3
$\begingroup$

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 ?

Improve this question
$\endgroup$
5
  • $\begingroup$ @whuber I corrected my tag.. Can you please help? At least give a direction of thinking why true or why not true.. $\endgroup$ – Qwerty Aug 29 '17 at 20:58
  • 1
    $\begingroup$ Do you know what the minimal sufficient statistics are for these two cases? That would be a good place to begin. $\endgroup$ – whuber♦ Aug 29 '17 at 21:10
  • $\begingroup$ @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 $\endgroup$ – Qwerty Aug 29 '17 at 21:17
  • 1
    $\begingroup$ For completeness, see en.wikipedia.org/wiki/…. For both situations, notice that you can compute the mean and the variance from the order statistics. $\endgroup$ – whuber♦ Aug 29 '17 at 21:20
  • 1
    $\begingroup$ @whuber So you are saying the order statistics are $\textbf{complete}$ ? $\endgroup$ – Qwerty Aug 30 '17 at 17:10
1
$\begingroup$

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?

Improve this answer
$\endgroup$

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.