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Rao-Blackwell Proof

Based on the Rao-Blackwell Theorem Proof shown in the image above, I have a question regarding the portion boxed in red:

Why is $Var(\hat\theta|T) = 0$ if $\hat\theta=E(\hat\theta|T)$?

I understand that in this case, $\hat\theta$ is a function of $T$, but I don't see why this implies $Var(\hat\theta|T)=0$.

Yaofeng Wang
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1 Answers1

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Concerning the section in red. We have $\mbox{Var}(\hat{\theta}) > \mbox{Var}(\tilde{\theta})$ or $\mbox{Var}(\hat\theta | T) = 0$. Suppose $\mbox{Var}(\hat{\theta}) = \mbox{Var}(\tilde{\theta})$ (there is no $\leq$ because of theorem B), then $\mbox{Var}(\hat\theta |T)=0$ implies $\hat\theta = \hat\theta (T)$, which in turn implies $\tilde\theta = \hat\theta$.

My guess is there is mis-understanding about the 'unless' operator. This is thoroughly dissected in, for instance, here.

Also, I think you are mixing up 'if' with 'only if'. For details, see, for instance, here.

AlvinL
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  • Hi Alvin, thank you for the reply! Based on your answer, I would like to ask why does $Var(\hat\theta|T)=0$ imply $\hat\theta = \hat\theta(T)$? – Yaofeng Wang Mar 19 '19 at 10:21