- #1

Cedric Chia

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- TL;DR Summary
- 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.

(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.

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