I know that if $X_k$ ~ $Unif(0,1)$ and is order statistics, then $E[X_k] = \frac{k}{n+1}$.
What's $E[X_k]$ for when $X_k$ ~ $Unif(a,b)$?
I think it's $a + \frac{k}{n+1}(b-a).$ Can someone confirm?
2026-03-26 04:50:24.1774500624
Expected value of general uniform order statistics
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