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...


1 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 $f(\alpha^*) = 2.$

  • $\begingroup$ Thanks! I actually figured the same thing out myself just a couple of minutes before checking if an answer had been given. If I am not mistaken, the result should be $\|X\|_{\psi_2} = \sqrt{2/3} \sigma$ (just for future reference). $\endgroup$
    – LSK21
    Aug 1 at 13:01
  • $\begingroup$ By the way, that is a pretty handy Wikipedia page, I'll have to bookmark that one :D $\endgroup$
    – LSK21
    Aug 1 at 13:03
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    $\begingroup$ Based on your comment, you are close probably doing everything correct but have a small arithmetic/algebra mistake. The correct answer is $||X||_{\psi_2} = \sqrt{8/3}\sigma$. $\endgroup$ Aug 1 at 18:33
  • $\begingroup$ Are you sure? I can't seem to find my mistake... the integral for $f(\alpha)$ solves to $\frac{1}{\sqrt{2\pi}\sigma} (\frac{\sqrt{2\pi} \alpha \sigma}{\sqrt{\alpha^2 - 2\sigma^2}}) = \frac{\alpha}{\sqrt{\alpha^2 - 2\sigma^2}}$ for me, am I good so far? $\endgroup$
    – LSK21
    Aug 2 at 7:41
  • $\begingroup$ You are correct so far, all you need to do is set the last expression equal to 2 and solve for $\alpha$. One way to quickly do this is $2 = \alpha/\sqrt{\alpha^2 - 2\sigma^2} \iff 4 = \alpha^2/(\alpha^2 - 2\sigma^2) \iff 4\alpha^2 - 8 \sigma^2 = \alpha^2 \iff 3\alpha^2 = 8\sigma^2 \iff \alpha = \sqrt{8/3}\sigma$ $\endgroup$ Aug 2 at 13:59

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