# Universal constant that characterizes sub-gaussian random variables

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?

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.