# Checking for the completeness for the M.S of $f(x) = \frac12 exp(-|x-\theta|)$

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? • The minimal sufficient statistic is not complete in this case. Consider the vector of the pairwise differences, with fixed expectation.. – Xi'an Sep 5 '20 at 13:16 • @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? – user1916067 Sep 5 '20 at 16:16
• Minimal sufficient statistic is the full set of order statistics $(X_{(1)},X_{(2)},\ldots,X_{(n)})$. – StubbornAtom Sep 5 '20 at 19:34
• @StubbornAtom: the set $\{X_1,\ldots,X_n\}$ is equally minimal sufficient. – Xi'an Sep 6 '20 at 8:03

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.