Subject: HW4 solutions

From: Brian Schroeder <[email-masked]>
Date: Tue, 30 Jan 2007 23:01:20 -0800

Here are some solutions to HW4.
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\begin{document}

\begin{center}
\textbf{MATH 2B HOMEWORK\ \#4 SOLUTIONS}

\bigskip
\end{center}

\textbf{(1)} 
Suppose the distribution of height over a large population of individuals
is approximately normal. Ten percent of individuals in the
population are over 6 feet tall, while the average height is 5 feet
10 inches. What, approximately, is the probability that in a group
of 100 people picked at random from this population there will be
two or more individuals over 6 feet 2 inches tall?

{\em Solution.}
Let $X$ denote a random variable that represents a
height of an individual. It is a normal distribution with mean $\mu
=70$ inches.
Standard deviation $\sigma $ can be computed from%
\[
P(X>72)=0.1.
\]%
This implies that%
\[
0.9=P\left( X\leqslant 72\right) =P\left( \frac{X-\mu }{\sigma
}\leqslant \frac{72-70}{\sigma }\right) =\Phi \left( \frac{2}{\sigma
}\right)
\]%
and since $\Phi \left( 1.285\right) =0.9$ we conclude $\sigma
=2/1.285=1.556. $

(If you used feet instead of inches you should get $\mu =5.8333$ feet and $%
\sigma =0.1297$.)

Let $X_{i}$ be the indicator that $i-$th person is taller than 74 inches, $%
i=1,\ldots ,100$. Then
\begin{eqnarray*}
P\left( X_{i}=1\right) &=&P(X>74)=1-P(X\leqslant 74) \\
&=&1-P\left( \frac{X-\mu }{\sigma }\leqslant
\frac{74-70}{1.556}\right) =1-\Phi \left( 2.57\right) =0.0051.
\end{eqnarray*}%
If we define $S_{100}=\sum\limits_{i=1}^{100}X_{i}$ then $S_{100}$
represents number of people that are taller than 74 inches. We can
use Poisson approximation for $S_{100}$ since $n=100$ is large and
$p=0.0051$ is
small. This approximation gives us%
\begin{eqnarray*}
P\left( S_{100}\geqslant 2\right) &=&1-P\left( S_{100}=0\right)
-P(S_{100}=1)
\\
&\approx &1-e^{-0.51}-e^{-0.51}\cdot 0.51=0.0932.
\end{eqnarray*}

\textbf{(2)} ({\em Pitman 4.4.6})  \textbf{Cauchy distribution.}
Suppose that a particle is fired from the origin in the $(x,y)$-plane
in a straight line in a direction at random angle $\Phi$ to the
$x$-axis, and let $Y$ be the $y$-coordinate of the place where the
particle hits the line $\{x = 1\}$. Show that if $\Phi$ has uniform
$(-\pi/2,\pi/2)$ distribution, then $f_Y(y) = \frac{1}{\pi(1+y^2)}$
This is called the {\em Cauchy distribution}. Show that the Cauchy
distribution is symmetric about $0$, but that the expectation of a
Cauchy random variable is undefined.

