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For independent variables $X_1$,$X_2$, $X_3$, $...$, $X_n$ with $n>2$, all uniformly distributed in the interval $[0,1]$ and if $X_1 < X_2 < ... < X_n$ I am trying to find the pdf of $Y_n$ = $\frac{X_n}{(X_1 X_2 \dots X_n)^\frac{1}{n-1}}$ How should I go about this?

I used $logY$ to break down the fraction into a sum of $log X_k$s but kind of stuck from here. Playing around with the following facts but feel like there could be a better way..

The log of a uniform random variable on [0,1] follows an exponential distribution with mean 1, and the sum of $n$ exponential random variables with mean 1 follows $Gamma(1, n)$

Any suggestions on how to tackle this?

user97019
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  • If $X_n$ is indeed present in the denominator, then nominator and denominator of $Y_n$ are not independent, and then I don't think you can go very far. Are you sure that the product of $X$'s does not stop at $X_{n-1}$? Because it looks like a geometric mean. – Alecos Papadopoulos Sep 26 '13 at 12:16
  • @AlecosPapadopoulos Pretty sure. I wasn't really thinking about independence while taking the log.. maybe I need a different approach? – user97019 Sep 26 '13 at 13:17

1 Answers1

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To make the nominator and denominator independent, write

$$Y_n = \frac{X_n}{(X_1 X_2 \dots X_n)^\frac{1}{n-1}}= \frac{X_nX_n^{-\frac{1}{n-1}}}{(X_1 X_2 \dots X_{n-1})^\frac{1}{n-1}}= \frac{X_n^{\frac{n-2}{n-1}}}{(X_1 X_2 \dots X_{n-1})^\frac{1}{n-1}}$$

Define $Z \equiv X_n^{\frac{n-2}{n-1}}$. Then by the change-of-variable formula we obtain that $$ f_{Z}(z) = \frac {n-1}{n-2}z^{\frac {1}{n-2}} \qquad [1]$$

which is a beta distribution $B(a,b)$ with $a= \frac {n-1}{n-2}$ and $b=1$.

Now the distribution of the product, call it $Z_{\pi}$, of $n-1$ i.i.d. $U(0,1)$ random variables is (see here for the n-case)

$$ f_{Z_{\pi}}(z_{\pi}) = \frac {1}{(n-2)!}\cdot \Big(-\ln z_{\pi}\Big)^{n-2} $$

Define $G \equiv \left(Z_{\pi}\right)^{\frac {1}{n-1}}$, i.e. the geometric mean. Applying here the change of variable formula we obtain

$$f_G(g) = \frac {(n-1)^{n-1}}{(n-2)!}\Big(-g\ln g\Big)^{n-2} \qquad [2]$$

Now your variable has a ratio distribution , $Y_n = Z/G$ with both variables having support in $(0,1)$, so

$$f_{Y_n}(y_n) = \int_0^1g\frac {n-1}{n-2}(gy_n)^{\frac {1}{n-2}}\frac {(n-1)^{n-1}}{(n-2)!}\Big(-g\ln g\Big)^{n-2}dg $$

$$=(y_n)^{\frac {1}{n-2}}\frac {1}{n-2}\frac {(n-1)^{n}}{(n-2)!}\int_0^1g^{\frac {n-1 + (n-2)^2)}{n-2}}\Big(-\ln g\Big)^{n-2}dg $$

...but I am just back from the dentist, so I am leaving this final integration to you.

Alecos Papadopoulos
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