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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 $(X_1, X_2, \ldots, X_n)$.)

whuber
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cyzyk
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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. – Corvus Mar 05 '13 at 17:19
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    @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 05 '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... – Corvus Mar 05 '13 at 18:12
  • @Corone The equality is as obvious as they get: the two sums are over the same collection of values; the second merely has been put in order. – whuber Mar 05 '13 at 18:15
  • @whuber Ouch! Clearly I'm thick but I still don't get it :-(. If $X_i=(4,3,2,1)$ then $X_{i:4} = (1,1,1,1)$? – Corvus Mar 05 '13 at 18:24
  • @Corone Oh, it's just a matter of notation: $X_{i:n}$ means the $i$th smallest of $n$ values, so that $X_{1:n}\le X_{2:n}\le\ldots\le X_{n:n}$. Whence if $X=(4,3,2,1)$ then $\sum X_i$ = $4+3+2+1$ while $\sum X_{i:4}$ = $1+2+3+4$. – whuber Mar 05 '13 at 19:04
  • Ah - I had read it as between "i and n". Makes sense now. I'll add that to the question incase anyone else is unfamiliar (I like to hope it's not just me) – Corvus Mar 05 '13 at 19:13
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    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 05 '13 at 20:28
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    $X_i \sim \mathcal{NE}(\mu,\sigma)$. Basically I need to show that $\frac{2}{\sigma} \sum_{i=1}^n (X_i - X_{1:n})$ is a random variable from $\chi^2$ dist. with 2(n-1) degrees of freedom. – cyzyk Mar 05 '13 at 21:06
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    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 05 '13 at 22:24
  • What is the distribution, and its CDF/PDF, symbolized by $\mathcal{NE}(\mu,\sigma)$? (cc @Glen_b) – Alecos Papadopoulos Nov 05 '14 at 02:01
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    @Alecos I wonder if OP means some form of shifted negative exponential. In any case, we should be given the explicit distributional form. – Glen_b Nov 05 '14 at 02:51
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    Does this answer your question? Suppose $Y_1, \dots, Y_n \overset{\text{iid}}{\sim} \text{Exp}(1)$. Show $\sum_{i=1}^{n}(Y_i - Y_{(1)}) \sim \text{Gamma}(n-1, 1)$ – StubbornAtom Jul 08 '20 at 08:16

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