Let $W_1, W_2$ be unbiased estimators of $\theta$ and let $T$ be a sufficient statistic for $\theta$. Is it true that $Var(E(W_1|T)) \leq Var(W_2)$?

I think that it may fails to be true.

If the statement is true then $Var(E(W_1|T))$ is the minimum among the variances $Var(W)$ of all unbiased estimator $W$ so $E(W_1|T)$ is a UMVUE of $\theta$. But Lehmann-Scheffe's theorem tells us that if $T$ is a complete sufficient statistic and $U$ is a unbiased estimator then $E(U|T)$ is a UMVUE of $\theta$. So we don't need the completeness of $T$ if the statement is true.

However I cannot prove it. Anyone can help me?


You can take a look at the proof of Rao-Blackwell theorem, \begin{align} var(E[W_1|T])&=E(E[W_1|T] - \theta)^2\\ &=E(E[W_1 - \theta|T])^2\\ &\le E(E(W_1 - \theta)^2|T])\\ &= E(W_1 - \theta)^2=var(W_1). \end{align} When the inequality can stem from Jensen's inequality or even from Cauchy-Schwarz inequality. Anyway, it tells you only how to improve a given estimator and without the condition of completeness of $T$, it is not necessarily gives you an UMVUE.

Thus, your statement is generally false, however it is not that easy to construct a counterexample as in regular cases a sufficient statistic is also complete and the first iteration of the Rao-Blackwell already gives you the UMVUE.

  • $\begingroup$ For the actual counterexample, see here. $\endgroup$ – bellcircle Jul 13 '18 at 12:29

No, the conjecture is not true. If $T$ is sufficient, the unbiased estimator $\mathrm{E}(W_1|T)$ is a function of $T$ and is hence an unbiased estimator of $\theta$ (trivially a statistic by definition of sufficiency). However, if $T$ is not complete, which means it might be correlated with some unbiased estimator of zero, named $U$. Let $W_2= \mathrm{E}(W_1|T) + a U$ be another unbiased estimator of $\theta$. We have \begin{align*} \mathrm{Var}(W_2) = \mathrm{Var}(\mathrm{E}(W_1|T)) + a^2\mathrm{Var}(U) +2a\mathrm{Cov}(\mathrm{E}(W_1|T),U). \end{align*} Hence, $\mathrm{Var}(W_2)\leq \mathrm{Var}(\mathrm{E}(W_1|T))$ if $a^2\mathrm{Var}(U) +2a\mathrm{Cov}(\mathrm{E}(W_1|T),U)\leq 0$. In other words, if $T$ is not complete, we can always find a constant $a$: \begin{align*} a\in\Bigg(0,\frac{-2\mathrm{Cov}(\mathrm{E}(W_1|T),U)}{\mathrm{Var}(U)}\Bigg]; &\quad \mbox{if }Cov(\mathrm{E}(W_1|T),U)<0,\\ a\in\Bigg[\frac{-2\mathrm{Cov}(\mathrm{E}(W_1|T),U)}{\mathrm{Var}(U)},0\Bigg); &\quad \mbox{if }Cov(\mathrm{E}(W_1|T),U)>0, \end{align*} such that $\mathrm{Var}(W_2)\leq \mathrm{Var}(\mathrm{E}(W_1|T))$, where $\mathrm{E}(W_1|T)$ and $W_2$ are all unbiased estimator of $\theta$ and $T$ is sufficient but not complete.


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