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\usepackage{fancyvrb}
\usepackage{relsize}
\usepackage{hyperref}

\usepackage{listings}
\usepackage{amsmath}

\begin{document}

\lstset{xleftmargin=5mm,framexleftmargin=10mm,basicstyle=\footnotesize}

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

Directions: {\bf \Large Work only on this sheet} (on both sides, if
needed).  MAKE SURE TO COPY YOUR ANSWERS TO A SEPARATE SHEET FOR SENDING
ME AN ELECTRONIC COPY LATER.

{\bf Important note:}  Remember that in problems calling for R code, you
are allowed to use any built-in R function, e.g. {\bf choose()}, {\bf
pnorm()}, etc.

{\bf 1.} Consider the class enrollment size example, starting on
p.97.  Suppose the distribution of enrollment size is Poisson, rather
than approximate normal.  Assume the mean is again 28.8.  

\begin{itemize}

\item [(a)] (20) Find Var(N).

\item [(b)] (20) Find $F_N(26)$.

\item [(c)] (15) Find $P(N \geq 30 | N \geq 25)$.

\end{itemize}

{\bf 2.} Consider the network intrusion example on p.97. 

\begin{itemize}

\item [(a)] (15) Let $Y = Z^2$.  Name the parametric family of densities
that Y's density belongs to, including the parameter values, if any.

\item [(b)] (15) Let G denote the indicator random variable for the event
$X \geq 535$.  Find Var(G).

\end{itemize}

{\bf 3.} (15) Suppose R didn't include the {\bf sample()} function.  We
could use the code below instead.  Here's an example of usage:

\begin{lstlisting}
> x <- samp(c(1,6,8),1000,c(0.2,0.5,0.3))
> sum(x==1)
[1] 224
> sum(x==6)
[1] 495
> sum(x==8)
[1] 281
\end{lstlisting}

Here we generate 1000 numbers from {1,2,3}, with probability 0.2, 0.5
and 0.2, respectively, and then count the numbers of 1s, 6s and 8s we
get.

The built-in R function {\bf cumsum()} finds cumulative sums, e.g.

\begin{lstlisting}
> cumsum(c(3,8,1))
[1]  3 11 12
\end{lstlisting}

\begin{lstlisting}[numbers=left]
# what we'd do if there were no sample() 
# ftn in R; sample n items (with
# replacement) from the vector nums, with 
# probabilities given by prob;
# the vectors nums and prob must have 
# the same length (not checked here);
# not claimed efficient
samp <- function(nums,n,prob) {
   samps <- vector(length=n)
   cumulprob <- cumsum(prob)
   for (i in 1:n) 
      samps[i] <- sample_one_item(nums,cumulprob)
   return(samps)
}

sample_one_item <- function(nums,cumulprob) {
   u <- runif(1)
   lc <- length(cumulprob) 
   for (j in 1:(lc-1)) {
      if (BLANKa) BLANKb
   }
   BLANKc
}
\end{lstlisting}

\onecolumn

{\bf Solutions:}

{\bf 1.a} Since N has a Poisson distribution, Var(N) = E(N) = 28.8.

{\bf 1.b} For a Poisson random variable M, $\lambda = EM$, so answer is

\begin{lstlisting}
ppois(26,28.8)
\end{lstlisting}

{\bf 1.c} (4.50) still holds, and evaluates to 

\begin{lstlisting}[numbers=left]
(1 - ppois(29,28.8)) / (1 - ppois(24,28.8))
\end{lstlisting}

{\bf 2.a}  Chi-square, 1 degree of freedom.

{\bf 2.b}  From Section 3.6:

\begin{equation}
Var(G) = P(G = 1) \cdot [1 - P(G=1)] = 0.01 \cdot 0.99
\end{equation}

{\bf 3.}

\begin{lstlisting}[numbers=left]
# what we'd do if there were no sample() ftn in R; sample n items (with
# replacement) from the vector nums, with probabilities given by prob;
# the vectors nums and prob must have the same length (not checked here);
# not claimed efficient
samp <- function(nums,n,prob) {
   samps <- vector(length=n)
   cumulprob <- cumsum(prob)
   for (i in 1:n) 
      samps[i] <- sample_one_item(nums,cumulprob)
   return(samps)
}

sample_one_item <- function(nums,cumulprob) {
   u <- runif(1)
   lc <- length(cumulprob) 
   for (j in 1:(lc-1)) {
      if (u < cumulprob[j]) return(nums[j])
   }
   return(nums[lc])
}
\end{lstlisting}

\end{document}