# How to improve an estimator for a Poisson sample

Given the statistical model $(\mathbb N_0^n, P(\mathbb N_0^n),\operatorname{Poi}(\vartheta)^{\otimes n}:\vartheta >0)$, $T(X)=X_1X_2$ is an unbiased estimator of $\vartheta^2$. I want to improve this estimator using the statistic $$U(X)=\sum_{i=1}^n X_i$$

It is well known that $U(X)$ is complete and sufficient. Thus by Lehmann-Scheffé $$E\left(X_1X_2\mid\sigma(\sum X_i)\right)$$ is UMVU. I calculated $$E(X_1X_2\mid \sigma(\sum X_i))=X_1E(X_2\mid \sigma(\sum X_i)=X_1E(X_1\mid \sigma(\sum X_i)=\frac{X_1}{n}\biggr(E(X_1\mid \sigma(\sum X_i)+X_1E(X_2\mid \sigma(\sum X_i)+\dots X_1E(X_n\mid \sigma(\sum X_i)\biggl)=X_1E(X_1+\dots +X_n \mid \sigma(\sum X_i)\biggl)=\frac {X_1(\sum_{i=1}^n X_i)}{n}$$. I used several times that $X_i$ is iid, in the first equation I used that $X_1X_2$ is $\sigma(\sum X_i)$ measurable, i.e. $E(X_1X_2 \mid \sigma (\sum X_i)=X_1E(X_2 \mid \sigma (\sum X_i))$

But this has to be wrong because I checked if this estimator is unbiased for $\vartheta^2$, but it is not! Therefore I conclude that I am missing something, probably my attempt is wrong. How is this done correctly?

• Are you sure $T(X) = X_1X_2$ is unbiased AND that the average is biased? If it is, then $X_1X_3$ is also unbiased, etc., and the average of a bunch of unbiased estimators is also unbiased. – jbowman May 17 '18 at 18:56
• $E(X_1X_2|\dots) \neq X_1E(X_2|\dots)$. – jbowman May 17 '18 at 19:05
• @jbowman I think yes, since $E(X_1 X_2)=E(X_1)E(X_2)$ by independency and $X_i$ is poisson distributed, i.e. $E(X_i)=\vartheta$$– user207460 May 17 '18 at 19:05 • X_1 and X_2 are independent, but not given \sum X_i. – Xi'an May 17 '18 at 19:07 • @jbowman why does this not hold since X_1 is \sigma \sum X_i measurable, i.e. see en.wikipedia.org/wiki/Conditional_expectation#Basic_properties – user207460 May 17 '18 at 19:07 ## 1 Answer Hint #1: You should start from the start, by considering the distribution of$$X_1,X_2,\ldots,X_{n-1},\overbrace{\textstyle{\sum_{i=1}^n} X_i}^{S_n}$$which has the density$$\theta^{x_1+x_2+\ldots+x_{n-1}+\underbrace{\sum_{i=1}^n x_i}_{s_n}-\sum_{i=2}^nx_i}e^{-n\theta}\Big/x_1!x_2!\cdots\left(\textstyle{\underbrace{\sum_{i=1}^n x_i}_{s_n}-\sum_{i=2}^nx_i}\right)!\, \mathbb{I}_{\underbrace{\sum_{i=1}^n x_i}_{s_n}-\sum_{i=2}^nx_i\ge 0}$$that is,$(X_1,X_2,\ldots,X_{n-1},S_n)has the density \begin{align}\dfrac{\theta^{s}e^{-n\theta}}{s!}\,\dfrac{s!}{x_1!\cdots x_{n-1}!(s-x_1-\ldots-x_{n-1})!}&\,\mathbb{I}_{s\ge x_1+\ldots+x_{n-1}}\end{align} Hint #2: SinceS_n\sim {\cal P}(n\theta)$, the conditional density of$(X_1,\ldots,X_{n-1})$given$S_n=sis \begin{align}\dfrac{n^{-n} s!} {x_1!\cdots x_{n-1}!(s-x_1-\ldots-x_{n-1})!}&\,\mathbb{I}_{s\ge x_1+\ldots+x_{n-1}}\end{align} which is a Multinomial{\cal M}_n(S_n;1/n,\ldots,1/n)$distribution Hint #3: Since the conditional distribution of$(X_1,\ldots,X_{n-1})$given$S_n=s$is given above, the marginal conditional distribution of$(X_1,X_2)$given$S_n=sis \begin{align}p(x_1,x_2) &= \dfrac{n^{-x_1-x_2} ((n-2)/n)^{s-x_1-x_2} s!} {x_1!x_2! (s-x_1-x_2)!}\,\mathbb{I}_{s\ge x_1+x_{2}} \end{align} which is a Multinomial{\cal M}_3(s;1/n,1/n,n-2/n)$Hint #4: Since the conditional distribution of$(X_1,X_2)$given$S_n=s$is given above, the conditional expectation of$X_1X_2$given$S_n=s\$ is derived from the covariance of this distribution

\begin{align}\mathbb{E}[X_1X_2|S_n=s] &= -\frac{s}{n^2}+\frac{s^2}{n^2}=\frac{s(s-1)}{n^2}\end{align}

and a simulation experiment shows this UMVUE is indeed unbiased.