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?


  • 1
    $\begingroup$ 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 -Reinstate Monica Sep 22 '14 at 0:25
  • $\begingroup$ In this case, with the definition here, it seems to be reasonably straightforward. $\endgroup$ – Glen_b -Reinstate Monica Sep 22 '14 at 0:33

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.

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.