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{\bf Question}

A firm of consultants wins contracts according to a Poisson
process with rate $\lambda$.  The contracts yield income $\pounds
Y_i\ \ i = 1, 2,\ldots$ which are independent and identically
distributed random variables with distribution $$p\{Y = y\} =
\frac{\alpha^y}{\lambda y} \hspace{.2in} {\rm for \ \ } y = 1,
2,..., {\rm where \ \ } \alpha = 1 - e^{- \lambda}.$$ Let $X(t)$
denote the total value of the contracts obtained in a time
interval of length $t.$  Prove that the probability generating
function for $X(t)$ is $$G(z) = e^{-\lambda t}(1 - \alpha z)^{-t},
\hspace{.2in} {\rm where \ \ \ } -1 \leq z \alpha < 1$$  Hence, or
otherwise, find the probability distribution of $X(t)$ and its
mean and variance.

[Note that $\ln(1 +x) = x - \frac{x^2}{2} + \frac{x^3}{3} -
\frac{x^4}{4} + ....$ for $-1 <  x \leq  1$.]



\vspace{.25in}

{\bf Answer}

$X(t)$ is a Compound Poisson Process.

The p.g.f. of each $Y_i$ is $$A(z) = \sum_{y=1}^\infty
\frac{\alpha ^y z^y}{\lambda y} = -\frac{1}{\lambda} \ln(1 -
\alpha z)$$

By the theory of compound Poisson Processes $X(t)$ has p.g.f.
\begin{eqnarray*}G(z) & = & e^{-\lambda t} \exp(\lambda t A(z)) \\
& = & e^{-\lambda t}\exp(-t \ln(1-\alpha z)) \\ & = & e^{-\lambda
t}(1 - \alpha z)^{-t} \end{eqnarray*} The probability distribution
of $X(t)$ is given by \begin{eqnarray*} P(X(t)) = j) & = & {\rm
coefficient\ of\ }z^j {\rm in\ the\ p.g.f.} \\ & = & e^{-\lambda
t} \left( \begin{array}{c} t+j-1 \\ j \end{array} \right) \alpha
^j \\ & = & e^{- \lambda t} \left( \begin{array}{c} t+j-1 \\ j
\end{array} \right) (1 - e^{-\lambda})^j \end{eqnarray*}

$\ds E(X) = \left( \frac{\pl G}{\pl z}\right)_{z=1} = t(e^\lambda
- 1) \hspace{.2in} (\alpha = 1 - e^{-\lambda})$

$\ds E(X(X-1)) = \left( \frac{\pl ^2 G}{\pl z^2} \right)_{z=1} = t
(t+1) (e^\lambda-1)^2$

$\ds Var X = E(X(X-1)) + E(X) - (E(X))^2 = t e^\lambda(e^\lambda -
1)$




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