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I am having a hard time trying to solve this problem. I don't know how to start it. Any help would be greatly appreciated.

Let T be any unbiased estimator of $\tau(\theta),$ and let $W$ be a sufficient statistic for theta. Define $\phi(W)=E[T|W]$. Show that $\phi(W)$ is an unbiased estimator of $\tau(\theta),$ and $\mathrm{Var}(T)=\mathrm{Var}\left(\phi(W)\right) + E\left[\mathrm{Var}(T|W)\right]$.

The only thing I know is that this may be a part of the Rao-Blackwell Thm.

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We want to show that $E(\phi(W))=\tau(\theta)$.
Start with the definition:
$$ \begin{align*} E(\phi(W))&=E[E(T|W)]\\ &=E[T]\\ &=\tau(\theta) \end{align*} $$

where the last equality holds, because $T$ is an unbiased estimator for $\tau(\theta).$ Hence $\phi(W)$ is unbiased for $\tau(\theta).$

For the second part, use the fact that for any two random variables, $X$ and $Y$,
$$\mathrm{Var}(X)=\mathrm{Var}\left[E(X|Y)\right]+E\left[\mathrm{Var}(X|Y)\right].$$


Added:
As noted by @guy, for the first part, we also need to show that $\phi(W)$ is an estimator of $\theta$. But this is obvious. See comments below.

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  • $\begingroup$ One thing to add: you have shown $\phi(W)$ is unbiased, but not that $\phi(W)$ is an estimator, i.e. we need $\phi(W)$ not to depend on $\theta$. It is trivial, of course, but this is where sufficiency comes in, and it is easy miss. $\endgroup$
    – guy
    Jul 9, 2011 at 22:08
  • $\begingroup$ @guy: I thought it was obvious since $W$ is a sufficient statistic for $\theta$, $\phi(\theta)$ is a function only of the sample and does not depend on $\theta$. Thanks for pointing it out though. I'll add to my answer. $\endgroup$
    – Nana
    Jul 9, 2011 at 22:18

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