Universal constant that characterizes sub-gaussian random variables

Question

Let $X$ be a centered (that is, $\mathbb{E}X=0$) sub-gaussian random variable. The sub-gaussian norm of $X$ is defined as $\|X\|_{\psi_2}=\inf\{t>0: \mathbb{E}[\exp(X^2/t^2 )]\leq 2\}$. In other words, $\mathbb{E}[\exp(X^2/\|X\|^2_{\psi_2} )]\leq 2$.

Sub-gaussian random variable $X$ can also be characterized by $\mathbb{E}[e^{\lambda X}]\leq e^{C\lambda^2\|X\|^2_{\psi_2}}$ for all $\lambda\in\mathbb{R}$. I am interested in finding the constant $C$.

How do I do it from first principle?

For example, we have the inequality $e^x\leq x+ e^{x^2}$ for all $x\in\mathbb{R}$. Then, taking expectation on both sides, $\mathbb{E}[e^{\lambda X}]\leq \mathbb{E}[e^{\lambda^2 X^2}]=\mathbb{E}\left[\exp\left(\lambda^2\|X\|^2_{\psi_2}\cdot\frac{X^2}{\|X\|^2_{\psi_2}}\right)\right]$.

How do I go from here?

Answer 1

Since $$ \int e^{\alpha x^2/2}\,e^{-x^2/2}\,\frac{\text dx}{\sqrt{2\pi}} =\int e^{-(1-\alpha)x^2/2}\,\frac{\text dx}{\sqrt{2\pi}} =\int e^{-(1-\alpha)x^2/2}\,\frac{\frac{1}{\sqrt{1-\alpha}}\,\text dx}{\sqrt{\frac{2\pi}{(1-\alpha)}}}=\{1-\alpha\}^{-1/2} $$ one can deduce that $$\|X\|_{\psi_2}^2=8/3$$ for a Gaussian variable. Computing it for a sub-Gaussian variable will depend on its density.