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Suppose $\{X_i\}_{i=1}^n\overset{i.i.d}{\sim}X,$ and $X$ has density $f(x)=\frac{1}{b}\exp\{-\frac{1}{b}(x-a)\},x>a$. What is the UMVUE of $\frac{a}{b}$?

Here is what I've done so far.

It can be shown that $(X_{(1)},\sum[X_i-X_{(1)}])$ is sufficient and complete statistic. Also, $X_{(1)}$ is independent with $\sum[X_i-X_{(1)}])$. We also have the UMVUE of $b$ and $a$ are $\frac{1}{n-1}\sum[X_i-X_{(1)}]$ and $X_{(1)} - \frac{1}{n(n-1)}\sum[X_i-X_{(1)}]$, respectively. Also, we know that $\mathbb{E}\left(\frac{n-2}{4}\frac{1}{\sum[X_i-X_{(1)}]}\right)=\frac{1}{b}$. However, I don't know how to get the UMVUE of $a/b$.

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Since nobody answered, I'll try to answer my own question. Please let me know if anything is not correct.

Denote $S=\frac{1}{n-1}\sum_{i=1}^n[X_i-X_{(1)}]$. We have $$S\sim \frac{b}{n-1}\text{Gamma}(n-2,1),$$ and $$\frac{1}{S}\sim \frac{n-1}{b}\text{inv-Gamma}(n-2,1)$$ These implies $$\mathbb{E}\frac{1}{S}=\frac{n-1}{b(n-3)}.$$

Also, $$\mathbb{E}\left(X_{(1)}-c_1S\right)=a,$$ thus we have $$\mathbb{E}\left[\left(X_{(1)}-c_1S\right)\frac{1}{S}\right]=\mathbb{E}\left[\frac{X_{(1)}}{S}-\frac{c_2}{n}\right].$$ Since $X_{(1)}$ is independent with $S$, we have $$\mathbb{E}X_{(1)}\frac{1}{S}=\mathbb{E}X_{(1)}\mathbb{E}\frac{1}{S}=c_3+\frac{a}{b}c_4$$ Since $X_{(1)}\text{ and }S$ complete and sufficient, with some additional work, we can get UMVUE of $\frac{a}{b}$. $c_1, c_2, c_3, c_4$ are some constant.

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    $\begingroup$ This works just fine. Using $\frac2b\sum_{i=1}^n (X_i-X_{(1)})\sim \chi^2_{2(n-1)}$, I get a slightly different unbiased estimator of $\frac1b$. But nonetheless the UMVUE is a linear function of the ratio $\frac{X_{(1)}}{\sum_{i=1}^n (X_i-X_{(1)})}$. $\endgroup$ Jan 27, 2021 at 13:19

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