Suppose $X_1$ and $X_2$ are independent observations from $$ f(x)=\frac{1}{\beta}e^{-x/\beta}, \quad x>0 $$ We know that a complete sufficient statistic for $\beta$ will be $X_1+X_2$. Now how do I find an unbiased estimator of $e^{-\frac{10}{\beta}}$? 10 is an arbitrary choice here. I tried many functions of $X_1+X_2$ and none of them have the desired expectation. There must be some cute trick I can't think of.
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Hint:
$\mathbb P(X_1>10)=e^{-\frac{10}{\beta}}$
So $\mathbb P(X_1>10)=\mathbb E(1_{\{X_1 >10 \}})$ is an unbiased estimator. It is enough to conditioning it by $\bar{X}$(that is complete and sufficient ).
$\mathbb P(X_1>10|\bar{X})=\mathbb E(1_{\{X_1 >10 \}}|\bar{X})=g(\bar{X})$ is UMVUE for $e^{-\frac{10}{\beta}}$.
$\mathbb E (g(\bar{X}))=\mathbb E \mathbb E(1_{\{X_1 >10 \}}|\bar{X})= E(1_{\{X_1 >10 \}})=e^{-\frac{10}{\beta}}$
Masoud
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1Thanks! That's refreshing! – Fireant Apr 12 '20 at 01:54
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You are welcome! – Masoud Apr 12 '20 at 01:54