Suppose $X_1....X_n$ is a random sample from a Poi($\theta$) population. Find the best unbiased estimator of $\theta^2e^{-\theta}$

My attempt:

Let $\sum_1^nX_i=T$. We know $T$ is complete and sufficient. So we seek an unbiased estimator of $\theta^2e^{-\theta}$ then condition it on T then find the expected value.

An unbiased estimator of $\theta^2e^{-\theta}$ is $2 \chi_{[X_1=2]}$

We calculate $E(2\chi_{[X_1=2]}\mid T=t) = 2\Pr(X=2\mid T=t)$

By Bayes this is

$$2(tC2) \left( 1-\frac{1}{n}\right)^{t-2} \left( \frac{1}{n} \right)^2$$

I'm very unsure of this result. It matches none of my classmates, but I cannot see where I'm making an error. I also would be curious to see alternative approaches.


1 Answer 1


\begin{align} & \Pr(X_1=2\mid T=t) \\[8pt] = {} & \frac{\Pr(X_1=2\ \&\ T=t)}{\Pr(T=t)} \\[8pt] = {} & \frac{\Pr(X_1=2\ \&\ X_2+\cdots+X_n = t-2)}{(n\theta)^t e^{-n\theta}/t!} \\[8pt] = {} & \frac{\big( \theta^2 e^{-\theta}/2! \big)\big( ((n-1)\theta)^{t-2} e^{-(n-1)\theta} /(t-2)! \big)}{(n\theta)^t e^{-n\theta}/t!} \\[8pt] = {} & \frac{t!}{2!(t-2)!}\cdot\frac{(n-1)^{t-2}}{n^t} = \binom t 2 \left( \frac 1 n \right)^2 \left( 1 - \frac 1 n \right)^{t-2} \end{align}

So what you got is right.


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