Given the random sample $X_1, \ldots, X_n$ of the random variable $X \sim \textrm{Poisson}(\lambda)$, find the UMVUE for $g(\lambda) = \lambda \exp(-\lambda)$.
We know through the exponential family that $\sum X_i$ is a sufficient and complete statistic for $\lambda$. Define $W(X)$ as
$$
W(X) =
\begin{cases}
1, & \text{if $X_1 = 1$} && \\
0, & \text{otherwise}
\end{cases}
$$
Then
$$ E[W(X)] = \sum_{w = 0}^{\infty} w P(W = w) = 1 \cdot P(X_1 = 1) = \lambda e^{-\lambda} $$
Using Lehmann-Scheffé
$$ T^{*}(X) = E[W(X) | T(X)] = \sum w P(W(X) = w | T(X) = t ) = P(X_1 = 1 | \sum_{i = 1}^{n} X_i )$$
we obtain
$$ \lambda \left(\frac{n-1}{n} \right)^{\sum_{i = 1}^{n} X_i} $$
Is the above reasoning correct?
Best Answer
The parameter should cancel out in the very last steps: \begin{align*} P[X_1 = 1 | \sum_{i = 1}^n X_i = t] &= \frac{P[X_1 = 1 \text{ and }\sum_{i = 2}^n X_i = t - 1]}{P[\sum_{i = 1}^n X_i = t]} \\ &= \frac{(\lambda^1 e^{-\lambda}/1!) \times [((n - 1)\lambda)^{t - 1} e^{-(n - 1)\lambda}/(t - 1)!]}{(n\lambda)^t e^{-n\lambda}/t!} \\ &= \frac{\lambda^t e^{-n\lambda} (n - 1)^{t - 1}t!}{n^t \lambda^t e^{-n\lambda} (t - 1)!} = \frac{t}{n}\Big(\frac{n - 1}{n}\Big)^{t - 1} \end{align*} So the UMVUE is $\frac{\sum_{i = 1}^n X_i}{n}\Big(\frac{n - 1}{n}\Big)^{\sum_{i = 1}^n X_i - 1}$