MLE and UMVUE for $P(X_1=2)$ when $X_i\sim \text{Poisson}(\lambda)$

I've been studying for my exam and came across the following problem:

Suppose that $$X_1,\ldots,X_n$$ is a random sample from a Poisson distribution with mean $$\lambda$$.(a) Find the maximum likelihood estimator (MLE) of $$\eta = P(X_1 = 2\mid \lambda)$$. (b) Find the UMVUE of $$\eta$$.

I am not sure about my solution, so I would like to ask for some feedback.

Sol: (a). Recall that the MLE for $$\lambda$$ is the mean of the sample, $$\overline{X}$$, thus $$\eta_{MLE}= \frac{e^{-\overline{X}}\overline{X}^2}{2!}$$.

(b). We see that the joint pmf $$f(x\mid\lambda)=e^{-\lambda}\prod_{i=1}^{n}\frac{1}{x_i!}e^{\sum_ix_i\ln\lambda}$$ is part of the exponential family. Thus, $$Y=\sum_ix_i$$ is a complete and sufficient statistics for $$\lambda$$. Therefore, $$Y^*=\frac{e^{-Y}Y^2}{2!}$$ is complete and sufficient for $$\eta$$. Further, let $$W=1$$, if $$X_1=2$$ and $$0$$ otherwise. $$E(W)=1\cdot P(X_1=2)=\eta$$. Thus, $$W$$ is unbiased for $$\eta$$.

We construct next the Rao-Blackwell estimator: $$η^*=E(W\mid Y^*=y^*)=E(1_{\{X_1=2\}}\mid Y^*=y^*)=P(X_1=2\mid \frac{e^{-Y}Y^2}{2!}=y^*)=\frac{P(X_1=2,\frac{e^{-Y}Y^2}{2!}=y^*)}{P(\frac{e^{-Y}Y^2=y^*}{2!})}$$

I am not sure if my thinking is on the right track. I would appreciate any help. Thanks!

• You are on the right track but it would be easier if you condition on $Y=\sum X_i$ to answer (b). Answer to (a) is of course correct because of invariance of MLE. – StubbornAtom Jun 20 at 16:05
• Thanks for the quick reply. I did condition on Y and got the same answer as John Wick got below. We conclude that the statistic $g(Y) = \frac{(n-1)^{Y-2}}{n^Y}\cdot \frac{Y(Y-1)}{2}$ is an UMVUE based on Lehmann-Scheffe Lemma, right? – murph Jun 20 at 16:33
• Absolutely..... – StubbornAtom Jun 20 at 16:50

In the answer of (b), you know that $$Y= \sum x_i$$ is complete and sufficient for the family. So, it is enough to find an estimator of the form $$\hat\eta=g(Y)$$ such that $$E\hat\eta = \eta = \frac{e^{-\lambda}\lambda^2}{2!}.$$ Now $$Y\sim \text{Poi} (n\lambda).$$ Hence, getting an expansion in the RHS we get $$\sum_{k=0}^\infty g(k)\frac{e^{-n\lambda}(n\lambda)^k}{k!} = \frac{e^{-\lambda}\lambda^2}{2!}$$
$$\sum_{k=0}^\infty g(k)\frac{(n\lambda)^k}{k!} = \frac{e^{(n-1)\lambda}\lambda^2}{2!}= \frac{\lambda^2}2 \sum_{l=0}^\infty \frac{((n-1)\lambda)^l}{l!}.$$
Hence $$g(0)=g(1) =0.$$ And for $$k\ge 2,$$ $$g(k) = \frac{(n-1)^{k-2}}{n^k}k(k-1)/2.$$