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 \(x\mapsto \lambda^2x e^{-\lambda x} \mathbf{1}_{\mathbb{R}_+}(x)\)).
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