How to check if its finite

Question

For $X \sim Pois(\lambda)$, find $E(2^X)$ if it is finite.

I know how to solve this (we use Law of Unconscious Statistician) but am doubtful as to how we specify the condition for which it is finite. Can someone tell me how we find the condition?

Answer 1

It should be finite for any $\lambda$. The MGF of the Poisson distribution is $\mathbb{E}[\exp(tX)] = \exp(\lambda(e^t-1))$, so for $t=\log(2)$, you have $\mathbb{E}[2^X] = e^\lambda$, which is finite for any $\lambda$ (although $\lambda$ should of course be positive).

Answer 2

The generating function for a Poisson distribution is $$ P(s) := \mathbb E\left[s^X\right] = \sum_{k=0}^\infty \mathbb P(X=k)s^k =\sum_{k=0}^\infty e^{-\lambda} \frac{(\lambda s)^k}{k!}, $$ where this series converges. Since $$ \sum_{k=0}^\infty \frac{z^k}{k!} = e^z,\quad \forall z\in\mathbb C $$ we see that $\mathbb E\left[2^X\right] = P(2)$ is finite, and compute $$ P(2) = \sum_{k=0}^\infty e^{-\lambda}\frac{(\lambda 2)^k}{k!} = e^{-\lambda}e^{2\lambda} = e^\lambda. $$