Expectation for a sum of Poisson random variables

Question

Let $Z = \sum_{r=1}^{\infty}X_{r}$, where $X_{r}$ is of Poisson Distribution with intensity rate $\frac{1}{r^{2}}$ for $r \geq 1$. Assume $Z$ is of some distribution (not necessarily Poisson itself).

I need to compute the expectation, $E(Z)$, and was given as a hint to use Monotone Convergence.

However, I am having a hard time setting up this problem as well as figuring out where Monotone Convergence comes into play (for example, what here is monotone)?

Since $X_{r}$ is of Poisson Distribution, we should have $E(Z) = E(\sum_{r=1}^{\infty}X_{r}) =\int_{-\infty}^{\infty} \sum_{r=1}^{\infty} X_{r}\frac{e^{1/r^2} \left(\frac{1}{r^2} \right)^{k}}{k!}dX_{r}$, I think. But, I do not know how to simplify the integral on the right (or even if I have written out the expectation correctly. Then, from that point, how do I apply the Monotone Convergence Theorem?

Thanks ahead of time for your time and patience!

Answer 1

The partial sums of the first $k$ terms, say $S_k$, are monotone. In particular. $$ ES_k=\sum_{r=1}^k\frac{1}{r^2} $$ Now let $k\to\infty$ and apply the MCT.

Answer 2

Let $Z_n=\sum_{r=1}^n X_r$. Then $$ \mathsf{E}Z=\lim_{n\to\infty}\mathsf{E}Z_n=\lim_{n\to\infty}\sum_{r=1}^n\mathsf{E}X_r=\sum_{r=1}^\infty r^{-2}=\frac{1}{6}\pi^2. $$

Answer 3

Monotone convergence theorem Let $(X_n)_{n\ge}$ be random variables such that $X_n\ge 0$ and $X_n\le X_{n+1}$ for all $n$. Then $X_n \uparrow X$ as $n\to\infty$ a.s. and $$ \Bbb E(X_n) \uparrow \Bbb E(X)\quad \text{as } n\to\infty. $$

Let $X_r$ be Poisson random variables with parameter $\frac{1}{r^2}$. Let $Z_n = X_1 +\cdots + X_n$ and $Z=\sum_{r=1}^\infty X_r$ so we have $Z_n \uparrow Z$ a.s. and $$ \Bbb E(Z_n)=\sum_{r=1}^n\Bbb E(X_r)=\sum_{r=1}^n \frac{1}{r^2}\to\Bbb E(Z)=\sum_{r=1}^\infty\frac{1}{r^2}=\frac{\pi^2}{6} $$