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I stumbled on a curious problem:

Choose $n \in \mathbb{Z}_{\geqslant 2}$, and consider independent identically distribution random variables, uniformly distributed on the unit interval. What is the distribution of $$ W_n = \log\left(\frac{\max(U_1,U_2,\ldots,U_n)^n}{U_1 U_2 \cdots U_n} \right) $$

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You wrote the question and answered yourself at the same time!? –  user17762 May 22 '12 at 20:39
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@Marvis Yes, I was trying out the new feature of SE site. I just could not resist sharing the demonstration. –  Sasha May 22 '12 at 20:40
    
Ok. Thanks for making me aware of it. –  user17762 May 22 '12 at 20:44
    
What is the source of the problem? –  Jonas Meyer May 22 '12 at 21:48
    
@JonasMeyer Actually I am not sure of exact source. I arrived at it en route to checking this statement connecting Pareto random variables to $\chi^2$ distribution. –  Sasha May 22 '12 at 22:26
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1 Answer 1

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Since the product is symmetric, $U_1 U_2 \cdots U_n = U_{1:n} U_{2:n} \cdots U_{n:n}$, where $U_{k:n}$ denotes $k$-th order statistics in a sample of size $n$. We use this to express $W_n$ as a function of order statistics: $$ W_n = \log\left(\frac{U_{n:n}^n}{ U_{1:n} U_{2:n} \cdots U_{n-1:n} U_{n:n} } \right) = \log\left(\frac{U_{n:n}^{n-1}}{ U_{1:n} U_{2:n} \cdots U_{n-1:n} } \right) $$ Recall that order statitics vector $(U_{1:n}, U_{2:n}, \ldots, U_{n,n})$ is equal in distribution to: $$ (U_{1:n}, U_{2:n}, \ldots, U_{n,n}) \stackrel{d}{=} \frac{( X_1, X_1+X_2, \ldots, X_1+X_2+\cdots+X_n )}{X_1 + X_2 + \cdots + X_n+X_{n+1}} $$ where $X_k$, for $k=1,2,\ldots,n+1$, are independent identically distributed exponential random variables with unit mean, $\mathcal{E}(1)$. Using this: $$ W_n = \log \frac{(X_1+X_2 + \cdots + X_n)^{n-1}}{X_1 (X_1+X_2) \cdots (X_1+X_2+\cdots + X_{n-1}) } \stackrel{d}{=} \log\frac{1}{V_{1:n-1} V_{2:n-1} \cdots V_{n-1:n-1}} $$ where $V_{k:n-1} = \frac{X_1+X_2+\cdots+X_k}{X_1+X_2+\cdots+X_{n-1}+X_{n}}$ are order statistics of size $n-1$ sample of uniform random variables.

Since the product is symmetric we have, once again, $V_{1:n-1} V_{2:n-1} \cdots V_{n-1:n-1} = V_1 V_2 \cdots V_{n-1}$. Now $$ W_n = \log\frac{1}{V_{1} V_{2} \cdots V_{n-1}} = \sum_{k=1}^{n-1} \left(-\log(V_k) \right) $$ Since $-\log(V_k)$ is equal in distribution to $\mathcal{E}(1)$, random variable $W_n$ is equal in distribution to the sum of $n-1$ independent identically distributed $\mathcal{E}(1)$ random variables which, in turn, equals in distribution to gamma random variable $\Gamma(n-1,1)$, also known as $\chi^2_{2n-2}$ random variable.

Thus we established that $W_n$ follows a chi-squared distribution with $2(n-1)$ degrees of freedom: $$ \log\left(\frac{\max(U_1,U_2,\ldots,U_n)^n}{U_1 U_2 \cdots U_n} \right) \stackrel{d}{=} \chi^2_{n-1} $$

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