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Let be $(X_{1},X_{2})$ a simple random sample, with $X_{i}\sim\textrm{Exp}(\frac{1}{\theta})$ with $\theta$ unknown. I need to find the UMVUE of $\eta=\mathbb{P}(x>t)=e^{\frac{-t}{\theta}}$, with $t>0$ fixed.

I know that $T=\sum_{i}X_{i}$ is a sufficient and complete statistic of $\theta$. Also I know that this type of exercises are usually resolved with the Rao-Blackwell theorem.

I'm having troubles with finding a unbiased estimator of $\mathbb{P}(x>t)$. I thought of $e^{-t/\hat{x}}$. I know that $\hat{x}\sim \Gamma(2,1/\theta)$ and $\mathbb{E}(g(X))=\int g(x)f(x)dx$ but how do I check that $$\int_{0}^{\infty} e^{-t/z}\frac{2^2ze^{-2z/\theta}}{2\theta^2}dz=e^{-t/\theta}?$$

Lancet S.
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2 Answers2

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You may need an ancillary statistic. Note $S=X_1/(X_1+X_2)\sim U[0,1]$. By Basu, $S$ is independent of $T=X_1+X_2$. So $$P(X_1>x|T=t)=P(S>x/t)=(1-x/t)\mathbf{1}_{\{t:x\leq t\}}(t)=h(t)$$ We check that $h(T)$ is the UMVUE: $$E[h(T)]=\theta^{-2}\int_x^\infty(1-x/t)te^{-t/\theta}dt=\theta^{-2}\theta^2e^{-x/\theta}=e^{-x/\theta}$$

Snoop
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The estimator $W(X_1,X_2) = W(X_1) = \mathbb{1}_{X_1 >t}$ should work since $$ \mathbb{E} W(X_1) = \mathbb{E}(\mathbb{1}_{X_1 >t}) = P(X_1 > t) = e^{-t/\theta}. $$

Now you can use the Lehman-Scheffe Theorem (see my post here) and your UMVUE is

$$ E[W|T] $$ where $T$ is as you defined in your post.

Generally, if you wanted an unbiased estimator of $e^{-mt/\theta}$ you could take $$ W(X_1,X_2,\dots, X_m) = \mathbb{1}_{X_1 >t \land X_2 > t \dots \land X_m> t} $$ and use independence to show unbiasedness.

WeakLearner
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