Completeness of order statistics for normal setup

Suppose we have iid $X_1,\cdots, X_n\sim N(\mu,\sigma^2)$ where $\mu$ is unknown.

Let $X_{(1)}, X_{(2)},\cdots,X_{(n)}$be the order statistics.

Is the order statistics $\textbf{complete sufficient}$ when

$A)$ $\sigma$ is known ?

$B)$ $\sigma$ is unknown ?

• @whuber I corrected my tag.. Can you please help? At least give a direction of thinking why true or why not true.. – Qwerty Aug 29 '17 at 20:58
• Do you know what the minimal sufficient statistics are for these two cases? That would be a good place to begin. – whuber Aug 29 '17 at 21:10
• @whuber For first case $\bar{X}$ , and for the second case, $(\bar{X}, \sum(X_i-\bar{X})^2$ . but my concern is about the order statistics.. I know these are sufficient , but not minimal sufficient (for the second case at least)... But i have no clue what to do about completeness – Qwerty Aug 29 '17 at 21:17
• For completeness, see en.wikipedia.org/wiki/…. For both situations, notice that you can compute the mean and the variance from the order statistics. – whuber Aug 29 '17 at 21:20
• @whuber So you are saying the order statistics are $\textbf{complete}$ ? – Qwerty Aug 30 '17 at 17:10

I will give a more extended hint, which should cover cases A) and B) simultaneously. To show that a sufficient statistic is not complete, we need to find a function of it with expectation identically zero. To this end let the usual sufficient and complete statistic be $$(\bar{X}_n, S_n^2) = \left(\frac1n \sum_1^n X_{(i)},\frac1{n-1}\sum_1^n (X_{(i)}-\bar{X}_n)^2 \right)$$ which also shows that this sufficient statistic can we written as a function of the order statistics.
Now consider the centered order statistics, given by $$( U_1=X_{(1)}-\bar{X}_n, \dotsc, U_n=X_{(n)}-\bar{X}_n )$$ Can you show that the expectation of $\sum_1^n U_i$ is zero? And what is then the conclusion about the completeness or not of the normal order statistics?