# Find UMVU estimator for $e^{-3 \theta}$ given a complete sufficient statistic $X \sim Pois(\theta)$ with $\theta>0$.

My attempt: We know, since $$X\sim Pois(\theta)$$ that $$\mathbb{P}_{\theta}(X=x)=e^{-\theta}\theta^{x}/x!$$. A given tip is that we must recall that $$e^{x}=\sum^{\infty}_{k=0}\frac{x^{k}}{k!}$$. I know that once we find an unbiased estimator that is a function of our complete sufficient statistic $$X$$, that this estimator must then automatically be UMVU. However, I'm not sure how to approach this question by even finding an expression for an unbiased estimator.

Question: How to approach/solve this exercise?

Thanks!

We need to find some function $$g(k)$$ s.t. $$\mathbb E[g(X)]=e^{-3\theta}$$ for all $$\theta>0$$. Write this expectation: $$\mathbb E[g(X)]=\sum_{k=0}^\infty g(k)\dfrac{\theta^k}{k!}e^{-\theta}=e^{-3\theta}.$$ Multiply both parts by $$e^\theta$$ and get $$\sum_{k=0}^\infty g(k)\dfrac{\theta^k}{k!}=e^{-2\theta} = \sum_{k=0}^\infty \frac{(-2)^k\theta^k}{k!}.$$ Since these sums are equal for each $$\theta>0$$, we get $$g(k)=(-2)^k$$ and $$g(X)=(-2)^X$$ is unique unbiased estimator for $$e^{-3\theta}$$.