# Conditional Expectation (Poisson) UMVUE

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:

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

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

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

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
$$E(X_1X_2X_3\mid T)=\frac{1}{n^3}T(T-1)(T-2)$$
• I do not undestand what is my error. Could you please note what was my mistake? because according to my calculations UMVUE for $\lambda^3$ is $E(X_1X_2X_3\mid T)=\frac{T^3}{n^3}$ – Pedros Jul 1 '19 at 4:22
• I doubt your work is correct, because you have apparently used linearity of expectation, but $X_1X_2X_3$ is not linear in $X_1,X_2,X_3$. – StubbornAtom Jul 1 '19 at 12:36
• @Pedros Since $T$ is $\mathsf{Poisson}(n\lambda)$, you have $E(T-n\lambda)^3=n\lambda$, which shows $E(T^3)\ne n^3\lambda^3$. And to reiterate, the step $E(X_1X_3X_3\mid T)=E(X_1\mid T)E(X_2\mid T)E(X_3\mid T)$ has no justification. Why do you think this is true? – StubbornAtom Jul 2 '19 at 5:20