Originally Posted by

**Moo** Hello,

If you know no formula for this, the best way is to partition the set of events :

(assuming that N and the sequence X_i are independent, and I guess there's an extra N in what you wrote... anyway, it would be the same reasoning)

$\displaystyle \mathbb{E}\left(\sum_{i=1}^N X_i\right)=\sum_{k=0}^\infty k \mathbb{P}\left(\sum_{i=1}^N X_i=k\right)$

And $\displaystyle P(\sum_{i=1}^N X_i=k)=\sum_{n=1}^\infty P(\sum_{i=1}^N X_i=k ~,~ N=n)$

$\displaystyle =\sum_{n=1}^\infty P(\sum_{i=1}^n X_i=k ~,~ N=n)=\sum_{n=1}^\infty P(\sum_{i=1}^{n}X_i=k)P(N=n)$

The sum of Poisson variables is a Poisson variable (whose parameter is the sum of the parameters)

And you have to see what happens if N is 0. This should be stated in your problem, or conventionally, the sum would be 0.

It's a bit messy, but you can get the information you need from here...