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The minimum sufficient statistics for $f(x) = \frac12 exp(-|x-\theta|)$ for $-\infty < \theta < +\infty$ is $ T(X) = \{X_{(1)},X_{(n)} \}$. I want to show that the above is complete.

$f(x) = \frac12 exp(-|x-\theta|)$ is location family therefore the range is an ancillary statistic which is free of $\theta$

Let $h((T(x))$ be $X_{(n)}-X_{(n)}$

$0 = E_\theta [ g(h(T(X))] = \int_{-\infty}^{+\infty} g(h(t)) f(h(t))dt$ $.

$f(h(t))$ is now $\frac12 exp(-|(t_{(n)}-\theta) - (t_{(1)}-\theta)|)$ which reduces to

$\frac12 exp(-|(t_{(n)}- t_{(1)}|)$.

When$ \int_{-\infty}^{+\infty} g(h(t)) \frac12 exp(-|(t_{(n)}- t_{(1)}|)dt = 0$, then $g(h(t)) = 0$ $\forall h(t)$

Now, I want to claim that since $h(T)$ is complete and there is a 1-1 function between $h(T)$ and $T$ therefore $T$ is also complete. Is this approach correct?

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  • $\begingroup$ The minimal sufficient statistic is not complete in this case. Consider the vector of the pairwise differences, with fixed expectation.. $\endgroup$ – Xi'an Sep 5 '20 at 13:16
  • $\begingroup$ @Xian, just to follow your suggestion. You would want me to apply a function on T s.t I have something like $( X_(1)-X_(2),,,X(n-1) - X(n)) as my new t correct? $\endgroup$ – user1916067 Sep 5 '20 at 16:16
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    $\begingroup$ Minimal sufficient statistic is the full set of order statistics $(X_{(1)},X_{(2)},\ldots,X_{(n)})$. $\endgroup$ – StubbornAtom Sep 5 '20 at 19:34
  • $\begingroup$ @StubbornAtom: the set $\{X_1,\ldots,X_n\}$ is equally minimal sufficient. $\endgroup$ – Xi'an Sep 6 '20 at 8:03
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Since the double exponential distribution is not part of an exponential family, there cannot exist a sufficient distribution with fixed (in $n$) dimension. The minimal sufficient statistic in this case is $(X_{(1)},\ldots,X_{(n)})$ or, equivalently, $\{X_1,\ldots,X_n\}$. Since $\mathbb{E}[X_i-X_j]=0$, this statistic cannot be complete.

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