{\em Solution.} We have that $Y=\tan \Phi .$ Uniform distribution on interval $%
\left( -\pi /2,\pi /2\right) $ is defined with%
\[
f_{\Phi }\left( \phi \right) =\left\{
\begin{array}{lll}
1/\pi , &  & \phi \in \left( -\pi /2,\pi /2\right)  \\
0, &  & \textrm{otherwise}%
\end{array}%
\right. .
\]%
Then%
\begin{eqnarray*}
f_{Y}\left( y\right)  &=&\frac{d\left( F_{y}\left( y\right) \right) }{dy}=%
\frac{d\left( \left( P(Y\leqslant y\right) \right)
}{dy}=\frac{d\left(
P\left( \tan \Phi \leqslant y\right) \right) }{dy} \\
&=&\frac{d\left( P\left( \Phi \leqslant \arctan y\right) \right) }{dy}%
=f_{\Phi }\left( \arctan y\right) \frac{d\left( \arctan y\right) }{dy}=\frac{%
1}{\pi \left( 1+y^{2}\right) }.
\end{eqnarray*}%
Obviously, $f_{Y}\left( y\right) =f_{Y}\left( -y\right) $, i.e.
Cauchy distribution is symmetric about 0. Expectation of this random
variable is
not defined because%
\[
\int\limits_{-\infty }^{\infty }\frac{y}{\pi \left( 1+y^{2}\right) }dy=\frac{%
1}{\pi }\left( \int\limits_{-\infty }^{0}\frac{y}{\left( 1+y^{2}\right) }%
dy+\int\limits_{0}^{\infty }\frac{y}{\left( 1+y^{2}\right)
}dy\right)
\]%
and each of these integrals does not converge since $y/\left(
1+y^{2}\right) \sim 1/y$ at infinity and $\int\limits_{1}^{\infty
}1/ydy$ does not converge.

\textbf{(3)} ({\em Pitman 5.3.8}) Peter and Paul agree to meet at a
restaurant at noon. Peter arrives at a time normally distributed with
mean 12:00 noon, and standard deviation $5$ minutes. Paul arrives at a
time normally distributed with mean 12:02 PM, and standard deviation
$3$ minutes. Assuming the two arrival times are independent, find the
chance that
\begin{itemize}
\item[a)] Peter arrives before Paul;
\item[b)] both men arrive within 3 minutes of noon;
\item[c)] the two men arrive within 3 minutes of each other.
\end{itemize}

{\em Solution.} Let $X$ and $Y$ represent times when Peter and Paul
arrive, respectively. We can take noon to be the origin of the
coordinate system and then $X$ and $Y$ are normal $\left(
0,25\right) $ and $\ $normal $\left( 2,9\right) $ distributions,
respectively. Using formula for linear
combination of normal variables given on page 363 we obtain $X-Y$ is normal $%
\left( -2,34\right) $ distribution.

a) Probability that Peter arrives before Paul is%
\[
P\left( X<Y\right) =P\left( X-Y<0\right) =P\left( \frac{X-Y+2}{\sqrt{34}}<%
\frac{0+2}{\sqrt{34}}\right) =\Phi \left( 0.34\right) =0.6331.
\]

b) Probability that both men arrive within 3 minutes of noon is%
\begin{eqnarray*}
P\left( \left\vert X\right\vert \leqslant 3\textrm{ }\&\textrm{
}\left\vert
Y\right\vert \leqslant 3\right) &=&P\left( -3\leqslant X\leqslant 3\textrm{ }%
\right) P\left( -3\leqslant \textrm{ }Y\leqslant 3\right) \\
&=&P\left( \frac{-3}{5}\leqslant \frac{X}{5}\leqslant \frac{3}{5}\textrm{ }%
\right) P\left( \frac{-5}{3}\leqslant \textrm{ }\frac{Y-2}{3}\leqslant \frac{1%
}{3}\right) \\
&=&\left( 2\Phi \left( 0.6\right) -1\right) \left( \Phi \left(
0.33\right)
+\Phi \left( 1.67\right) -1\right) \\
&=&0.2626.
\end{eqnarray*}

c) Probability that two men arrive within 3 minutes of each other is%
\begin{eqnarray*}
P\left( \left\vert X-Y\right\vert <3\right) &=&P\left(
-3<X-Y<3\right)
=P\left( \frac{-3+2}{\sqrt{34}}<\frac{X-Y+2}{\sqrt{34}}<\frac{3+2}{\sqrt{34}}%
\right) \\
&=&\Phi \left( 0.86\right) +\Phi \left( 0.17\right) -1=0.3726.
\end{eqnarray*}

\textbf{(4)} ({\em Pitman, p. 336, problem 21}) Suppose $R_1$ and $R_2$
are two independent random variables with the same density function
$f(x) = x\exp(-\frac{1}{2}x^2)$ for $x\ge 0$. Find
\begin{itemize}
\item[(a)] the density of $Y = \min\{R_1,R_2\}$;
\item[(b)] the density of $Y^2$;
\item[(c)] $E(Y^2)$.
\end{itemize}

