I know there have been a lot of questions asked on this forum relating to order statistics, so, hopefully, this is not going to be a duplicate. I am trying to understand how I should go about conducting a hypothesis test with order statistics.

Say, we have a set $\{x_{(i)}\}_{i=1}^n$ of order statistics for a sample draw from a distribution believed to be uniform $(0,1)$, i.e. $H_0:X \sim U(0,1)$. I want to test this hypothesis by examining the first $k$ order statistics. Now, it is not quite clear to me which is the best way to proceed as there seem to be several options for how to test (i.e. how to calculate p-values for) this :

$(1) \space p = P(X_{(k)} \leq x_{(k)})$

$(2) \space p = P(X_{(1)} \geq x_{(1)} , X_{(k)} \leq x_{(k)} )$

$(3)\space p = P(X_{(1)} \leq x_{(1)} , X_{(2)} \leq x_{(2)} , \ldots , X_{(k)} \leq x_{(k)})$

My thought on the above options so far: $(1)$ - ignores information contained in the first $k-1$ observations, $(3)$ - captures all information but is too complex to calculate, $(2)$ - happy (?) middle. Am I right in my thinking? Is there a better way to do it? Is there an efficient way to calculate $(3)$? Many thanks.

Added 1. I suppose there is a variant of $(2)$ above, namely

$(4) \space p = P(X_{(1)} \leq x_{(1)} , X_{(k)} \leq x_{(k)} )$

This now confuses me even further: how to interpret the results of $(2) $ and $(4)$? Which one of them is more appropriate (powerful ?) for testing the hypothesis?


One thing that comes to mind is Kolmogorov–Smirnov, suitably modified. The empirical distribution function has the value $j/n$ on the interval from $x_{(j-1)}$ to $x_{(j)}$. You have its values only up to $x_{(k)}$. What the distribution of the maximum-discrepancy statistic is when the data are truncated like that is something I haven't thought about; one would need to deal with that.

But I wouldn't be too surprised if more powerful tests can be done.

  • $\begingroup$ Thank you for your input. Any thoughts on the three options presented in the OP? Thanks. $\endgroup$ – Confounded Jul 16 '15 at 18:31
  • 1
    $\begingroup$ Kolmogorov–Smirnov uses all of the available order statistics, not just the first and last ones. $\endgroup$ – Michael Hardy Jul 16 '15 at 18:33

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