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Let $X_1,...,X_n$ be a random sample of i.i.d. exponential distribution with probability density function $$f(x|\theta)=\frac{1}{\theta}exp(-\frac{x}{\theta}), \ x\geq0$$ Let $S_n=\sum_{i=1}^nX_i$ and $X_{(1)}=\text{min}_{1\leq i\leq n}X_i$. Find the closed form expression for $W_n=E(nX_{(1)}|S_n)$ and verify $W_n$ is the UMVUE for $\theta$.

My attempt is that first noticed that the pdf for $X_{(1)}=\frac{n}{\theta}exp(-\frac{nx}{\theta}) \sim \text{Exp}(\frac{\theta}{n})$, and $E(X_{(1)})=\frac{\theta}{n}$. Thus $nX_{(1)}$ is unbiased for $\theta$.

Also $S_n=\sum_{i=1}^nX_i$ is complete sufficient statistic for $\theta$ and $S_n \sim \Gamma(n,\frac{1}{\theta})$. So by Lehmann–Scheffé, $W_n$ is the UMVUE for $\theta$.

But how can I find the closed form expression for $W_n$?

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    $\begingroup$ See stats.stackexchange.com/q/153016/119261 $\endgroup$ Commented Dec 19, 2021 at 5:10
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    $\begingroup$ The trick is in proving that $(X_1,\ldots,X_n)/S_n$ is ancillary and independent of $S_n$. $\endgroup$
    – Xi'an
    Commented Dec 19, 2021 at 11:38

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