac{f(x,y)}{f_{X}(x)f_{Y}(y)}-1 \right]\end{displaymath}


\begin{displaymath}-I= E\left[ log\frac{f_{X}(x)f_{Y}(y)}{f_{XY}(x,y)} -1 \right] \leq E\left[ \frac{f_{X}(x)f_{Y}(y)}{f_{XY}(x,y)} -1 \right] =0 \end{displaymath}

so $I\geq 0$
Copyright 2008, Todd Moon. Cite/attribute Resource. admin. (2006, June 13). Homework Solutions. Retrieved November 24, 2009, from Free Online Course Materials — USU OpenCourseWare Web site: http://ocw.usu.edu/Electrical_and_Computer_Engineering/Stochastic_Processes/hw3solgs.html. This work is licensed under a Creative Commons License. Creative Commons License
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