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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.

{\bf \Large SHOW YOUR WORK!}  

{\bf 1.} (10) In the ALOHA example, pp.9-12, find $p_{X_1,X_2}(1,2)$. Your
answer must consist of a single number found in those pages.

{\bf 2.} Find the following quantities for the density (5.17).  {\bf In
all cases, do NOT evaluate any integrals; leave your answer in integral
form.}

\begin{itemize}

\item [(a)] (10) $E(X^2+XY^{0.5})$

\item [(b)] (10) $P(Y > 0.5X)$

\item [(c)] (15) $F_{X,Y}(0.6,0.2)$

\end{itemize}

{\bf 3.} Find the following quantities for the dice example on p.115.
{\bf In all cases, leave your answers as numerical expressions, e.g.
$5/2.3 + 6^{1+\sqrt{2}}$.}  Feel free to cite (3.29).

\begin{itemize}

\item [(a)] (10) Cov(X,2S)

\item [(b)] (10) Cov(X,S+Y)

\item [(c)] (10) Cov(X+2Y,3X-Y)

\item [(d)] (10) $p_{X,S}(3,8)$

\end{itemize}

{\bf 4.} (15) Consider the ``Senthi'' example, pp.118-119.  Let R denote
the time it takes to go from state 1 to state 3.  Find $f_R(v)$.  {\bf
Leave your answer in integral form.}

{\bf Solutions:}

{\bf 1.} 0.20 (see just below (2.14))

{\bf 2a.} 

$$
 \int_{0}^{1} \int_{0}^{s} (s^2 + st^{0.5}) ~ 8st ~ dt ~ ds
$$

{\bf 2b.} 

$$
 \int_{0}^{1} \int_{0.5s}^{s} 8st ~ dt ~ ds
$$

{\bf 2c.}  By definition,

$$
F_{X,Y}(0.6,0.2) = P(X \leq 0.6 \textrm{ and } Y \leq 0.2) 
$$

The region in question has an irregular shape, so the so the answer is
a two-part integral:

$$
\int_{0}^{0.2} \int_{0}^{s} 8st ~ dt ~ ds
+
\int_{0.2}^{0.6} \int_{0}^{0.2} 8st ~ dt ~ ds
$$

{\bf 3a.} This and the next two parts make use of (5.22) and other
mailing tubes.

$$
Cov(X,2S) = 2 Cov(X,S) = 2 \cdot 0.707
$$

using (5.72).

{\bf 3b.} 

$$
Cov(X,S+Y) = Cov(X,S) + Cov(X,Y) = 0.707 + 0
$$

{\bf 3c.} In (5.22), take a = 1, b = 2, c = 3 and d = -1.  Then use the
fact that Cov(X,X) = Var(X), etc.  The result is 

$$
3 Var(X) - 2 Var(Y) = Var(X) = 2.92
$$

{\bf 4.} Write $R = R_1 + R_2$, where $R_i$ is the time to go from state
i to state i+1.  $R_1 \textrm{ and } R_2$ are independent, due to the
Markov/memoryless property, and as the example points out, $R_i$ has an
exponential distribution with parameter i(g-i).  So, we are basically in
the same situation as the backup battery example in Sec. 5.5.4, and

$$
f_R(v) = \int_{0}^v \lambda_1 e^{-\lambda_1 s} \lambda_2 e^{-\lambda_2(v-s)}~ ds
$$

where $\lambda_i = i(g-i)$.

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