{\em Solution.}
a) Using formula for density of the minimum given on
page 317
we obtain%
\begin{eqnarray*}
F_{Y}\left( y\right) &=&P\left( Y\leqslant y\right) =1-\left(
1-P\left( R_{1}\leqslant y\right) \right) \left( 1-P\left(
R_{2}\leqslant y\right)
\right) \\
&=&1-\left( 1-F_{R_{1}}\left( y\right) \right) \left(
1-F_{R_{2}}\left( y\right) \right) .
\end{eqnarray*}%
If $y\leqslant 0$ then $F_{Y}\left( y\right) =0.$ Now, for
$y\geqslant 0.$
Since%
\[
F_{R_{1}}\left( y\right) =F_{R_{2}}\left( y\right) =\int\limits_{0}^{y}xe^{-%
\frac{1}{2}x^{2}}dx=\left. -e^{-\frac{1}{2}x^{2}}\right\vert _{0}^{y}=1-e^{-%
\frac{1}{2}y^{2}}
\]%
we get%
\[
F_{Y}\left( y\right) =1-e^{-y^{2}}.
\]%
>From this follows%
\[
f_{Y}\left( y\right) =\left( 1-e^{-y^{2}}\right) ^{\prime
}=2ye^{-y^{2}},\qquad y\geqslant 0.
\]

b) $Z=Y^{2}$ is an increasing function on $\left( 0,\infty \right) $
and
using formula on page 304 we obtain%
\[
f_{Z}\left( z\right) =f_{Y}\left( \sqrt{z}\right) \cdot \frac{1}{2\sqrt{z}}%
=e^{-z},\qquad z\geqslant 0.
\]%
This is the exponential $\left( 1\right) $ distribution.

c) Expectation of the exponential $\left( 1\right) $ distribution is
1 (see page 279).

\textbf{(5)} Suppose $X$ and $Y$ are two independent random
variables. $X$ has density $$f(x) = \left\{\begin{array}{ll} 6x(1-x) &
0 < x < 1 \\ 0 & \mathrm{otherwise}
\end{array}
\right.$$
and $Y$ has density
$$g(y) = \left\{\begin{array}{ll} 2y & 0 < y < 1 \\
0 & \mathrm{otherwise}.\end{array}\right.$$
Use the convolution formula to calculate the density function $h(t)$
of $X+Y$.

{\em Solution.}
$X$ and $Y$ are two independent and non-negative
variables.
Then $X+Y$ is also non-negative and by convolution formula on page 372%
\[
h_{X+Y}\left( t\right) =\int\limits_{0}^{t}f\left( x\right) g\left(
t-x\right) dx,\qquad t\geqslant 0.
\]%
We divide cases

a) $0\mathbf{\leqslant }t<1$ where%
\begin{eqnarray*}
h_{X+Y}\left( t\right) &=&\int\limits_{0}^{t}6x\left( 1-x\right)
\cdot
2\left( t-x\right) dx \\
&=&12\left. \left( \frac{1}{2}x^{2}t-\frac{1}{3}x^{3}\left( 1+t\right) +%
\frac{1}{4}x^{4}\right) \right\vert _{0}^{t} \\
&=&2t^{3}-t^{4};
\end{eqnarray*}

b) $1\mathbf{\leqslant }t<2$ where%
\begin{eqnarray*}
h_{X+Y}\left( t\right) &=&\int\limits_{t-1}^{1}6x\left( 1-x\right)
\cdot
2\left( t-x\right) dx \\
&=&12\left. \left( \frac{1}{2}x^{2}t-\frac{1}{3}x^{3}\left( 1+t\right) +%
\frac{1}{4}x^{4}\right) \right\vert _{t-1}^{1} \\
&=&-8+16t-6t^{2}-2t^{3}+t^{4};
\end{eqnarray*}

c) $2\mathbf{\leqslant }t$ where%
\[
h_{X+Y}\left( t\right) =0.
\]%
We conclude that%
\[
h_{X+Y}\left( t\right) =\left\{
\begin{array}{lll}
2t^{3}-t^{4}, &  & 0\mathbf{\leqslant }t<1 \\
-8+16t-6t^{2}-2t^{3}+t^{4}, &  & 1\mathbf{\leqslant }t<2 \\
0, &  & \textrm{otherwise}%
\end{array}%
\right. .
\]

