Sub-gaussian Norm of Normal Distribution

Question

Let $X \sim \mathcal{N}(0, \sigma^2)$. In Vershynin's "High-dimensional Probability," it is observed that the sub-gaussian norm of $X$ is
$$ \|X\|_{\phi_2} = C\sigma $$
for some absolute constant $C$. I understand this is because of the tail bound of Gaussian random variables. However, I am wondering if there is an explicit result for the sub-gaussian norm of a normally distributed random variable. I assume there would be, but I am unsure of how to go about calculating it using only the definition of the sub-gaussian norm…

Answer 1

Probably the best way to compute the Sub-gaussian norm is to define the function $f:[0,\infty) \rightarrow (0,\infty]$ defined by \begin{align} f(\alpha) \triangleq \mathbb{E}[e^{X^2/\alpha^2}] &= \int_{-\infty}^\infty \frac{1}{\sqrt{2\sigma^2\pi}}e^{-x^2/2\sigma^2}e^{x^2/\alpha^2}dx \\ &= \frac{1}{\sqrt{2\sigma^2\pi}}\int_{-\infty}^\infty \exp \left ( \left (\frac{1}{\alpha^2} - \frac{1}{2\sigma^2}\right ) x^2\right )dx \end{align} The inner integral is finite as long as $1/\alpha^2 < 1/2\sigma^2$, and its value is a standard integral. I leave it to you to simplify this function.

It is also the case that $f(\alpha)$ is decreasing on $[0,\infty)$. To get the exact sub-gaussian norm, it is given by $||X||_{\psi_2} = \alpha^*$ where $\alpha^*$ is the unique value such that $f(\alpha^*) = 2.$