2
$\begingroup$

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} $

$\endgroup$
2
1
$\begingroup$

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

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

$\endgroup$
4
  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ @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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.