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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.

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  • $\begingroup$ What does "$\ldots$ let $S_n=Z_1+…+Z_n$ for all $n$. For fixed $n$ let $S_n=Z_1+…+Z_n$ for all $n$." mean? In the last sentence, is $n$ fixed or not? $\endgroup$ – Dilip Sarwate Oct 27 '11 at 13:38
  • $\begingroup$ A related question is here. Perhaps the two questions might be merged? $\endgroup$ – Dilip Sarwate Oct 27 '11 at 13:40
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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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  • $\begingroup$ I am not quite clear about what $\mathrm d\mathbf s$ mean. $\endgroup$ – Fan Zhang Oct 27 '11 at 14:34
  • $\begingroup$ The notation $\mathrm P(\mathbf S\in\mathrm d\mathbf s)$ is a shorthand for $\mathrm P(S_1\in\mathrm ds_1,S_2\in\mathrm ds_2,\ldots,S_{n+1}\in\mathrm ds_{n+1})$. $\endgroup$ – Did Oct 27 '11 at 14:42
  • $\begingroup$ From the above statement, so we can use the $U_j$ to simulate the distribution of order statistics from uniform distribution? $\endgroup$ – Fan Zhang Oct 27 '11 at 15:28
  • $\begingroup$ And is it the connection between Poisson process and order statistics? $\endgroup$ – Fan Zhang Oct 27 '11 at 15:32
  • $\begingroup$ @FanZhang Yes, there is a connection with Poisson processes. See here for one way of thinking about it. $\endgroup$ – Dilip Sarwate Oct 27 '11 at 16:41

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