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I saw a paper which says that:

Let $Z_i$ be i.i.d. exponential random variables with mean $1$, and let $S_n = Z_1 + \dots + Z_n$ for all $n$. For a fixed $n$, let $U_j = S_j/S_{n+1}$, then $(U_1,\dots,U_n)$ has the same distribution as the order statistics of a sample of size $n$ from the uniform distribution on $[0,1]$.

So my question is how to prove this.

  • 0
    A related question is [$h$ere](http://math.stackexchange.com/q/74218/15941). Perhaps the two questions might be merged?2011-10-27

1 Answers 1

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The short answer is: by coming back to the definitions. This is done in two steps.

First step: distribution of $\mathbf S=(S_1,S_2,\ldots,S_{n+1})$

For $\mathbf s=(s_1,s_2,\ldots,s_{n+1})$ in $\mathbb R^{n+1}$, $ \mathrm P(\mathbf S\in\mathrm d\mathbf s)=\mathrm P(Z_1\in\mathrm ds_1,s_1+Z_2\in\mathrm ds_2,\ldots,s_n+Z_{n+1}\in\mathrm ds_{n+1}). $ The independence of the random variables $(Z_i)$ implies that $ \mathrm P(\mathbf S\in\mathrm d\mathbf s)=\prod\limits_{i=1}^{n+1} \mathrm e^{-(s_i-s_{i-1})}[s_i\geqslant s_{i-1}]\mathrm ds_i, $ where $s_0=0$, hence $ \mathrm P(\mathbf S\in\mathrm d\mathbf s)=\mathrm e^{-s_{n+1}}[0\leqslant s_1\leqslant s_2\leqslant\cdots\leqslant s_{n+1}]\mathrm ds_1\mathrm ds_2\cdots\mathrm ds_{n+1}. $ Second step: distribution of $\mathbf U=(U_1,U_2,\ldots,U_n)$

For $\mathbf u=(u_1,u_2,\ldots,u_{n})$ in $\mathbb R^n$, $ \mathrm P(\mathbf U\in\mathrm d\mathbf u)=\int_{s\geqslant0} \mathrm P(S_1\in s\mathrm du_1,S_2\in s\mathrm du_2,\ldots, S_n\in s\mathrm du_n,S_{n+1}\in\mathrm ds), $ hence $ \mathrm P(\mathbf U\in\mathrm d\mathbf u)=\int_{s\geqslant0}\mathrm e^{-s}[0\leqslant su_1\leqslant su_2\leqslant\cdots\leqslant su_n\leqslant s]s\mathrm du_1s\mathrm du_2\cdots s\mathrm du_n\mathrm ds, $ which is $ \mathrm P(\mathbf U\in\mathrm d\mathbf u)=[0\leqslant u_1\leqslant u_2\leqslant\cdots\leqslant u_n\leqslant1]\mathrm du_1\mathrm du_2\cdots \mathrm du_n\int_{s\geqslant0} s^n\mathrm e^{-s}\mathrm ds. $ The last integral does not depend on $\mathbf u$ (and its value is $n!$) hence $\mathbf U$ is uniform on the simplex $ \{\mathbf u\in\mathbb R^n\mid0\leqslant u_1\leqslant u_2\leqslant\cdots\leqslant u_n\leqslant1\}. $ This is the distribution of the order statistics of an i.i.d. sample of size $n$ from the uniform distribution on the interval $(0,1)$.

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    Fan: Try imitating [this](http://math.stackexchange.com/questions/30938/given-pdf-of-i-and-r-both-i-and-r-are-independent-rvs-how-to-find-cdf-of-w/30966#30966).2011-11-01