Expected Value of 2^X and 2^-X for Geometric and Poisson Distributions?

[Cedric Chia]

TL;DR

Help find E[2^X] and E[2^-X] if X~Geom(p) and if X~Pois(lambda)

For the following distributions find $$E[2^X]$$ and $$E[2^{-X}]$$ if finite. In each case,clearly state for what values of the parameter the expectation is finite.

(a) $$X\sim Geom(p)$$
(b) $$X\sim Pois(\lambda)$$

My attempt:

Using LOTUS and $$E[X]=\sum_{k=0}^{\infty}kP(X=k)=\frac{1-p}{p}$$
thus
$$E[2^X]= \frac{1-p}{p} \sum_{k=0}^{\infty}\frac{2^k}{k}$$
and
$$E[2^{-X}]= \frac{1-p}{p} \sum_{k=0}^{\infty}\frac{1}{k2^k}$$
for $$X\sim Geom(p)$$
but the expected value in either case seems to be divergent, I don't know how to continue, please help.

 

[tnich]

$$E[2^X]= \frac{1-p}{p} \sum_{k=0}^{\infty}\frac{2^k}{k}$$

I don't think this step is correct. I suggest that you start from $E[f(X)] = \sum_k {f(k)p(k)}$.

 

[WWGD]

Along what tnich said, maybe you can use a transform like:

$P[2^X >a]=P[ e^{Xln2}>a]=...$

A simpler example: Given X, if you want to find the distribution of $X^2$ , we use:

$P[X^2>a]= P[ X > a^{1/2}]$