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\begin{document}

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

Directions: {\bf \Large 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.

{\bf \Large Unless otherwise stated, give numerical answers as
expressions, e.g. $\frac{2}{3} \times 6 - 1.8$.  Do NOT use
calculators.}

{\bf 1.} (15) Using Equation (1.64), give a numerical expression for
$Var(X_1)$.

{\bf 2.} Suppose X and Y are independent random variables with standard
deviations 3 and 4, respectively.

\begin{itemize}

\item [(a)] (10) Find Var(X+Y).

\item [(b)] (10) Find Var(2X+Y).

\end{itemize}

{\bf 3.} (30) Fill in the blanks in the following simulation, which finds
the approximate variance of N, the number of rolls of a die needed to
get the face having just one dot.

\begin{Verbatim}[fontsize=\relsize{-2}]
onesixth <- 1/6
sumn <- 0
sumn2 <- 0
for (i in 1:10000) {
   n <- 0
   while(TRUE) {
      ________________________________________
      if (______________________________ < onesixth) break
   }
   sumn <- sumn + n
   sumn2 <- sumn2 + n^2
}
approxvarn <- ____________________________________________
cat("the approx. value of Var(N) is ",approx,"\n")
\end{Verbatim}

{\bf 4.} (20) Jack and Jill keep rolling a four-sided and three-sided die.
The first player to get the face having just one dot wins, except that
if they both get a 1, it's a tie, and play continues.  Let N denote the
number of turns needed.  Find $p_N(1)$.

{\bf 5.} (15) Let X be the total number of dots we get if we roll three
dice.  Find an upper bound for $P(X \geq 15)$, using our course
materials.

{\bf Solutions:}

{\bf 1.}  $Var(X_1) = E(X_1)^2 - (EX_1)^2$.  The last term is $1.52^2$,
and the next-to-last is $1^2 \cdot 0.48 + 2^2 \cdot 0.52$.

{\bf 2.}  By (1.61), $Var(X+Y) = Var(X) + Var(Y) = 3^2 + 4^2$.  By
(1.48), $Var(2X) = 2^2 Var(X) = 2^2 \cdot 3^2$, so again by (1.61),
$Var(2X+Y) = 2^2 \cdot 3^2 + 4^2$.

{\bf 3.}  
\begin{Verbatim}[fontsize=\relsize{-2}]
n <- n + 1
runif(1)
sumn2/10000 - (sumn/10000)^2
\end{Verbatim}

{\bf 4.}  

\begin{eqnarray*}
p_N(1) &=& P(N = 1) \\
&=&  P(\textrm{Jack gets 1 and Jill doesn't or vice versa}) \\
&=& \frac{1}{4} \cdot \frac{2}{3} + \frac{3}{4} \cdot \frac{1}{3}
\end{eqnarray*}

{\bf 5.}  Use Markov's Inequality:

$$
P(X \geq 15) \leq \frac{EX}{15} = \frac{3(3.5)}{15}
$$

(Of course, it's a very poor bound in this case.)

\end{document}

