1. Combined Poisson

I am preparing for my test which is on Friday. There is this problem which is killing me and I don't have any idea how to solve it! It is in conditional probabilities chapter. The question is

X ~ Poi (a)
Y ~ Poi (b)

Z = X^Y

What is the Expected value of Z.

I will be so thankful if anyone could help me.

2. Originally Posted by Olivia787878
I am preparing for my test which is on Friday. There is this problem which is killing me and I don't have any idea how to solve it! It is in conditional probabilities chapter. The question is

X ~ Poi (a)
Y ~ Poi (b)

Z = X^Y

What is the Expected value of Z.

I will be so thankful if anyone could help me.
Is it $\displaystyle E[X^Y]$?

You know that $\displaystyle E[s^Y]=e^{b(s-1)}$, this is the moment generating function. Or maybe you know $\displaystyle E[e^{tY}]=e^{b(e^t-1)}$, which is the same with $\displaystyle s=e^t$. Anyway, you can compute this, even if you don't know it already.

On the other hand, since (you should have mentioned it) X and Y are independent, the conditional distribution of Y given X=x is the same as the distribution of Y. Therefore, for all $\displaystyle x\in\mathbb{N}$, $\displaystyle E[X^Y|X=x] = E[x^Y]=e^{b(x-1)}$. Then $\displaystyle E[X^Y]=E[E[X^Y|X]] = E[e^{b(X-1)}]$, and the last expectation can be computed using the moment generating function again.

The last inequalities can also be written $\displaystyle E[X^Y]=\sum_{x=0}^\infty E[X^Y|X=x] P(X=x) = \sum_{x=0}^\infty e^{b(x-1)} P(X=x)$ $\displaystyle =E[e^{b(X-1)}]$.