Subject: Meetings on Tuesday

From: <[email-masked]>
Date: Tue, 30 Jan 2007 13:11:30 -0800 (PST)

Hello,
I wanted to remind you that we have meetings on Tuesday. Today only 
Ben, Dan and I were present together with Prof. Lorden. I haven't 
heard from any of you saying that you are not able to come.
Please note Tuesday noon in your calendars.

Midterm will be given out on Feb 2(Friday) and will be due Feb 12
(Monday). We will grade on Monday. Prof. Lorden will send a draft of 
midterm that we are supposed to read, look for mistakes and give 
suggestions. There will be a midterm review given by me. The time is 
not determined yet. Prof. Lorden will ask students what suits them 
best.

On Thursday recitation it would be good to do an example where you use 
the derivative of the distribution function to get the density 
function rather than some formula.

See you on Tuesday,
Mirjana

---

From: Brian Schroeder <[email-masked]>
Date: Wed, 31 Jan 2007 17:32:47 -0800

Following some suggestions, I have updated the solutions to problems 2, 
4, and 7.

Brian
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\begin{document}

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

\bigskip
\end{center}

\noindent \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?

\noindent{\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*}

\noindent\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.

\noindent{\em Solution.} We have that $Y=\tan \Phi .$ Uniform distribution on
$\left( -\pi /2,\pi /2\right)$ 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%
\[
\int\limits_{-\infty }^{\infty }\frac{y}{\pi \left( 1+y^{2}\right) }dy=\frac{%
1}{\pi }\int\limits_{-\infty }^{\infty }\frac{y}{\pi \left( 1+y^{2}\right) }dy
\]%
is not absolutely convergent, as $\left|\frac{y}{1+y^2}\right| \sim 1/|y|$
as $y\to \pm\infty$.

\noindent\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}

\noindent{\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*}

\noindent\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}

\noindent{\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) Since $Y\ge 0$, we have $F_Y(y) = P(Y \le y) = P(Y^2 \le y^2) = P(Z
\le y^2)$. So $F_Z(y^2) = F_Y(y)$. So for $z\ge 0$, we have $F_Z(z) =
F_Y(\sqrt{z})$. Therefore for $z\ge 0$, \[f_Z(z) = \frac{dF_Z(z)}{dz} =
\frac{dF_Y(\sqrt{z})}{dz} =
\frac{dF_Y(\sqrt{z})}{d\sqrt{z}}\cdot\frac{d\sqrt{z}}{dz} =
f_Y(\sqrt{z}) \cdot \frac{1}{2\sqrt{z}} = e^{-z}\]
So 
\[
f_{Z}\left( z\right) = e^{-z},\qquad z\geqslant 0.
\]

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

\noindent\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 $X+Y$.

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

\noindent\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}

\noindent{\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 $x
\geq 1$, $f_X(x)=0$.

\noindent\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?

\noindent{\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.)

(Another way to find $q$ is to find the probability
$q_0 = P(D\mathrm{~is~not~hit~by~}H)$ for a given hunter $H$. Then by
independence of the hunters, $q = 1 - q_0^8$. We have $q_0 = 0.9 +
0.1\cdot 0.6 = 0.96$ (either $H$ aims at a duck other than $D$,
with probability $0.9$, or $H$ aims at $D$ and misses, with
probability $0.1\cdot 0.6$). So $q = 1 - (0.96)^8 \approx 0.28$.)
\bigskip

\end{document}

---

From: Aliekber Gurel <[email-masked]>
Date: Thu, 1 Feb 2007 11:38:36 -0800 (PST)

Fortunately, the grade is recorded but she doesnt have the set.
I dont have it either and I couldnt see it in any of the slots.
If anyone has this set, can he/she please give it to me or leave it in my
mailbox.

Thanks in advance

-Ali