Evaluating the expected product of Poisson and exponential random variables. Consider a bank with 1000 customers. On average there are 60 withdrawal requests per month, while the number of withdrawals in a single month is Poisson distributed. On average, the amount of each withdrawal is 700 euro and the amounts are exponentially distributed. Calculate the probability that the sum total of withdrawals in a given month exceeds 50,000 euro.
My approach was to use the pdfs for Poisson and exponential random variables to evaluate the expectation of the product of the two variables:
\begin{equation}
f_p(x) = \frac{e^{-\lambda}\lambda^x}{x!} \\
f_e(y) = \lambda e^{-\lambda x} \\
\int_{x=0}^{1000}\int_{y=\frac{50000}{x}}^{\infty}xyf_p(x)f_e(y) dy dx
\end{equation}
But this integral is unwieldy, and I suspect incorrectly specified.
Any hints on a better approach are appreciated.
 A: Let $N(t)$ be a Poisson process with rate $\lambda>0$, and  $W_n$ an i.i.d. sequence of $\mathrm{Pois}(\mu)$ random variables independent of $N(t)$, and $C>0$. Define $Z(t) = \sum_{m=1}^{N(t)}W_m$. Then the probability we seek is
\begin{align}
\mathbb P(Z(t)>C) &= \mathbb P\left(\sum_{m=1}^{N(t)}W_m>C\right)\\
&= \sum_{n=1}^\infty \mathbb P\left(\sum_{m=1}^{N(t)}W_m>C\mid N(t)=n \right)\mathbb P(N(t)=n)\\
&= \sum_{n=1}^\infty \mathbb P\left(\sum_{m=1}^n W_m >C\right)\mathbb P(N(t)=n)\\
&= \sum_{n=1}^\infty\sum_{m=n}^\infty e^{-\mu C}\frac{(\mu C)^m}{m!}e^{-\lambda t}\frac{(\lambda t)^n}{n!}\\
&= e^{-(\mu C+\lambda t)}\sum_{n=1}^\infty \frac{(\lambda t)^n}{n!}\sum_{m=n}^\infty \frac{(\mu C)^m}{m!}\\
&= e^{-(\mu C+\lambda t)}\sum_{n=1}^\infty \frac{(\lambda t)^n}{n!}\frac{e^{C \mu } (\Gamma (n)-\Gamma (n,C \mu ))}{\Gamma (n)}\\
&= e^{-\lambda t}\sum_{n=1}^\infty\frac{(\lambda t)^n}{n!(n-1)!}((n-1)! - \Gamma(n,C\mu)),
\end{align}
but Mathematica will not evaluate this sum. We can however compute the mean of $Z(t)$:
\begin{align}
\mathbb E[Z(t)] &= \sum_{n=1}^\infty \mathbb E[Z(t)\mid N(t)=n]\mathbb P(N(t)=n)\\
&=\sum_{n=1}^\infty \mathbb E\left[\sum_{m=1}^{N(t)} W_m\mid N(t)=n \right]\mathbb P(N(t)=n)\\
&= \sum_{n=1}^\infty \mathbb E\left[\sum_{m=1}^n W_m\right]\mathbb P(N(t)=n)\\
&= \sum_{n=1}^\infty n\mu e^{-\lambda t}\frac{(\lambda t)^n}{n!}\\
&= \lambda\mu t.
\end{align}
