1
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – jbowman May 17 '18 at 18:56
  • 1
    $\begingroup$ $E(X_1X_2|\dots) \neq X_1E(X_2|\dots)$. $\endgroup$ – jbowman May 17 '18 at 19:05
  • $\begingroup$ @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$$ $\endgroup$ – user207460 May 17 '18 at 19:05
  • 2
    $\begingroup$ $X_1$ and $X_2$ are independent, but not given $\sum X_i$. $\endgroup$ – Xi'an May 17 '18 at 19:07
  • $\begingroup$ @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 $\endgroup$ – user207460 May 17 '18 at 19:07
2
$\begingroup$

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: Since $S_n\sim {\cal P}(n\theta)$, the conditional density of $(X_1,\ldots,X_{n-1})$ given $S_n=s$ is

\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 $S_n=s$ is

\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.

$\endgroup$

Your Answer

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