Expectation of Poisson ^ -1

**Zithuan** · May 14th 2009, 09:01 AM

I've been able to work through most of the problem so far, so I'm only posting what's relevant:

$X_1, X_2, ..., X_n \sim Poisson(\theta)$ and $Y = \sum{X_i}$

What I need help with is calculating: $E(\frac{n}{Y+1})$

It seems like it ought to be easy, but I'm drawing a blank. Is using the mgf the strategy I should be looking at? If so, how?

**Moo** · May 14th 2009, 09:19 AM

Hello,

If they follow a Poisson distribution, it means that $\mathbb{P}(X_1=k)=\frac{e^{-\lambda}\lambda^k}{k!}$ , $k\in\mathbb{N}$

By the mgf method, you can see that $Y$ follows a Poisson distribution with parameter $n\theta$

Then, recall that for any measurable and bounded function h, $\mathbb{E}(h(Z))=\sum_{z\in D} h(z)\mathbb{P}(Z=z)$ , where D is the domain of the rv Z.

So here, we have :
$E=\mathbb{E}\left(\frac{n}{Y+1}\right)=\sum_{k=0}^ \infty \frac{n}{k+1}\cdot \mathbb{P}(Y=k)$

$E=\sum_{k=0}^\infty \frac{n}{k+1}\cdot \frac{(n\theta)^k}{k!} \cdot e^{-n\theta}$

$E=e^{-n\theta}\sum_{k=0}^\infty \frac{n^{k+1} \theta^k}{(k+1)!}$

$E=e^{-n\theta}\cdot\frac1\theta \sum_{k=0}^\infty \frac{(n\theta)^{k+1}}{(k+1)!}$

Change the indice :
$E=e^{-n\theta}\cdot\frac1\theta \sum_{k=1}^\infty \frac{(n\theta)^k}{k!}$

Now recall this : $e^x=\sum_{k=0}^\infty \frac{x^k}{k!}=1+\sum_{k=1}^\infty \frac{x^k}{k!} \Rightarrow \sum_{k=1}^\infty \frac{x^k}{k!}=e^x-1$