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

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

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    $\begingroup$ 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 $E(X_1X_2X_3\mid T)=\frac{T^3}{n^3}$ $\endgroup$ – Pedros Jul 1 '19 at 4:22
  • $\begingroup$ I doubt your work is correct, because you have apparently used linearity of expectation, but $X_1X_2X_3$ is not linear in $X_1,X_2,X_3$. $\endgroup$ – StubbornAtom Jul 1 '19 at 12:36
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    $\begingroup$ @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? $\endgroup$ – StubbornAtom Jul 2 '19 at 5:20
  • $\begingroup$ stats.stackexchange.com/q/265821/119261 $\endgroup$ – StubbornAtom May 9 '20 at 8:19

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