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Given $(X, Y) \sim \mathbb{U}([0, 1]^2)$ a uniformly distributed random vector, I have to calculate the PDF of $XY$. I've seen about product distribution, but the deal here is that $X$ and $Y$ aren't independent. I tried using a test function (ie computing $\mathbb{E}(\phi(XY))$ for any $\phi$ positive measurable) but I found myself stuck having to compute the integral of $\frac{1}{x}$ over $[0, 1]$, which is ... not right.

The only thing I know is that the PDF of $(X, Y)$ is $f_{X,Y}(x, y) = \mathbb{1}_{[0, 1]^2}(x, y)$, so I'm really stuck.

EDIT : as it turns out, $X$ and $Y$ are indeed independent. Although that doesn't really help, since I still somehow have to integrate $\frac{1}{x}$ over $[0, 1]$ ...

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  • $\begingroup$ Why do you think that $X$ and $Y$ aren't independent? $\endgroup$ Nov 23, 2016 at 23:22
  • $\begingroup$ In the wording of the exercise, it doesn't explicitly say that $X$ and $Y$ are independent, which often means they are not, and I've been losing my head over this for something like 3 hours, so I really didn't want to say anything stupid. Unless this is ... ? It does seem like $f_{X,Y} = f_X f_Y$, I'll give you that ... $\endgroup$ Nov 23, 2016 at 23:25
  • $\begingroup$ Exactly: $f_{X,Y}(x,y)=f_X(x)f_Y(y)$, so $X$ and $Y$ are independent. $\endgroup$ Nov 23, 2016 at 23:26
  • $\begingroup$ Okay, I'll edit that in. It doesn't fix my issue of having to integrate $\frac{1}{x}$ over $[0, 1]$ though ... $\endgroup$ Nov 23, 2016 at 23:26

2 Answers 2

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Start by finding the CDF of $XY$: if $0<z<1$, then $$ \mathbb{P}(XY\leq z)=\int_0^1\int_0^11_{xy\leq z}\;dydx $$ If $x\geq z$ then $\frac{z}{x}\leq 1$, while if $0\leq x<z$ then $\frac{z}{x}>1$. Therefore $$ \int_0^1\int_0^11_{xy\leq z}\;dydx=\int_0^z\int_0^1\;dydx+\int_z^1\int_0^{\frac{z}{x}}\;dydx=z+z\int_z^1\frac{dx}{x}=z(1-\log z)$$

The pdf is the derivative of the CDF, hence is $$ 1-\log z-\frac{z}{z}=-\log z$$ for $z\in (0,1)$.

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  • $\begingroup$ Alright I see what I've been doing wrong. I did not take the bounds into account properly it seems, that plus actually not trying to find the CDF but focusing too much of the PDF. Thanks a lot ! $\endgroup$ Nov 24, 2016 at 0:07
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Let $U$ and $V$ be an independent uniform random variables, and let $X = U \cdot V$. Almost surely $1>X>0$, and therefore define an almost surely positive random variable $Z = -\log X$.

Clearly $Z = -\log U - \log V$. Random variables $S = -\log U$ and $T = -\log V$ are independent, and follow a standard exponential distributions with PDFs $f_S\left(s\right) = \mathrm{e}^{-s} \mathbf{1}_{s > 0}$, and $f_T\left(t\right) = \mathrm{e}^{-t} \mathbf{1}_{t > 0}$.

The sum of independent standard exponential random variables follows gamma distribution with shape parameter $\alpha = 2$, with pdf $f_{Z}\left(z\right) = z \mathrm{e}^{-z} \mathbf{1}_{z > 0}$.

From here, it is easy to find $f_{X}\left(x\right)$: $$ f_{X}\left(x\right) = f_{Z}\left(-\log x\right) \cdot \left| \partial_x \left(- \log x\right) \right| = \left(-\log x\right) x \cdot \left| \frac{1}{x} \right| \mathbf{1}_{0<x<1} = \left(-\log x\right) \mathbf{1}_{0<x<1} $$

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  • $\begingroup$ While this does work, it is way too intricate for my liking. By this I mean, I do understand it, but there is no way I would have been able to come up with it. Thanks anyway for your answer ! $\endgroup$ Nov 24, 2016 at 0:09

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