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\begin{document}


{\bf Question}

\begin{description}
\item[(a)] Give a set of axioms for a Poisson process with
constant rate $\lambda$.  State the distribution of the number of
events, $N(t),$  in a time interval of length $t$ units.

Show that for the overlapping time intervals $(0,t]$ and $(0,s]$
where $s < t,$ $$P(N(s) = i | N(t) = j) = \left( \begin{array}{c}
j
\\i
\end{array} \right)\left( \frac{s}{t}
\right)^i \left( 1 - \frac{s}{t} \right) ^{j-i},$$ where $$\left(
\begin{array}{c} j \\i \end{array} \right) = \frac{j!}{i!(j-i)!}$$

\item[(b)] An insurance company receives claims at times $t_1, t_2,
t_3,\ldots$ which are points in a Poisson process with rate
$\lambda$ per week such that $0 < t_1 < t_2 <\ldots$.  Find the
probability distribution of the time until the $n$-th claim after
the start of the financial year.

The claims at times $t_1, t_2, t_3,\ldots$ are $Y_1, Y_2,
Y_3,\ldots$ respectively, which are independent random variables
each having the same probability generating function $A(z).$ Show
that the probability generating function for the total amount of
claims which the company has to meet in the first $t$ weeks of the
financial year is $$e^{\lambda t A(z)- \lambda t}.$$
\end{description}



\vspace{.25in}

{\bf Answer}

\begin{description}
\item[(a)] Axioms for a Poisson process.
\begin{description}
\item[(i)] The numbers of events in disjoint time intervals are
independent.
\item[(ii)] $P$(1 event in $(t, t+\delta t]) = \lambda \delta t + o(\delta
t)$ as $\delta t \to 0$

$P$(0 events in $(t, t+\delta t]) = 1- \lambda \delta t + o(\delta
t)$ as $\delta t \to 0$
\end{description}

N(t) is Poisson($\lambda t).$ So $\ds P(N(t) = n) = \frac{(\lambda
t)^n e ^{-\lambda t}}{n!}\ \  n = 0, 1, 2, \ldots$

\begin{eqnarray*} P(N(s)= i | N(t) = j) & = & \frac{P(N(s) = i \,
\ \&\  \, N(t) = j)}{P(N(t) = j)} \\ & = & \frac{P(N(s) = i \, \
\&\ \, N(9s,t]) = j-i)}{P(N(t))=j} \\ & = & \frac{ \ds
\frac{(\lambda s)^i e^{-\lambda s} }{i!}\times
\frac{(\lambda(t-s))^{j-i}e^{-\lambda (t-s)}}{(j-i)!}}{\ds
\frac{(\lambda t)^j e^{-\lambda t}}{j!}}
\\ & = & \frac{j!}{i!(j-i)!} \cdot \frac{s^i (t-s)^{j-i}}{t^j} \\ & = & \left(
\begin{array}{c}  j  \\ i \end{array} \right) \left( \frac{s}{t}
\right)^i \left( 1 - \frac{s}{t} \right)^{j-i} \end{eqnarray*}


\item[(b)] Let T = time until $n$-th claim. \begin{eqnarray*} P(T
\leq t) & = & 1 - P(T>t) \\ & = & 1 - P(N(t) <n) \\ & = & 1 -
\sum_{j=0}^{n-1} p_j(t) \end{eqnarray*} To find the p.d.f. we need
\begin{eqnarray*} \frac{d}{dt} P(T \leq t) & = &  -
\sum_{j=0}^{n-1}p_j'(t) \\ & = & - \lambda e ^{-\lambda t} -
\sum_{j=1}^{n-1} \lambda e^{-\lambda t} \left( \frac{(\lambda
t)^{j-1}}{(j-1)!} - \frac{(\lambda t)^j}{j!} \right) \\ & = &
\lambda e^{-\lambda t} \frac{(\lambda t)^{n-1}}{(n-1)!}
\hspace{.2in} {\rm i.e.\ \ } \Gamma(n,\lambda) \end{eqnarray*}
\end{description}

The total amount of claims in the first $t$ weeks is

$$X(t) = Y_1 + Y_2 + .. + Y_{N(t)}$$

Its p.g.f. is
\begin{eqnarray*} \sum_{j=0}^\infty P(X(t) = j)z^j & = &
\sum_{j=0}^\infty \sum_{n=0}^\infty z^j P(X(t) = j | N(t) =
n)P(N(t) = n) \\ & = & \sum_{j=0}^\infty \sum_{n=0}^\infty z^j
P(Y_1 + ... + Y_n = j) \frac{(\lambda t)^n e^{-\lambda t}}{n!} \\&
= & \sum_{n=0}^\infty \left( \sum_{j=0}^\infty z^j P(Y_1 + ... +
Y_n = j) \right) \frac{(\lambda t)^n e^{-\lambda t}}{n!}\\ & = &
e^{-\lambda t} \sum_{n=0}^\infty (A(z))^n \cdot \frac{(\lambda
t)^n}{n!} \\ & = & \exp( \lambda t A(z) - \lambda t)
\end{eqnarray*}



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