# UMVUE $g(\lambda)$ = $e^\lambda$ when $x_i \sim Pois(\lambda)$

Let $x_1 ... x_n$ be $Pois(\lambda)$ Find UMVUE of $e^\lambda$

From a previous question, I found the UMVUE of $e^{-\lambda}$ to be $(\frac{n-1}{n})^{t}$ where $t = \sum_{i=0}^n(x_i)$. $\sum_{i=0}^n(x_i)$ is our complete and sufficient statistic for $\lambda$ and has the distribution $Pois(n\lambda)$.

So I considered the estimator $(\frac{n-1}{n})^{-t}$ and took its expected value to see how close I got.

$$E((\frac{n-1}{n})^{-t}) = \sum_{i=0}^\infty\frac{(\frac{n-1}{n})^{-i}e^{-n\lambda}(n\lambda)^{i}}{i!}.$$

Which my calculator says is $$e^{\frac{n\lambda}{n-1}}$$ This seems close.

I have three questions regarding this result,

1) How did my calculator come up with this answer?
2) Can I get to the umvue by modifying my estimator? How?

3) Is there a different estimator I should consider instead? ETA: Not necessary, answering 1 and 2 were sufficient!

For 1, You're interested in $E[a^t]$ where $t$ is poisson with rate $n\lambda$. Explicitly:

$$E[a^x]=\sum_{x=0}^\infty a^x\frac{1}{e^{n\lambda}}\frac{n\lambda^x}{x!}=\frac{1}{e^{\lambda}}\sum_{x=0}^\infty \frac{(na\lambda)^x}{x!}=e^{na\lambda-n\lambda}=e^{n\lambda(a-1)}.$$

Now plug in $a=\left(\frac{n-1}{n}\right)^{-1}$. You should get $e^{n\lambda/(n-1)}$.

1) How did my calculator come up with this answer?

Alex R. has already explained this.

2) Can I get to the umvue by modifying my estimator? How?

Using Alex R. 's answer, clearly you don't have an unbiased estimator. However, if you can find $a$ such that,

$$E[a^t] = e^{\lambda},$$

then you will have an unbiased estimator. So you solve for $a$ in the equation $e^{n\lambda (a-1)} = e^{\lambda}$. This should lead you to a UMVUE.

3) Is there a different estimator I should consider instead?

Don't know if there can be a different estimator or not here.