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For a sample of i.i.d. random variables $X_1,X_2,\cdots,X_n$, the expectation of the sum of all order statistics ($X_{(1)}+X_{(2)}+\cdots+X_{(n)}$) are igual to $nE[X_1]$? Is there a general form for the distribution of $X_{(1)}+X_{(2)}+\cdots+X_{(n)}$?

I was thinking that if I have the sum of all order statistics is equivalent to have the original distribution of $X$ evaluated in $nx$, but I have no idea if it's correct or if there is some theorem abuot that result.

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    $\begingroup$ $X_{(1)}+X_{(2)}+\cdots+X_{(n)}$ is just a reordering of $X_{1}+X_{2}+\cdots+X_{n}$ $\endgroup$ – Daniel Nov 28 '16 at 14:40
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Since the order statistic of a sequence of random variables $X_1,\ldots,X_n$ is just a rearrangement of the random variables, the equation $\sum \limits_{i=1}^n X_{(i)} = \sum \limits_{i=1}^n X_{i}$ holds in all generality.

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