Double exponential

We have already mentioned in a previous post an amusing property of the exponential distribution. Here is another one: if $X$ and $Y$ are two independent exponential random variables with mean $1/\lambda$ and $1/\mu$ respectively then $X-Y$  follows the double exponential distribution on $\mathbb{R}$ with  density

$$x\in\mathbb{R}\mapsto \frac{\lambda\mu}{\lambda+\mu}\left(e^{\mu x}\mathbf{1}_{\mathbb{R}_-}(x)+e^{-\lambda x}\mathbf{1}_{\mathbb{R}_+}(x)\right).$$

In other words, we have the mixture $$\mathcal{L}(X-Y)=\mathcal{L}(X)*\mathcal{L}(-Y)=\frac{\mu}{\lambda+\mu}\mathcal{L}(X)+\frac{\lambda}{\lambda+\mu}\mathcal{L}(-Y).$$

In particular, when $\lambda=\mu$ we get the symmetric double exponential (Laplace distribution) with density $x\mapsto \frac{\lambda}{2} e^{-\lambda|x|}$.  Another way to state the property is to say that the double exponential is the image of the product distribution $\mathcal{E}(\lambda)\otimes\mathcal{E}(\mu)$ by the linear map $(x,y)\mapsto x-y$. Note that the density of $X+Y$ is $$x\mapsto \lambda\mu\frac{e^{-\mu x}-e^{-\lambda x}}{\lambda-\mu}\mathbf{1}_{\mathbb{R}_+}(x)$$

(when $\lambda=\mu$ we recover by continuity the Gamma density $x\mapsto \lambda^2x e^{-\lambda x} \mathbf{1}_{\mathbb{R}_+}(x)$).