I want to compute the density function of the following process: I sample a value $z$ following the Laplace distribution with mean $0$ and deviation $\sigma$ (Lap($0,\sigma$)). Therefore, I have the random variable $Z\sim Lap(0,\sigma)$. Independently, I sample $y_1,y_2$ following a uniform distribution U$[0,1]$, and I compute $x_1=z\cdot y_1$, $x_2=z\cdot y_2$ so I have the random variables $X_1=Z*Y_1$ and $X_2=Z*Y_2$ where $Z\sim Lap(0,\sigma)$ and $Y_1,Y_2\sim U[0,1]$. My goal is to obtain explicitly $f_{X_1,X_2}(x_1,x_2)$ (the joint density function).

Since $f_{X_1,X_2}(x_1,x_2)=f_{X_2|X_1}(x_2|x_1)\cdot f_{X_1}(x_1)$ I first computed $f_{X_1}(x_1)$:

$$f_{X_1}(x_1)=\begin{cases} \int_{-\infty}^{x_1}\frac{1}{s}f_{Z}(s)ds & x_1\leq 0\\ \int_{x_1}^{\infty}\frac{1}{s}f_{Z}(s)ds & x_1> 0\\ \end{cases} $$ where $f_{Z}$ is the Laplace density function. However, I'm struggling in the computation of $f_{X_2|X_1}(x_2|x_1)$

So far my strategy was to separate the case in which $x_1\neq 0$ and therefore, since $x_1=z\cdot y_1$, we have: $$f_{X_2|X_1}(x_2|x_1)=\int_{-\infty}^{\infty}f_{Y_2}(\frac{x_2}{z})\cdot f_{Z|X_1}(z|x_1)dz$$ But I did not so much progress from here. Any hint or help would be appreciated.


1 Answer 1


Since the variables $Z,Y_1,Y_2$ are sample independently it is clear that their joint density function is given by $f(z,y_1,y_2)=L(z)U(y_1)U(y_2)$. Then one can show that a convenient way to express the joint density of the derivative variables $X_1,X_2$ is

$$g(x_1,x_2)=\int_{\mathbb{R}^3} dz dy_1 dy_2\delta(zy_1-x_1)\delta(zy_2-x_2)f(z,y_1,y_2)$$

Isolating the arguments $y_1, y_2$ in the delta functions and performing the trivial integrals yields

$$g(x_1,x_2)=\int_{-\infty}^\infty \frac{dz}{z^2}L(z)U(\frac{x_1}{z})U(\frac{x_2}{z})$$

The uniform distributions impose the constraints $0\leq\frac{x_1}{z},\frac{x_2}{z}\leq 1$ which can equivalently be written as

$$\{x_1, x_2\geq 0,z>\max(x_1,x_2)\}\&\{x_1, x_2 < 0,z<\min(x_1,x_2)\}$$

Then we can write down the following expression for the joint density


where $\theta(x)$ is the Heaviside function. Since $L(-z)=L(z)$ we can simplify this further by changing variables in the second integral $z\to -z$. The expression simplifies to

$$g(x_1,x_2)=\theta(x_1x_2)\frac{1}{2\sigma}\int_{\max(|x_1|, |x_2|)}^\infty\frac{e^{-z/\sigma}}{z^2}dz$$

Finally, the remaining integral can be expressed in terms of special functions in the form

$$\int_{a}^\infty \frac{e^{-z/\sigma}}{z^2}dz=\frac{e^{-a/\sigma}}{a}-\frac{1}{\sigma}E_1(a/\sigma)$$

where $E_1(z)=\int_z^\infty \frac{e^{-x}}{x}dx$ is the exponential integral.


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