The distribution of the sum of sub-exponential random variables

Question

I know the sum of sub-gaussian random variables is sub-gaussian, but I want to ask whether the sum of sub-exponential random variables is still sub-exponential random variables.

Answer 1

Yes.

We use the definition that $X$ is sub-exponential if there exist $\nu, \alpha > 0$ such that, for all $|\lambda| \leq 1/\alpha$, $$ \mathbf{E} \exp\{\lambda(X - \mathbf{E} X)\} \leq e^{\frac{1}{2} \nu^2 \lambda^2}. $$

Now, let $X_1, X_2$ be sub-exponentials, we can apply Cauchy-Schwarz to compute that, when $|\lambda| \leq 1/(2(\alpha_1 \vee \alpha_2))$, \begin{align*} \mathbf{E}\exp\{\lambda[(X_1 + X_2) - \mathbf{E}(X_1 + X_2)]\} &\leq [\mathbf{E}\exp\{2\lambda(X_1 - \mathbf{E} X_1)\} \cdot \mathbf{E}\exp\{2\lambda(X_2 - \mathbf{E} X_2)\}]^{1/2} \\ &\leq \Bigl[\exp\Bigl\{\frac{1}{2} \nu_1^2 (2\lambda)^2\Bigr\} \cdot \exp\Bigl\{\frac{1}{2} \nu_2^2 (2\lambda)^2\Bigr\}\Bigr]^{1/2} \\ &\leq \exp\Bigl\{\frac{1}{2} \cdot 2(\nu_1^2 + \nu_2^2) \lambda^2\Bigr\}, \end{align*} so that $X_1 + X_2$ is subexponential. This can then be easily generalised to finite sums.