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\begin{document}


\begin{center}
\bf \large
 M.\ S.\ Comprehensive Exam 
(Part I of Written Exam)\\
Saturday,  January 3, 1998
\end{center}


\vsp
You have three hours.  Do as many problems as you can.  No one is
expected to answer all the problems correctly.  Partial credit will be
given.  All problems are worth an equal amount of credit.

\vsp
{\bf Put your solution to each problem on a separate sheet of paper.}


\vsp

\QQ
%Meeter -- Applied Statistics
\begin{parts}
\PP
I often recommend to clients using multiple regression that, if
they have two predictor variables
which are highly correlated, they consider using instead the sum
and the difference of the
variables, because their correlation will be low.  Is this true?

\PP
Find a condition for two linear combinations of two variables to be
uncorrelated.  How would
you modify the consultant's advice?
\end{parts}

\QQ
%Meeter -- Applied Statistics
In a multiple regression of $y$ on $x_1, x_2, x_3,$ and $x_4$, with 30
observations, suppose you wanted
to test whether it was reasonable to assume $\beta_2$, the regression
coefficient of $x_2$, equalled $\beta_3$, the coefficient of $x_3$.

\begin{parts}
\PP
 Describe how you would test this hypothesis, identifying the
sums of squares and specifying the degrees of freedom.

\PP
 How would you test the hypothesis that $\beta_2 = 2 \beta_3$?
\end{parts}

\QQ
%Lin -- Math. Statistics
Suppose that $Y$ is a gamma random variable with the density function
\[ f(y) =\left\{ \begin{array}{ll}
\lambda^r y^{r-1}e^{-\lambda y}/\Gamma(r),     &  \mbox{if  $y>0$},  \\
0,   &  \mbox{if $y\leq 0$}, 
\end{array} \right. \]
where $\lambda>0$ and $r>0$.

\begin{parts}
\PP
Show that the moment generating function of $Y$ is 
$M(t)=(\frac{\lambda}{\lambda-t})^r$ for $t<\lambda$.

\PP
Find the mean  E($Y$) and  the variance Var($Y$).
\end{parts}


\QQ  
%Hollander -- M.S. Math.  Statistics
Let $X$   have a Poisson distribution with parameter $\lambda$. Suppose
that we want to estimate $g(\lambda)=e^{-3\lambda}$.
\begin{parts}
\PP
Show that $T(X)=(-2)^X$ is an unbiased estimate of $g(\lambda)$.

\PP
Explain why $T(X)$ is not a good estimate of $g(\lambda)$ even though
it is unbiased.
\end{parts}

\QQ
%Hollander-- M.S. Statistics
Let $\{X, Y\}$ be jointly distributed with joint density function
\[ f(x, y) =\left\{ \begin{array}{ll}
e^{-y},     &  \mbox{if  $0<x<y<\infty$},    \\
0,   &  \mbox{Otherwise}  
\end{array} \right. \]

\begin{parts}
\PP
What is the marginal density function of $X$?

\PP
What is the conditional density function of $Y$ given $X$?

\PP
What is $\mbox{E}(Y|X=x)$?
\end{parts}

\QQ
%Song--- MS Probability
Tom has $\alpha$ batteries in a box of which $\beta$ are dead. He tests them
randomly and one by one. Every time that a good battery is drawn, he will
return it to the box; every time that a dead battery is drawn, he will
replace it by a good one.

\begin{parts}
\PP
What is the average number of good batteries in the box after $n$ of them
are checked?

\PP
What is the probability that on the $n$th draw Tom draws a good battery?
\end{parts}

\QQ 
%Song--- MS Probability
In a contest, contestants $A$, $B$, and $C$ are each asked, in turn, a general
scientific question. If a contestant gives a wrong answer to a question, he
goes out of the game. The remaining two will continue to compete until one of
them goes out. The last person remaining is the winner. Suppose that a
contestant knows the answer to a question independent of the other contestants
with probability $p$. What is the probability that $A$ goes out first, $B$
next, and $C$ wins?

\QQ
%Anuj--MS Probability
Independent trials, each resulting in a success with probability
$p$, are successively performed. Let $N$ be the time of the
first success. Find  $\var(N)$ given that $E(N) = {1 \over p}$.

\newpage
\QQ
%Huang--MS Mathematical Statistics
\begin{parts}
\PP
If $X$, $Y$, and $Z$ are any random variables with finite variances, and the conditional
covariance of $X$ and $Y$ given $Z$ is defined as
\[ \cov((X,Y)|Z)= E((X-E(X|Z))(Y-E(Y|Z))|Z),\]
show that
\[ \cov(X,Y)= E[\cov((X,Y)|Z)]+ \cov[ E(X|Z), E(Y|Z)].\]
\PP
Use (a) to derive the double expectation theorem
\[ \var(X)= E[\var(X|Z)] + \var[E(X|Z)], \]
where $\var(X|Z)= E((X-E(X|Z))^2|Z)$ is the conditional variance given $Z$.
\end{parts}


\end{document}


