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?
$\begingroup$
$\endgroup$
Add a comment
|
2 Answers
$\begingroup$
$\endgroup$
3
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).
-
$\begingroup$ Small nitpick: "non-negative," not "positive." $\endgroup$ Oct 25, 2017 at 5:31
-
1$\begingroup$ The pmf of the Poisson distribution needs a strictly positive value for $\lambda$. $\endgroup$– Ken WeiOct 25, 2017 at 6:28
-
$\begingroup$ My bad. I've always considered the case $\lambda=0$ (Dirac on the point $0$) as a degenerate case of Poisson. $\endgroup$ Oct 25, 2017 at 12:50
$\begingroup$
$\endgroup$
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 $$ P(2) = \sum_{k=0}^\infty e^{-\lambda}\frac{(\lambda 2)^k}{k!} = e^{-\lambda}e^{2\lambda} = e^\lambda. $$