Expectation of Poisson Random Variables

Question

I need help for this particular question because there's this particular step where I am unable to grasp! I've marked it in bold with the word "WHY??". I can understand every other part but this step annoys me. Please do help if you are to do so! It'll be greatly appreciated!

Let X and Y be i.i.d Poisson random variables with parameter λ.

Compute the expectation of $(1 + X + Y )^{-1}$

SOLUTION:

X ~ Poi($\lambda_1$)

Y ~ Poi($\lambda_2$)

X + Y ~ Poi($\lambda_1+\lambda_2$)

X + Y ~ Poi(2$\lambda$) (i.i.d)

Hence,

E[$\frac{1}{1+X+Y}$ ]

= E[$\frac{1}{1+W}$], where W ~ Poi(2$\lambda$)

= $\sum_{k=0}^{infinite}\frac{1}{1+k}$ * $e^{-2\lambda}$ * $\frac{{2\lambda}^k}{k!}$ ** WHY??

...

the rest are just simple manipulations

I thought that only E[X] applies for this definition. How is it that E[$\frac{1}{1+W}$] applies for this too?

Thank you so very much!

*edited power 2 to -1, my bad, apologies!

Answer 1

The Law of the Unconscious Statistician, as applied to the discrete integer-valued random variable $W$:

$$\mathsf E(f(W)) ~=~ \sum_{w=0}^\infty f(w)\,\mathsf P(W=w)$$

So $$\mathsf E\left(\tfrac 1{1+W}\right) ~=~\sum_{w=0}^\infty \frac{\mathsf P(W=w)}{1+w}$$

Answer 2

Substitute $W' = \frac 1{1+W}$, then by definition

$$ E[W'] = \sum_{k=0}^\infty k P(W' = k) = \sum_{k=0}^\infty k P(\frac 1{1+W} = k)\, , $$ and now $P(\frac1{1+W} = k) = P(1 = k+kW) = P(W = \frac{1-k}{k})$. Make the substitution $l = \frac{1-k}{k} \Leftrightarrow k = \frac{1}{1+l}$ to get precisely the sum $\sum_{l=0}^\infty \frac1{1+l} P(W = l)$.

In more generality, this is true. That is, if $X$ is a random variable, and $g$ is a measurable function, then

$$ E[g(X)] = \int g(x) f_X(x) dx\, , $$ where $f_X$ is the probability density function of $X$. (Or replace the integral with a sum in the discrete case.)