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Let $F_{k:n}(x)$ denote the distribution function of $k$th order statistic, i.e. $k$th lowest of $n$ i.i.d. draws from a smooth distribution $F$ with support $[0,\bar{x}]$. Then $F_{k+1}(x)-F_k(x) \leq 0$, because of stochastic dominance of higher order statistics. But what is the sign of $[F_{k+1:n}(x)-F_{k:n}(x)]-[F_{k:n}(x)-F_{k-1:n}(x)]$?

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Depends on the relationship between $ x $, $ k $, and $ n $ and the distribution. For example consider whether $ x $ is close to $ k/n $, if the distribution is uniform.

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  • $\begingroup$ The answer to my question turned out way less useful than I hoped, it indeed depends on the details, so the answer you provided is correct. An example: uniform[0,1] distribution, n=3, k=2, gives $-x^2 (4x-3)$, which is obviously positive for $x<\frac{3}{4}$ and negative otherwise. $\endgroup$ – TomH Apr 26 '14 at 8:33

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