Discussion
Suppose \(\{X_{1}, X_{2}, \ldots, X_{n}\}\), \(n \ge 2\) is a random sample from Poisson distribution with mean \(\theta\). If \(T = \sum_{i=1}^{n} X_{i}\), then the MVUE of \(e^{-\theta}\) is :
Answer: B
The MVUE of \(e^{-\theta}\) is \(\left(\frac{n - 1}{n}\right)^{T}\).
\(e^{-\theta} = P(X_{1} = 0)\), so the indicator \(I(X_{1} = 0)\) is unbiased. T is complete and sufficient, so by Rao-Blackwell and Lehmann-Scheffe the MVUE is \(E[I(X_{1} = 0) \mid T]\). Given \(T = t\), \(X_{1}\) is binomial \((t, 1/n)\), so
\[ \begin{aligned} P(X_{1} = 0 \mid T = t) &= \left(1 - \frac{1}{n}\right)^{t} \\ &= \left(\frac{n - 1}{n}\right)^{t} \end{aligned} \]
Check of unbiasedness: \(E[s^{T}] = e^{n\theta(s - 1)}\) with \(s = (n - 1)/n\) gives \(e^{-\theta}\).
\((n/(n - 1))^{T}\) has expectation \(e^{n\theta/(n - 1)}\), and the ratio forms involving \(T/(n - T)\) are not unbiased and are undefined for some values of T.
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