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

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.

StubbornAtom
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DONDON
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  • First show that your sufficient statistic is a complete statistic and then find unbiased estimators of $\mu$ and $\sigma$ based on this statistic as an application of Lehmann-Scheffe theorem. Find $E(\min X_i)$ and $E(\sum X_i)$ for starters. // Please take care in writing the correct pdf. – StubbornAtom Aug 15 '19 at 20:28
  • To show completeness of the equivalent sufficient statistic $(\min_{1\le i\le n}X_i,\sum (X_i-\min_{1\le i\le n} X_i))$, see https://math.stackexchange.com/questions/3505396/complete-sufficient-statistic-for-double-parameter-exponential/. – StubbornAtom Jan 12 '20 at 18:58

1 Answers1

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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.

Fitoor
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  • How can we show that the two distributions are independent? I can't get a joint probability distribution. So you can't show completeness; – DONDON Sep 07 '19 at 09:27