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$(X_1,...,X_n)$ is a random sample, $V_n$ is an unbiased estimator of the population parameter $\theta$ and $T_n$ is a sufficient statistic for $\theta$. Then by Rao-Blackwell theorem the rv

$\varphi(T_n):=\mathbb E_\theta[V_n|T_n]$

is an unbiased estimator of $\theta$ whic is uniformly better than $V_n$.

I know the definitions of conditional expectation and distribution, but I can't get how $\varphi(T_n)$ is an estimator (a function of the random sample that does't depend on $\theta$) using sufficiency

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For simplicity, consider the discrete case:

$$\mathbb E_\theta[V_n|T_n=t]=\sum_{\lbrace x:T_n(x)=t\rbrace}\mathbb P_\theta(X=x|T_n=t)V_n(x)$$

By sufficiency of $T_n$, $\mathbb P_\theta(X=x|T_n=t):=g(x,t)$ is independent of $\theta$. So,

$$\mathbb E_\theta[V_n|T_n=t]=\sum_{\lbrace x:T_n(x)=t\rbrace}g(x,t)V_n(x)$$

is independent of $\theta$ and is a statistic.

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  • $\begingroup$ $\mathbb{E}_\theta[V_n|T_n=t]=\sum_{x:T_n(x)=t} \mathbb{P}_\theta (V_n=V_n(x)|T_n=t)V_n(x)= \sum_{x:T_n(x)=t} \mathbb{P}_\theta (X=x|T_n=t)V_n(x)$ where the last equality follows from $V_n=V_n(x) \iff X=x$? $\endgroup$ – Louis Feb 11 '15 at 8:34
  • $\begingroup$ No, $\mathbb E_\theta[V_n|T_n=t]=\sum_{\lbrace x:T_n(x)=t\rbrace}\mathbb P_\theta(X=x|T_n=t)V_n(x)$ follows by definition of expectation. We sum over all possibilities of $x$ given $T_n=t$. It is not true in general that $\mathbb E_\theta[V_n|T_n=t]=\sum_{\lbrace x:T_n(x)=t\rbrace}\mathbb P_\theta(V_n=V_n(x)|T_n=t)V_n(x)$ $\endgroup$ – Sarastro Feb 11 '15 at 17:51
  • $\begingroup$ Ok! I forgot that $V_n$ is a function of $X$ $\endgroup$ – Louis Feb 12 '15 at 9:31

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