\textbf{(6)} ({\em 5.2.6 parts (a) and (b)}) Let $X$ and $Y$ have joint
density $$f(x,y) = \left\{ \begin{array}{ll}
90(y-x)^8 & 0<x<y<1 \\
0 & \mathrm{otherwise}
\end{array}
\right.
$$

\begin{itemize}
\item[a)] Find $P(Y>2X)$.
\item[b)] Find the marginal density of $X$.
\end{itemize}

{\em Solution.}
a) {\setlength\arraycolsep{2pt}
\begin{eqnarray*} P(Y > 2X) &=& \int\!\!\!\int_{y>2x} f(x,y) dx dy =
\int_{-\infty}^{\infty} \int_{2x}^{\infty} f(x,y) dy dx =\\
&=& \int_{0}^{1/2} \int_{2x}^{1} 90 {(y-x)}^8 dy dx =
\int_{0}^{1/2} 90 \frac{{(y-x)}^9}{9} \, \bigg|_{2x}^{1} dx = \\
&=& \int_{0}^{1/2} 10 \left({(1-x)}^9 - x^9\right) dx = 10\left(
\frac{-{(1-x)}^{10}}{10} - \frac{x^{10}}{10}\right)\, \bigg|_0^{1/2}
= \\ &=& 1- {\left(\frac{1}{2}\right)}^9 \approx
0.998.\end{eqnarray*} }

b) For $0<x<1$, the marginal density of $X$ is
{\setlength\arraycolsep{2pt} \begin{eqnarray*} f_X(x) &=&
\int_{-\infty}^{\infty} f(x,y) dy = \\ &=& \int_{x}^{1} 90 {(y-x)}^8
dy  \ \ \  \textrm {(as }\,  f(x,y) \, \, \textrm{is defined to be 0
when} \ x \geq y \ \textrm{)}\\ &=& 90 \frac{{(y-x)}^9}{9}
\bigg|_x^1 = 10 {(1-x)}^{9}. \end{eqnarray*} } For $x\leq 0$ or $x
\geq 1$, $f_X(x)=0$.

\textbf{(7)} A flock of 10 ducks flies over a group of $8$
hunters. All at the same time, each hunter picks a duck at random and
fires at it, hitting it with probability $.4$. Assuming that the
hunters' choices and successes (or failures) are independent, what is
the expectation of the number of ducks that get hit?

{\em Solution.}
Pick a duck and call it $D$. The
probability that an individual hunter hits $D$ is $p=\frac{1}{10}
\cdot \frac{4}{10}$ (the probability of his picking $D$ out of the
ten ducks times the probability of hitting it.) Denote by $X$ the
number of hunters who hit $D$. $X$ is binomially distributed and we
have:
\[ q = P(D \mathrm{~is~hit}) = P(X>0) = 1 - P(X=0) = 1 - (1-p)^8 \approx 0.28. \]

Let now for each $i=1, \dots, 10$, $Y_i$ be the indicator of the event
``duck number $i$ gets hit'', i.e. $Y_i$ is 1 if duck number $i$ 
gets hit and 0 otherwise. Then the total number of killed ducks is
$\sum_i Y_i$ and its expectation is: 
\[ E\sum_i Y_i = \sum_i EY_i = \sum_i q = 10q \approx 2.8. \]
(Note that the $Y_i$'s above are identically distributed but not
independent and in particular their sum does not have the binomial
distribution.)

\bigskip

\end{document}