\documentclass[twocolumn]{article}

\setlength{\oddsidemargin}{-0.5in}
\setlength{\evensidemargin}{-0.5in}
\setlength{\topmargin}{0.0in}
\setlength{\headheight}{0in}
\setlength{\headsep}{0in}
\setlength{\textwidth}{7.0in}
\setlength{\textheight}{9.5in}
\setlength{\parindent}{0in}
\setlength{\parskip}{0.05in}
\setlength{\columnseprule}{0.3pt}
\usepackage{fancyvrb}
\usepackage{relsize}

\begin{document}

Name: \_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_

Directions: {\bf Work only on this sheet} (on both sides, if needed); do not
turn in any supplementary sheets of paper. There is actually plenty of room
for your answers, as long as you organize yourself BEFORE starting writing.
In order to get full credit, {\bf SHOW YOUR WORK}.


{\bf \Large NOTE:  ALL PROBLEMS HERE REQUIRE CLOSED-FORM SOLUTIONS.}

{\bf 1.}  Suppose $f_W(t) = ct^{c-1}$ for t in (0,1), with the density
being 0 elsewhere, for some unknown $c > 0$.  We have a random sample
$W_1,...,W_n$ from this density.

\begin{itemize}

\item [(a)] (10) Find $F_W$.

\item [(b)] (15) Find the Method of Moments Estimator of c.  It should
work out to be a function of $\bar{W}$.

\item [(c)] (15) Suppose c = 2 and n = 50.  Find the value of d for which
$F_{\bar{W}}(d) \approx 0.975$.

\end{itemize}

{\bf 2.} Suppose $f_W(t) = \frac{1}{c}$ on (0,c), 0 elsewhere, for some
$c > 0$.  We have a random sample $W_1,...,W_n$ from this density.

\begin{itemize}

\item [(a)] (15) Find the Maximum Likelihood Estimator (MLE).

\item [(b)] (15) Find the bias of the MLE.

\end{itemize}

{\bf 3.}  (15) Suppose the Y = 1 or 0, with probability 0.6 and 0.4.  If Y =
1, X has the density $3t^2$ on (0,1); if Y = 0, then X's density is 2t.
Find $m_{Y;X}(t)$.

{\bf 4.} (15) Suppose we assume that $Var(Y|X=t) = \sigma^2$, independent
of t, but do not assume linearity.  Which of the statements below is
true?  {\bf BE SURE TO JUSTIFY YOUR ANSWER, USING MATH.}

\begin{itemize}

\item [(i)] $Var(Y) \leq \sigma^2$

\item [(ii)] $Var(Y) \geq \sigma^2$

\item [(iii)] Either (i) or (ii) could be true, depending on the
specific distributions.

\end{itemize}

{\bf Solutions:}

{\bf 1.a} For t in (0,1),

\begin{equation}
F_W(t) = P(W \leq t) = \int_{0}^{t} cs^{c-1} ~ ds = t^c
\end{equation}

$F_W(t)$ is 0 for t less than 0, and is 1 for t bigger than 1.

{\bf 1.b}  

\begin{equation}
EW = \int_{0}^{1} t ct^{c-1}~ dt = \frac{c}{c+1}
\end{equation}

So, set 

\begin{equation}
\bar{W} = \frac{\hat{c}}{\hat{c}+1}
\end{equation}

yielding

\begin{equation}
\hat{c} = \frac{\bar{W}}{1-\bar{W}}
\end{equation}

{\bf 1.c}  By the Central Limit Theorem $\bar{W}$ is approximately
normally distributed with mean EW and variance $EW/n^2$.

We already found that EW = c/(c+1) = 2/3 in this case.  Also $E(W^2)$
turns out to be 1/2, so Var(W) = 1/18.  Thus

\begin{equation}
Var(\hat{W}) = \frac{1/18}{50}
\end{equation}

By the CLT, the 97.5th percentile is about 1.96 standard deviations
above the mean, i.e.

\begin{equation}
d \approx \frac{2}{3} + 1.96 \sqrt{\frac{1}{900}} \approx \frac{2}{3} +
\frac{1}{15} = 11/15
\end{equation}

{\bf \Large  NOTE THAT YOU DID NOT NEED A CALCULATOR!} 

{\bf 2.a}  The likelihood is

\begin{equation}
\left ( \frac{1}{c} \right )^n
\end{equation}

as long as

\begin{equation}
c \geq max_i W_i
\end{equation}

So, 

\begin{equation}
\hat{c} = max_i W_i
\end{equation}

just as in our discrete example in the PLN (conference paper numbers).

{\bf 2.b}  The bias is $E\hat{C} - c$.  To get $E\hat{c}$ we need the
density of that estimator, which we get as follows:

\begin{equation}
P(\hat{c} \leq t) = P(\textrm{all} ~ W_i \leq t) = \left ( \frac{t}{c}
\right )^n
\end{equation}

So, 

\begin{equation}
f_{\hat{c}}(t) = \frac{n}{c^n}  t^{n-1}
\end{equation}

Integrating against t, we find that

\begin{equation}
E\hat{C} = \frac{n}{n+1} c
\end{equation}

So the bias is c/(n+1), not bad at all.

{\bf 3} If Y takes on only the values 0 and 1, then $m_{Y;X}(t) = P(Y = 1
| X = t)$.  From our PLN, the latter quantity is 

\begin{equation}
\frac{1}
{1 + \frac{(1-q) f_{X|Y=0}(t)}{q f_{X|Y=1}(t)}} 
\end{equation}

The information given in the problem implies that this last expression
is

\begin{equation}
\frac{1}
{1+ \frac{0.4}{0.6}\frac{2t}{3t^2}} =
\frac{1}{1+\frac{4}{9t}}
\end{equation}

{\bf 4.}  From the ``law of total variance'' in our PLN,

\begin{equation}
Var(Y) = E[Var(Y|X)] + Var[E(Y|X)]
\end{equation}

That first term on the right is

\begin{equation}
E[\sigma^2] = \sigma^2
\end{equation}

The second term is a variance, and is thus nonnegative.  Therefor

\begin{equation}
Var(Y) \geq \sigma^2
\end{equation}

\end{document}

