How to calculate UMVUE of $\mu,\sigma$ in a two-parameter exponential distribution?

Question

$$f(x;\mu,\sigma)= 1/\sigma\times \exp((x-\mu)/\sigma)\times I(\mu<x<\infty)$$

first, I calculated the C.S.S

$$f(x_1, \ldots, x_n;\mu,\sigma)= (1/\sigma)^n\times \exp(\sum(x-\mu)/\sigma)\times I(\mu<\min(x)<\infty)$$

$$f(x_1, \ldots, x_n;\mu,\sigma)= (1/\sigma)^n\times \exp(-n\mu/\sigma)\times \exp(\sum x_i/\sigma)\times I(\mu<\min(x))$$

therefore, $\min(x), \sum x_i$ C.S.S

I want to calculate the UMVUE of $\mu, \sigma$. I can't calculate after C.S.S comes out in two dimensions. I would appreciate it if you give me a hint.

Answer 1

Proceeding as suggested in the comments establish the following.

$U = min(X)$ has density $f(x;\mu,\frac{\sigma}{n})$

$V = \sum_{i=1}^n X_{i}$ is $Gamma(n,\sigma)$ shifted by $\mu$

Both are complete.

$E[U] = \mu + \frac{\sigma}{n}$

$E[V] = n(\mu + \sigma)$

Switch things around to get the unbiased estimators. Using Lehmann-Scheffe these estimators are proved to be UMVUEs.