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Since I'm interested in the simulation of probability distributions on the computer. I want to simulate the Laplace distribution. When I was looking on the laplace distribution wikipedia page (http://en.wikipedia.org/wiki/Laplace_distribution#Related_distributions) my I fell on $log(\frac{X}{Y}) \sim Laplace(0,1)$ for Uniformly distributed X and Y.

Before Using this I wanna ofcourse prove it. So I started writing $P(log(\frac{X}{Y}) \leq x)$, I divided in the cases x<0, x=0 and x>0. But i didn't get any further. Help would be appreciated.

Oh and I ofcourse googled this result, but I only found pages mentioning this, not proving it.

Krasdo
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1 Answers1

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For $x \in \mathbb R$, we have \begin{align*} P\left(\log\frac XY \le x\right) &= P\left(\frac XY \le \exp x\right)\\ &= \int_0^1 P\left(\frac Xy \le \exp x\right)\; dy\\ &= \int_0^1 P\left(X \le y\exp x\right)\;dy\\ &= \int_0^1 \min\{y\exp x, 1\}\, dy\\ \end{align*} For $x \le 0$, $y\exp x \le 1$ for all $y \in [0,1]$, hence $$ P\left(\log \frac XY\le x\right) = \int_0^1 y\exp x\, dy = \frac 12 \exp x $$ for $x \ge 0$: \begin{align*} P \left(\log \frac XY \le x \right)&= \int_0^{\exp (-x)} y\exp x\,dy +\bigl (1 - \exp(-x)\bigr)\\ &= \exp x \cdot \frac 12 \exp(-2x) + 1 -\exp(-x)\\ &= 1 - \frac 12 \exp(-x) \end{align*} Hence $$ P\left( \log\frac XY \le x\right) = \begin{cases} \frac 12 \exp x & x \le 0 \\ 1 - \frac 12 \exp(-x) & x \ge 0\end{cases} $$ So $\log\frac XY \sim \text{Laplace}(0,1)$.

martini
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