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Let $ X_1, ... , X_n $ be i.i.d random variables with pdf given by

$$f(x;\theta) = \exp(-(x-\theta))I_{(\theta, \infty)}(x)$$

It is asked to find a sufficient statistics for $ \theta $ and to verify if it is complete too. Since

$$L(\theta;x)=\exp(-\sum x_i) \exp(n\theta) I_{(\theta, \infty)}(x_{(1)}),$$

by the factorization theorem, $X_{(1)} $ is sufficient. But I could not prove (or disprove) that it is complete, using the definition. Is there another way of doing it? Or how we show it, by the definition?

Thanks!

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    $\begingroup$ You've posted quite a few self-study questions already - please add the tag each time. What definition of completeness do you have? $\endgroup$ – Glen_b Sep 22 '14 at 0:25
  • $\begingroup$ In this case, with the definition here, it seems to be reasonably straightforward. $\endgroup$ – Glen_b Sep 22 '14 at 0:33
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By the definition the pag. 42 the book Theory of Point Estimation - Lehmann and Casella, say T is said that it is complete if $E_{\theta}[f(T)]=0$ for all $\theta \in \Omega$ implies $f(t)=0\, (a.e \,\mathcal{P})$.

Well, $P(X_{(1)} \leq x) = P(min\{X_1,\ldots,X_n \} \leq x)=1 - P(min\{X_1,\ldots,X_n \} > x) = 1 - P(X_1>x) \ldots P(X_n>x) = 1 - [P(X_1>x)]^n = 1 - [1 - P(X_1<x)]^n = 1 - [1 - F_{X_1}(x)]^n.$

Then $F_{X_1}(x) = 1 - [1 - F_{X_1}(x)]^n \Rightarrow f_{X_{(1)}}=n[1 - F_{X_1}(x)]^{n-1}f_{X_1}(x).$

$E_{\theta}[g(T)]=0 \Rightarrow \int_{\theta}^{\infty}g(t)exp(-(t-\theta))dt = 0$ then $\frac{\partial }{\partial \theta}\int_{\theta}^{\infty}g(t)exp(-(t-\theta))dt = 0$ then $g(\theta)=0$ for all $\theta \in \Omega=R^{+}$.

In conclusion $T(\textbf{X})=X_{(1)}\, (a.e \,\mathcal{P})$ is a sufficient statistics and complete.

I hope to have answered your question.

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  • $\begingroup$ But you didn't use $f_{X_{(1)}}$ anywhere why did u calculate it ? $\endgroup$ – Ronald Oct 27 '18 at 8:25

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