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 $[F\_{k+1:n}(x)-F\_{k:n}(x)]-[F\_{k:n}(x)-F\_{k-1:n}(x)]$?



A question about the distributions of order statistics
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. 


