# Join density function of $X_1=Z*Y_1$ and $X_2=Z*Y_2$ with $Y_1,Y_2$ follow a uniform distribution and $Z$ Laplace distribution

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.

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

$$g(x_1,x_2)=\theta(x_1)\theta(x_2)\int_{\max(x_1,x_2)}^\infty\frac{L(z)}{z^2}dz+\theta(-x_1)\theta(-x_2)\int^{\min(x_1,x_2)}_{-\infty}\frac{L(z)}{z^2}dz$$

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.