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!

  • $\begingroup$ 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. $\endgroup$ – StubbornAtom Jun 20 at 16:05
  • $\begingroup$ 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? $\endgroup$ – murph Jun 20 at 16:33
  • $\begingroup$ Absolutely..... $\endgroup$ – StubbornAtom Jun 20 at 16:50

The answer to (a) is right.

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. $


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