# Minimal sufficiency of smallest order statistic

I am attempting to show that the smallest order statistic T is minimally sufficient for the mean of a distribution when the variance is known. In particular, iid random variables $X_1,\ldots,X_n$ have a pdf that is given by $f(x;θ)=(1/\sigma)e^{(-(x-\theta)/\sigma)}$.

I know that the two necessary conditions for minimal sufficiency are: i.) sufficiency; and ii.) that the statistic cannot be further reduced to another sufficient statistic. After having shown that T is sufficient, I failed to prove the second condition.

Since the statistic is clearly minimally sufficient (I wouldn't have been told to prove it, otherwise), it must be true that $T$ cannot be reduced to another arbitrarily defined, sufficient statistic $T'$. In other words, $T$ must be a defined function of $T'$. However, if $T' = \sum_i X_i$, for example, i.e. $T'$ is sufficient, then $T$ would only be a defined function of $T'$ if $n=1$. Hence, the second condition of minimal sufficiency would not hold. I must be missing something...

• Given that the dimension of your proposed minimally sufficient statistic is (fill in the blank) ___, can it be REDUCED to another sufficient statistic, i.e., to one with smaller dimension? May 28, 2018 at 0:26
• Clearly I misunderstood the definition of reduction. By your logic, it then follow that every one-dimensional, sufficient statistic is minimally sufficient, correct? May 28, 2018 at 19:04
• That's correct. It's all about the dimensionality of the sufficient statistic; once you've found a minimal sufficient statistic, any 1-1 transform of it is also a minimal sufficient statistic. May 28, 2018 at 19:18
• @jbowman: I don't think that is true! (while it might be a good initial guess). An example: $X \sim \cal{U}(-\theta, \theta)$. Then clearly $X$ is sufficient for $\theta$, but so is $\mid X \mid$, so $X$ is not minimally sufficient. You could construct similar examples based on the $\cal{N}(0,\sigma^2)$ family. Jul 12, 2018 at 1:24
• @kjetilbhalvorsen So, you're saying that $|X|$ is a reduction of $X$ because there are two $X$'s for every $|X|$? Jul 12, 2018 at 1:42

Some hints: The question about sufficiency is asked and answered before, for instance at UMVUE of two-parameter exponential family distribution For minimal sufficiency, use the same method as at Finding a minimal sufficient statistic and proving that it is incomplete, that is, look at the ratio of the likelihood functions for two different samples, and show it is independent of the parameter iff the two samples have the same value of the statistic $$T$$.