# Expectation and variance of $(\frac{n-1}{n})^{\sum_{i=1}^{n}X_i}$

Let $(X_1,..., X_n)$ be a random sample from a Poisson distribution where $X_i$'s are iid .Prove that $T=(\frac{n-1}{n})^{\sum_{i=1}^{n}X_i}$ is the MVUE for $q(\theta) = P(x=0)=e^{-\theta}$ and find its variance .

Based on Lehmann-Scheffe's theorem I can prove that $\sum_{i=1}^{n}X_i$ is a complete sufficient statistic since the joint density belongs to the exponential class of densities but I can not prove that $T$ is an unbiased estimator for $q(\theta)$ , all what I know that $\sum_{i=1}^{n}X_i \sim$ Poisson $(n\theta)$ . Any help or hint please ?

• No need to shout in the title of your question using huge :) – Daniel W. Farlow May 24 '17 at 0:35
• Are you sure the question is correct? T is not unbiased as $E[T]=E[\sum_{i=1}^{n}X_i](n-1)/n=n\theta(n-1)/n=(n-1)\theta$ – A-B-izi May 24 '17 at 0:50
• @A.Brizzi It is raised to the power $\sum X_i$ – Bahgat Nassour May 24 '17 at 0:53

Look at the expectation of $T$, following the definition. We get: $E[T]=\sum_{k=1}^{\infty} (\frac{n-1}{n})^{k} e ^{-n\theta} \frac{n\theta^{k}}{k!}= e ^{-n\theta}\sum_{k=1}^{\infty} \frac{((n-1)\theta)^{k}}{k!}$

By definition of exponential,it follows that:

$E[T]=e ^{-n\theta}e ^{(n-1)\theta}=e ^{-\theta}$.

This is an unbiased estimator for $q(\theta)$.

Hope this helps

• Check power of first term inside the first sum. – NCh May 24 '17 at 1:23

$$\mathbb Ec^{\sum_{i=1}^n X_i} = \sum_{k=0}^\infty c^k \frac{(n\theta)^k}{k!}e^{-n\theta}=e^{cn\theta}e^{-n\theta}=e^{n\theta(c-1)}.$$ Substitute $c=\frac{n-1}{n}$ and conclude.