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\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.

{\bf Note:  Some problems may require answers to be in the form of
numerical expression.  An example is:}

\begin{equation}
2/7 \cdot 1.39 + \sqrt{29.002} + \int_0^10.8 2t ~ dt  +
(1,2)  
\left (
   \begin{array}{cc}
   1 & 4 \\
   6 & 1   
   \end{array}
\right )     
\left (
   \begin{array}{c}
   6  \\
   6    
   \end{array}
\right )     
\end{equation}

No variables are allowed in numerical expressions.  Also, infinite
series do not count either.

{\bf 1.}  There is a town with two social groups.  Everyone is in
exactly one group People arrive from outside town, with exponentially
distributed interarrival times at rate $\alpha$, and join one of the
groups with probability 0.5 each.  Each person will occasionally switch
groups, with one possible ``switch'' being to leave town entirely.  A
person's time before switching is exponentially distributed with rate
$\sigma$; the switch will either be to the other group or to the outside
world, with probabilities q and 1-q, respectively.  Let the state of the
system be (i,j), where i and j are the number of current members in
groups 1 and 2, respectively.  

Answer in terms of $\alpha$, $\lambda$, $\tau$ and $\pi$:

\begin{itemize}

\item [(a)] Give the balance equation for the state (8,8).

\item [(b)] We have just entered state (5,0).  What is the mean time
until Group 2 becomes nonempty?

\item [(c)] The president of Group 1 tells reporter, ``We've found over
the years that \_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\% of our
members come from transfers.''


\end{itemize}

{\bf 2.}  Consider a renewal process in which lifetimes have the values
1, 2, 3 or 4, with probability 1/4 each.

\begin{itemize}

\item [(a)] Find P[N(3) = 2].

\item [(b)] For large integer t, find the probability that the current
renewal period began at t-2.

% \item [(c)] Suppose the lifetimes are continuous random variables, with
% a uniform distribution on (0,1).  Give a numerical expression for the
% density of $R_2$.

\end{itemize}

{\bf Solutions:}

{\bf 1.}

\begin{itemize}

\item [(a)]

\begin{equation}
\pi_{(8,8)} (\alpha + 2 \cdot 8 \cdot \sigma ) =
( \pi_{(9,8)} + \pi_{(8,9)}) \cdot \sigma (1-q) +
( \pi_{(9,7)} + \pi_{(7,9)}) \cdot \sigma q +
( \pi_{(8,7)} + \pi_{(7,8)}) \cdot 0.5 \alpha
\end{equation}

\item [(b)] .  The rate of transitions into that group
from outside is $0.5 \alpha$.  When the system is in state (i,j), the
rate of transitions into group 1 from group 2 is $j \sigma q$, so the
overall rate is $\sum_{i,j} \pi_{(i,j)} j \sigma q$.  Thus the fraction
of new members coming in to group 1 from transfers is

\begin{equation}
\frac{\sum_{i,j} \pi_{(i,j)} j \sigma q}
{\alpha + \sum_{i,j} \pi_{(i,j)} j \sigma q}
\end{equation}

By the way, note that $\sum_{i,j} \pi_{(i,j)} j \sigma q = \sigma q EN$, where N
is the number of members of group 1.

\end{itemize} 

{\bf 2.}

\begin{itemize}

\item [(a)] 

\begin{equation}
P[N(3) = 2] = P(R_3 \leq 3, R_4 > 4) ~  \textrm{(note latter condition!)}
\end{equation}

Then

\begin{eqnarray}
P(R_3 \leq 3, R_4 > 4) &=& P(R_1=1, R_2=2, R_3 > 3 \textrm{ or }
R_1=1, R_2=3  \textrm{ or }
R_1=2, R_2=3) \\
&=& P(L_1=1, L_2=1, L_3 > 1 \textrm{ or }
L_1=1, L_2=2  \textrm{ or }
L_1=2, L_2=1) \\
&=& 
\frac{1}{4} \cdot \frac{1}{4} \cdot \frac{3}{4} +
\frac{1}{4} \cdot \frac{1}{4} +
\frac{1}{4} \cdot \frac{1}{4} \\
&=& \frac{11}{64}
\end{eqnarray}

\item [(b)]

This is from p.244.  We have

\begin{equation}
\pi_2 = \frac{1-F_L(3)}{EL} = \frac{1-0.75}{2.5} = \frac{1}{10}
\end{equation}

\end{itemize}

\end{document}