Conditional Expectation (Poisson) UMVUE

Question

Suppose $X_1,X_2,\ldots,X_n$ is a random sample from a Poisson distribution with mean $λ$. How can I find the conditional expectation $E \left( X_1\times X_2\times X_3 \mid \sum_{i=1}^n X_i= z \right)$?

And $X_1,X_2,\ldots,X_n$ are independent identically distributed Random variables.

My thought:

$E \left( X_1\times X_2\times X_3 \mid \sum_{i=1}^n X_i=z \right)$

$= E \left( X_1 \mid\sum_{i=1}^n X_i=z \right)\times E \left( X_2 \mid\sum_{i=1}^n X_i=z \right)\times E \left( X_3 \mid\sum_{i=1}^n X_i=z \right)$

$= \frac{z}{n}\times \frac{z}{n}\times\frac{z}{n} $

Answer 1

An indirect way of obtaining the conditional expectation is to use the Lehmann-Scheffe theorem, which says that an unbiased estimator of a parametric function $g(\lambda)$ based on a complete sufficient statistic is the uniformly minimum variance unbiased estimator (UMVUE) of $g(\lambda)$.

Since $X_1X_2X_3$ is unbiased for $\lambda^3$ and $T=\sum\limits_{i=1}^n X_i$ is complete sufficient, by Lehmann-Scheffe the UMVUE of $\lambda^3$ is the quantity you are after: $E(X_1X_2X_3\mid T)$.

At the same time it can be verified that an unbiased estimator of $\lambda^3$ based on $T$ is $\frac{1}{n^3}T(T-1)(T-2)$. This is also UMVUE. As UMVUE is unique, we must have

$$E(X_1X_2X_3\mid T)=\frac{1}{n^3}T(T-1)(T-2)$$