If x has Poisson distribution with parameter $λ,$ how to find MLE and UMVE of $g(λ)=e^λ (1+λ)$ Given: A Poisson distribution with parameter $\lambda$, how to find MLE and UMVE of $g(\lambda)=e^{\lambda}(1+\lambda)$. I was trying to use the invariance property of MLE and then Lehmann-Scheffe's and Rao-Blackwell theorems. But I am not sure if I am taking the right route. Any suggestion would be appreciated.
Thanks
 A: I will not give anything away, I'll just make some plots I wish someone made for me when I first studied this.
Let $X_1,...,X_n \sim \text{Poisson}(\lambda)$.  For demonstration purposes let's say that our sample size is $n=50$ and unbeknownst to us the true fixed $\lambda$ is $3$.  The histogram below shows the sampling distribution of $\hat{\lambda}_{MLE}=\bar{X}$ based on 10,000 Monte Carlo simulations.  The mean of these 10,000 Monte Carlo samples is 3.003.  Since the maximum likelihood estimator is unbiased and a function of the complete sufficient statistic it is also the UMVUE for $\lambda$.

You are correct that based on the invariance property, $\hat{g}(\lambda)_{MLE}=g(\hat{\lambda}_{MLE})$.  Below is the histogram for the sampling distribution of this estimator over the same 10,000 Monte Carlo simulations.

The unknown fixed true $g(\lambda)$ is $g(3)=e^3(1+3)=80.3$, while our Monte Carlo mean is 84.3.  The maximum likelihood estimator of $g(\lambda)$ is a function of the complete sufficient statistic, but it is not unbiased so the MLE is not the UMVUE.  It looks like you'll have to find a suitable transformation of the MLE to make it unbiased.
