If $X_1,X_2,\ldots,X_n\sim \mathrm{Pois}(\lambda)$, find an unbiased estimator of $(1+\lambda)e^{-\lambda}$.
I am actually supposed to find the UMVUE of $(1+\lambda)e^{-\lambda}$. but I first have to find its unbiased estimator.
My Aproach:
I tried using the MLE of $\lambda$ which is $\hat{\lambda}:= \frac{1}{n}\sum_{i=1}^n X_i = \overline{X}$. Then I use the invariance property and it follows that $(1+\overline{X})e^{-\overline{X}}$will be the MLE of $(1+\lambda)e^{-\lambda}$ but I'm not sure if it is also unbiased.
Also I tried using that $(1+\lambda)e^{-\lambda} = e^{-\lambda}+\lambda e^{-\lambda} =P(X=0)+P(X=1)$ but I could not get anything.
Any ideas?
We use Lehmann-Scheff theorem here.
If $T$ is a complete sufficient statistic for $\lambda$ and $\mathbb{E}[g(T)] = f(\lambda)$ then $g(T)$ is the uniformly minimum-variance unbiased estimator (UMVUE) of $f(\lambda)$.
We just need to find $g$ such that $\mathbb{E}[g(T)] = f(\lambda)$.
$f(\lambda) = (1+\lambda)e^{-\lambda}$.
$\mathbb{E}[e^{-\hat{\lambda}}] = \mathbb{E}[e^{-\frac{1}{N}\sum_{i=1}^{N}X_i}] = \prod_{i=1}^N M_{x_{i}} (-1/N) = e^{(\lambda (e^{-1/N}-1))} $. Here we have used the Poisson MGF. So we will need to adjust our estimator in order to make it unbiased, as you can see. You can see the $e^{-\lambda}$ term in there.
Next we will also compute $\mathbb{E}[\hat{\lambda} e^{-\hat{\lambda}}]$. We can compute this as $\mathbb{E}[\frac{d}{dt} M_{\hat{\lambda}}(t)| (t= -1)]$. We are allowed to take the $\frac{d}{dt}$ out of the braces here.