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Let $X_1,...,X_n$ be iid Poisson($\lambda$) with $n\geq 4$. We are given the unbiased estimator $T(X)=I(X_1=0 \cap X_2=0 \cap X_3=0)$ for $f(\lambda)=e^{-3\lambda}$, and my task is to derive the Rao-Blackwellized version of this estimator.

I know that $U(X)=\sum_{i=1}^nX_i$ is sufficient for $\lambda$, so for the final estimator, I get the following: $$E[T|U=u]=\frac{(n-3)^u}{n^u}.$$

I got this answer by checking the cases when $U=1$, then $U=2$, etc. and this seemed to be the solution. Does this look correct? Frankly, I’m not used to seeing such an odd looking statistic.

StubbornAtom
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1 Answers1

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Due to independence of $X_1,X_2,\ldots,X_n$, we have $\sum\limits_{i=1}^n X_i\sim \text{Poisson}(n\lambda)$.

Hence for all $u\in\{0,1,2,\ldots\}$,

\begin{align} E(T\mid U=u)&=E(\mathbf 1_{X_1,X_2,X_3=0}\mid U=u) \\\\&=P\left(X_1,X_2,X_3=0\,\bigg | \sum_{i=1}^n X_i=u\right) \\\\&=\frac{P\left(X_1=0,X_2=0,X_3=0,\sum\limits_{i=4}^n X_i=u\right)}{P\left(\sum\limits_{i=1}^n X_i=u\right)} \\\\&=\frac{(P(X_1=0))^3 P\left(\sum\limits_{i=4}^n X_i=u\right)}{P\left(\sum\limits_{i=1}^n X_i=u\right)} \\\\&=\frac{e^{-3\lambda} e^{-(n-3)\lambda}((n-3)\lambda)^u}{e^{-n\lambda}(n\lambda)^u} \\\\&=\left(\frac{n-3}{n}\right)^u \end{align}

So the Rao-Blackwellised estimator is as you say

$$E\left(T\,\,\bigg |\, \sum\limits_{i=1}^n X_i\right)=\left(1-\frac{3}{n}\right)^{\sum\limits_{i=1}^n X_i}$$

StubbornAtom
  • 17,052