Expectation of e^xbar

Question

I trying to find the expectation of $(1-\bar X)e^{\bar X}$, where $X_1,...X_n$ are i.i.d poisson random variables with mean $\lambda$. I am stuck because I have no idea how to find the mean of $e^{\bar x}$. Any tips or hints to get going would be appreciated.

Answer 1

If $X\sim Po(\lambda)$ then $Y=\overline{X}_n\sim Po\left(n\lambda\right)$, even if it has a modified support.

$Y=0,\frac{1}{n},\frac{2}{n},...$

Remember also that $E(e^Y)$ is the definition of MGF with $t=1$

Answer 2

The moment generating function for the underlying Poisson distributions is

$$\mathbb E\left[e^{X_i t}\right] = \exp\left(\lambda (e^t-1)\right)$$ so for the sum of $n$ i.i.d. copies is $$\mathbb E\left[e^{\sum X_i t}\right] = \exp\left(n\lambda (e^t-1)\right)$$ and letting $t=\frac1n$ gives $$\mathbb E\left[e^{\bar{X}}\right] = \mathbb E\left[e^{\frac1n \sum X_i }\right] = \exp\left(n\lambda (e^{\frac1n}-1)\right)$$

Answer 3

In order to calculate $E(\overline{X} e^\overline{X})$ I think it is better to do it in an analytical way

Set $Y=\Sigma_i X_i\sim Po(n\lambda)$ you get

$$\mathbb{E}(\overline{X} e^\overline{X})=\frac{1}{n}\mathbb{E}(Ye^{Y/n})=\frac{e^{-n\lambda}}{n}\sum_{y=0}^{\infty}ye^{y/n}\frac{(n\lambda)^y}{y!}=$$

$$=\frac{e^{-n\lambda}}{n}\cdot e^{1/n}n\lambda\sum_{y-1=0}^{\infty}\frac{(e^{1/n}n\lambda)^{y-1}}{(y-1)!}=e^{1/n}\lambda e^{-n\lambda}e^{e^{1/n}n\lambda}=\lambda e^{1/n+n\lambda(e^{1/n}-1)}$$

With the same easy procedure, you can calculate also $E(e^{\overline{X}})$ in a couple of passages and without invoking MGF

$$\mathbb{E}\left[e^{\overline{X}}\right]=\sum_{y=0}^{\infty}e^{y/n}\cdot\frac{e^{-n\lambda}(n\lambda)^y}{y!}=e^{-n\lambda}\sum_{y=0}^{\infty}\frac{(e^{1/n}n\lambda)^y}{y!}=e^{-n\lambda}\cdot e^{e^{1/n}n\lambda}=e^{n\lambda(e^{1/n}-1)}$$

as previously showed by @Henry using MGF