# Finding UMVUE for exponential sample [duplicate]

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 $$W_n$$?

• Commented Dec 19, 2021 at 5:10
• The trick is in proving that $(X_1,\ldots,X_n)/S_n$ is ancillary and independent of $S_n$. Commented Dec 19, 2021 at 11:38