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?


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.


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 $n>2.$

  • $\begingroup$ (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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