# 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.

• soa.org/globalassets/assets/files/edu/… see example 1.12 it is pretty much exactly this. You would use a normal approximation here. You can find the mean and variance using the independence of the exponentials and poissons and using the tower law for expectation and variance. – George Dewhirst Dec 3 at 0:05
• I've removed the "stochastic-integral" and "stochastic-processes" tags because those don't apply to this question. – Math1000 Dec 3 at 2:04