$X\sim\frac{1}{b}\exp\{-\frac{1}{b}(x-a)\},x>a$. Find UMVUE of $\frac{a}{b}$

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 $$a/b$$.

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.
• 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 $\frac{X_{(1)}}{\sum_{i=1}^n (X_i-X_{(1)})}$. Jan 27, 2021 at 13:19