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How to find the distribution of $$\sum_{i=1}^n (X_i - X_{1:n}),$$ where $X_i$ are i.i.d. random variables and $X_{1:n} = \min(X_1,X_2,...,X_n)$?

I need to find the distribution in a particular case, but I would be grateful for a general proof.

(As a matter of notation, $X_{i:n}$ generally means the $i$th smallest of the $n$ values $(X_1, X_2, \ldots, X_n)$.)

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Can you make any assumptions about the distribution of $X_i$? Gaussian? Unimodal? Discrete? Support everywhere? Distributions of min/max are generally pretty annoying to work with, so anything might help. – Corone Mar 5 '13 at 17:19
@Corone Despite that there potentially is a simple elegant expression for this in terms of the common distribution $F$, because $\sum(X_i - X_{1:n}) = \sum(X_{i:n}-X_{1:n})$ and the joint distribution of the order statistics $X_{i:n}$ has a closed expression in terms of $F$. – whuber Mar 5 '13 at 18:06
@whuber sounds fun - I eagerly await your answer! And be sure to explain that equality - I may be being dense, but it is not obvious to me... – Corone Mar 5 '13 at 18:12
What bothers me here is that the notation $X_{i:n}$ is introduced but never used. The desired sum may be written just as $\sum_{i=1}^n(X_i-X_{(1)})$, and it is equal to $\sum_{i=1}^n(X_{(i)}-X_{(1)})$. Are there any typos on this question? – Zen Mar 5 '13 at 20:28
The fact that it's for exponential random variables makes life much easier both when dealing with order statistics and when subtracting the minimum ... but trying to deal with such things the general case isn't so easy. You should put the additional information in your question. Is this for some course of study? – Glen_b Mar 5 '13 at 22:24

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