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The random variables $X_1,\ldots,X_n$ are i.i.d and have the distribution $f(x)=(a+1) x^a$ on $[0,1]$.

I want to find a formula for $$F(t)=\operatorname{Pr}[X_1\cdots X_n \leq t]= {(a+1)^n}\int_{x_1,\ldots,x_n \in [0,1] \;\land \; x_n\leq \frac{t}{x_1\cdots x_{n-1}}} x_1^a\cdots x_n^a \ dx_1\cdots dx_n$$

where $t \in [0,1]$. This integral does not seem to split easily to an iterative integral because of the interleaved dependence between the variables. Any trick here?

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Perhaps the easiest way of doing this is to show that $Y = \log X_i$ is an exponential random variable, or rather it is the negative of an exponential random variable. One way to get this is to calculate the density given by $$ \begin{align*} f_{Y}(y) &= (a+1)\left( e^y \right)^a \left| \frac{d e^{y}}{d y} \right| \\ &= (a+1) e^{(a+1)y}, \end{align*} $$ and noting that $y \leq 0$. You can now use the fact that the sum of $n$ exponential random variables with common rate parameter $\lambda$, is a Gamma$(n, \lambda^{-1})$ random variable (using the shape/scale parameterisation). In particular let $$ S = \sum_{i=1}^{n} \log X_i, \qquad -S \sim \mbox{Gamma}(n,1/(a+1)) $$ and so in particular we get $$ \begin{align*} \mathbb{P}\left[ \prod_{i=1}^n X_i \leq z \right] &= \mathbb{P}\left[\sum \log X_i \leq \log z \right] \\ &= 1 - \frac{1}{\Gamma(n)}\gamma\left(n,-\frac{\log z}{a + 1}\right), \end{align*} $$ where $\gamma$ is the usual lower incomplete Gamma function.

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Here's something that works when $n=2.$ \begin{align} u & = x_1 x_2 \\ v & = x_1/x_2 \\[10pt] \text{So } x_1 & = \sqrt{uv} \\ \text{and } x_2 & = \sqrt{u/v \, } \\[10pt] \text{and } dx_1\,dx_2 & = \left| \dfrac{\partial(x_1,x_2) }{\partial(u,v)} \right| = v+1. \end{align} $$ \iint\limits_{\text{region}} x_1^a x_2^a \, dx_1 \, dx_2 = \int_0^1 \left( \int_u^{1/u} u^a (v+1) \, dv \right) \, du = \text{etc.} $$ Next I would try to figure out whether that can be adapted to $n>2.$

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  • $\begingroup$ (1) There is something wrong here: $\int_0^1 f(x)\dx\ne 1$. (2) Have you looked at the distribution of the logarithm of the $X_i$? With a view to using the fact that the log of a product is the sum of the logs, of course. $\endgroup$ – kimchi lover Aug 28 '17 at 21:34
  • $\begingroup$ @kimchilover : That's not something wrong. Look closely. I discarded the normalizing constant because it doesn't contribute to finding the integral. After you've found the integral, you bring the normalizing constant back in. $\endgroup$ – Michael Hardy Aug 28 '17 at 22:26
  • $\begingroup$ Whoops: I misdirected the comment. I meant it to go to the OP. $\endgroup$ – kimchi lover Aug 28 '17 at 22:29
  • $\begingroup$ @kimchilover : I see. What is needed is $\displaystyle \int_0^1(a+1)x^a \, dx. \qquad $ $\endgroup$ – Michael Hardy Aug 28 '17 at 22:42
  • $\begingroup$ Oh yes, thank you for pointing this out. Luckily it does not make a big difference. I edited. $\endgroup$ – Emolga Aug 29 '17 at 3:40

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