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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.

Srivatsan
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Fan Zhang
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1 Answers1

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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 $(0,1)$.

Did
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  • I am not quite clear about what $\mathrm d\mathbf s$ mean. – Fan Zhang Oct 27 '11 at 14:34
  • The notation $\mathrm P(\mathbf S\in\mathrm d\mathbf s)$ is a shorthand for $\mathrm P(S_1\in\mathrm ds_1,S_2\in\mathrm ds_2,\ldots,S_{n+1}\in\mathrm ds_{n+1})$. – Did Oct 27 '11 at 14:42
  • From the above statement, so we can use the $U_j$ to simulate the distribution of order statistics from uniform distribution? – Fan Zhang Oct 27 '11 at 15:28
  • And is it the connection between Poisson process and order statistics? – Fan Zhang Oct 27 '11 at 15:32
  • @FanZhang Yes, there is a connection with Poisson processes. See here for one way of thinking about it. – Dilip Sarwate Oct 27 '11 at 16:41
  • Fan: I do not understand your question. Obviously $(S_n)$ realizes a Poisson process (with constant intensity $1$), so the connection between Poisson processes and order statistics is exactly what my post explains. Could you reformulate your question? – Did Oct 27 '11 at 16:50
  • @DidierPiau Actuall my question comes from a question you know. I re-posted at http://mathoverflow.net/questions/78822/distribution-of-a-maximum, and Johan Wästlund gives an answer which uses $U_j$ to simulate the order statistics, I want to make sure that method is right. – Fan Zhang Oct 27 '11 at 17:02
  • Actually the present question is about continuous distributions and the other one about discrete distributions. Ex aequos simply do not occur in the continuous setting hence the present post can tell you nothing about the problem of ex aequos. // Note that I find disrespectful (to Johan Wästlund, in the present case) the tactics you use of reposting again and again very closely related questions. If you have a problem with Johan's method in his answer, ask him! Finally, I wonder why you accepted his answer if (you think) it needs some checking. – Did Oct 27 '11 at 17:37
  • @DidierPiau Thank you for your advice! I will not do this kind of things again. – Fan Zhang Oct 28 '11 at 02:45
  • Fan: Good. While we are dealing with questions of etiquette, let me mention that it is customary to accept an answer if and when it answers the question. Otherwise, one may explain why it does not. – Did Oct 30 '11 at 09:27
  • @DidierPiau I have another question that is: why $Z_i \in ds_i - s_{i-1} $ not $Z_i \in d(s_i - s_{i-1}) $ – Fan Zhang Oct 30 '11 at 16:20
  • Fan: Revised, to use an equivalent formulation. – Did Oct 30 '11 at 16:25
  • @DidierPiau Thank you, I understand. – Fan Zhang Oct 30 '11 at 17:28
  • @DidierPiau I am re-checking your approach. I wonder how you get the following equation in second step: $\mathrm P(\mathbf U\in\mathrm d\mathbf u)=\int_{s\geqslant0} \mathrm P(S_1\in s\mathrm du_1,S_2\in s\mathrm du_2,\ldots, S_n\in s\mathrm du_n,S_{n+1}\in\mathrm ds)$ – Fan Zhang Nov 01 '11 at 14:08
  • Fan: Try to compute the density of the distribution of $(U_1,U_2)$, you should see the general pattern emerge. – Did Nov 01 '11 at 14:11
  • @DidierPiau Is it possible derive the equation from: $P(u_1 < U_1 \le u_1 + d u_1, u_2 < U_2 \le u_2 + d u_2, \dots, u_{n} < U_{n} \le u_{n} + d u_{n})$, and let $du_1 \rightarrow 0, du_2 \rightarrow 0, ... , du_n \rightarrow 0$. – Fan Zhang Nov 01 '11 at 14:11
  • @DidierPiau Actually I find a bit hard to deal with the situation when a r.v divide another r.v. like $S_1/S_{n+1}$, how to get the pdf of the above one? – Fan Zhang Nov 01 '11 at 14:18
  • Fan: Try imitating this. – Did Nov 01 '11 at 16:22