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{\bf Question}

Suppose events occur in a Possion process with rate $\lambda$.
\begin{description}
\item[(i)] Find the conditional probability that there are $m$ events
in the first $s$ units of time, given that there are $n$ events in
the first $t$ units of time, where $0 \leq m \leq n$ and $0 \leq s
\leq t$.
\item[(ii)] If $N(t)$ denotes the number of events in $(0,t]$ and
$T$ denotes the time until the first event, find $P\{ T \leq s |
N(t) = n\}$ for $0 \leq s \leq t$ and $n$ a positive integer.
\end{description}

\vspace{.25in}

{\bf Answer}

\begin{description}
\item[(i)]
$\ds P(N(s) = m | N(t) = n)$
\begin{eqnarray*} &amp; = &amp; \frac{P(N(s) = m \,
{\rm and} \,  N(t) = n)}{P(N(t) = n)} \\ &amp; = &amp; \frac{P(N(0,s)=m)
\times P(N(s,t) =n-m)}{P(N(t) = n)} \\ &amp; &amp; {\rm since\ no.\ of\
events\ in\ } (0,s]{\rm\ and\ }(s,t]{\rm\ are\ independent.}\\ &amp; =
&amp; \frac{P(N(s) = m) \times P(N(t-s) = n-m)}{P(N(t) = n)} \\ &amp; = &amp;
\frac{e^{- \lambda s}(\lambda s)^m}{m!} \times
\frac{e^{-\lambda(t-s)}(\lambda(t-s))^{n-m}}{(n-m)!}  \times
\frac{n!}{e^{-\lambda t}(\lambda t)^n} \\ &amp;  = &amp;
\frac{n!}{m!(n-m)!} \left( \frac{s}{t} \right) ^m \left( 1 -
\frac{s}{t} \right) ^{n-m} \\ &amp; &amp; {\rm (note\ that\ this\ is\ a\
binomial\ probability)} \end{eqnarray*}


\item[(ii)]
$T \leq s$ is equivalent to $N(s) \geq 1$
\begin{eqnarray*} P(T \leq s | N(t) = n) &amp; = &amp; \sum_{m=1}^{n}
P(N(s) = m | N(t) = n) \\ &amp; = &amp; \sum_{m=1}^n \left(
\begin{array}{c} n \\ m \end{array} \right) \left( \frac{s}{t}
\right)^m \left(1 - \frac{s}{t} \right)^{n-m} \, {\rm by\ (i)} \\
&amp; = &amp; 1 - \left( 1 - \frac{s}{t} \right)^n \end{eqnarray*}
\end{description}



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