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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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    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?2011-10-27
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    A related question is [here](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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    I am not quite clear about what $\mathrm d\mathbf s$ mean.2011-10-27
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    The notation $\mathrm P(\mathbf S\in\mathrm d\mathbf s)$ is a shorthand for $\mathrm P(S_1\in\mathrm ds_1,S_2\in\mathrm ds_2,\ldots,S_{n+1}\in\mathrm ds_{n+1})$.2011-10-27
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    From the above statement, so we can use the $U_j$ to simulate the distribution of order statistics from uniform distribution?2011-10-27
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    And is it the connection between Poisson process and order statistics?2011-10-27
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    @FanZhang Yes, there is a connection with Poisson processes. See [here](http://math.stackexchange.com/questions/74218/relations-between-order-statistics-of-uniform-rvs-and-exponential-rvs/74296#74296) for one way of thinking about it.2011-10-27
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    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?2011-10-27
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    @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.2011-10-27
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    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.2011-10-27
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    @DidierPiau Thank you for your advice! I will not do this kind of things again.2011-10-28
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    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.2011-10-30
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    @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}) $2011-10-30
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    Fan: Revised, to use an equivalent formulation.2011-10-30
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    @DidierPiau Thank you, I understand.2011-10-30
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    @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)$2011-11-01
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    Fan: Try to compute the density of the distribution of $(U_1,U_2)$, you should see the general pattern emerge.2011-11-01
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    @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 $du_1 \rightarrow 0, du_2 \rightarrow 0, ... , du_n \rightarrow 0$.2011-11-01
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    @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?2011-11-01
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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