How to prove that the Laplace distribution is a sub-exponential distribution random variable

Question

How to prove that a random variable obeying the Laplace distribution is a sub-exponential distribution random variable?

Answer 1

The definition of a sub-exponential random variable $X$ with parameters $\nu, \alpha$ states that $\mathbb{E}[e^{\lambda X}]\leq e^{\frac{\nu^2\lambda^2}{2}}$ for $|\lambda|< \frac{1}{\alpha}$. In your case, this means that we essentially need to upper-bound the moment-generating function (MGF) of a Laplace distribution accordingly.

To this end, consider first a centered random variable $X\sim\mathrm{Laplace}(1)$. Its MGF is $\mathbb{E}[e^{\lambda X}]=\frac{1}{1-\lambda^2}$ for $|\lambda| < 1$. From there, you can find many upper-bounds on the MGF of the desired form. For example, you can check that $\mathbb{E}[e^{\lambda X}] \leq e^{2\lambda^2}$ for $|\lambda|<\frac12$, meaning that $X$ is sub-exponential with parameters $\nu=2, \alpha=2$.

If you want to extend the result to the Laplace distribution with parameter $b>0$, you just need to use the fact that if $X\sim\mathrm{Laplace}(1)$, then $bX\sim\mathrm{Laplace}(b)$. As such the upper-bound on the MGF becomes $$\mathbb{E}[e^{b\lambda X}] \leq e^{2b^2\lambda^2}\text{ for }|b\lambda|<\frac12 \iff \mathbb{E}[e^{\lambda bX}] \leq e^{2b^2\lambda^2}\text{ for }|\lambda|<\frac{1}{2b},$$ from which we conclude that the $\mathrm{Laplace}(b)$ distribution is sub-exponential with parameters $\nu=2b, \alpha=2b$.