Let $X_1,\ldots,X_n \sim$ Poisson$(\lambda)$. I wish to find UMVUE of $e^{-\lambda}$. Here $\overline{X}_n$ is complete and sufficient for $\lambda$ (hence for $e^{-\lambda}$??). Define $Y_i=1$ if $X_i=0,$ $Y_i=0$ otherwise. Clearly (I checked!) each $Y_i$ and hence $\overline{Y}_n$ is unbiased for $e^{-\lambda}$. Then I should be able to use the Lehmann Scheffe theorem to deduce the UMVUE. The estimator I get looks like $$\frac{1}{n} E\left\{\#X_i:X_i=0 \mid \sum_{i=1}^n X_i\right\}$$ whereas the correct answer is $$\left(\frac{n-1}{n}\right)^{\sum_{i=1}^n X_i}$$ I'd appreciate if anyone lets me know where I'm going wrong and guides me into the right path. Thank you.

  • $\begingroup$ @AdamO Thank you for pointing out the insight. Can you please give any clue about how I can achieve the "correct" estimator as mentioned? $\endgroup$ Commented Jun 5, 2018 at 16:20
  • $\begingroup$ Converted my comment to an answer. $\endgroup$
    – AdamO
    Commented Jun 5, 2018 at 16:30

1 Answer 1


Yes, Lehman Scheffe gives you the sufficiency of $\bar{X}$ for $e^{-\lambda}$. The problem is that $\bar{Y}$ does not actually condition on the sufficient statistic. Although the $Y$ are a function of the $X$, $\bar{Y}$ is not a function of the $\bar{X}$. The problem is that your expression isn't reduced. What, in fact, is the distribution of $\bar{Y}$ given $\bar{X}$? This is discussed in the Wiki page on the Rao Blackwell theorem.


Essentially, let $f_{\lambda}(x)$ be the poisson density $\exp(-\lambda)\lambda^x/x!$

\begin{eqnarray} E(Y_1 | \bar{X} = s) &=& P(X_1 = 0| \bar{X}=s) \\ &=& P(X_1 = 0, \sum_{i=2}^n X_i = ns) /P(\bar{X} = s)\\ &=& f_{\lambda}(0) f_{(n-1)\lambda}(ns) / f_{n\lambda}(ns) \\ \end{eqnarray}

Which some algebra reduces the last expression to $(1-\frac{1}{n})^{ns}$


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