What is $\ E[3^X]=?$ $\ E[1/(X+1)]=?$

Question

Let $\ X\sim\text{Pois}(λ)$
1) What is $\ E[3^X]=?$

Sol: $\ E[X]=λ$ , $\ E[3^X ]=3^λ$

2) What is $\ E[1/(X+1)]=?$

Sol: $\ 1/(1+λ)$

Is my sol true?

Answer 1

Recall that the PMF for a poisson distribution with parameter $\lambda$ is

$$Pr(X=k) = \frac{\lambda^ke^{-\lambda}}{k!}$$

Now... if we were to try to find $E[X]$ here, that would correspond to the summation:

$$\sum\limits_{k=0}^\infty k\cdot Pr(X=k) = \sum\limits_{k=0}^\infty k\cdot \frac{\lambda^k e^{-\lambda}}{k!} = \lambda$$

Finding $E[f(X)]$ will instead be: $\sum\limits_{k=0}^\infty f(k)\cdot Pr(X=k)$

(Recall that the limits on the summation are a result of this being the support of the distribution)

So, for your exact problem being asked, you are being tasked with trying to find the following two summations:

$$\sum\limits_{k=0}^\infty 3^k\cdot\frac{\lambda^k e^{-\lambda}}{k!}$$ and $$\sum\limits_{k=0}^\infty \frac{1}{k+1}\cdot\frac{\lambda^k e^{-\lambda}}{k!}$$

As for hints on how to evaluate these, recall what the Taylor series expansion for $e^z$ looks like.