Let $Y$ be a real-valued random variable with a distribution equal to that of $|X|$ where $X$ is Gaussian with mean 0 and variance $\sigma^2$. Is $Y$ a subgaussian random variable? This Related question. was asked, but I don't follow the comments-- in particular, we can't directly apply the characterization of subgaussian random variables, because $Y$ is not zero-mean:

A random variable $X$ with $E[X] = 0$ is subgaussian iff there exists constants $c\geq 0$ and a Gaussian random variable $Z \sim N(0, \tau^2)$ for some $\tau > 0$, such that for all $s\geq 0$, $$\Pr(|X|\geq s) \leq c \Pr(|Z| \geq s).$$

This is from Theorem 2.6 in Wainright's High-Dimensional Statistics.

I can't seem to get around the nonzero mean of $Y$ in trying to prove sub-gaussianity. If there is any intuition or proof for why $Y$ is or is not subgaussian, that would be great.

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    $\begingroup$ The answer is positive as implied by most of the equivalent characterizations listed at en.wikipedia.org/wiki/…. The point is that when both tails of $X$ decay no more slowly than a Gaussian distribution, the right tail of $|X|$ -- which combines the two tails of $X$ -- cannot decay any more slowly, either. (The left tail of $|X|$ is bounded at $0$ and presents no problem at all.) $\endgroup$
    – whuber
    Commented Aug 30, 2023 at 14:41

1 Answer 1


From Theorem 2.6 of the same book, sub-Gaussianity can be equivalently defined as follows :

Definition : $Y$ is sub-Gaussian if and only if there exists $\sigma>0$ such that $$\mathbb E[e^{\lambda Y}]\le e^{\lambda^2 \sigma^2/2}\,\,\forall\lambda\in\mathbb R$$

From that definition, one can prove that sub-Gaussianity of $Y$ implies $\mathbb E[Y]=0$ as follows :

By Taylor expansion, we have $$\sum_{k=0}^\infty\frac{\lambda^k}{k!}\mathbb E[Y^k] =\mathbb E[e^{\lambda Y}] \le e^{\lambda^2 \sigma^2/2} = \sum_{k=0}^\infty\frac{\sigma^{2k} \lambda^{2k}}{2^k k!} $$ Which implies that $$\lambda\mathbb E[Y] +\frac{\lambda^2}{2}\mathbb E[Y^2] \le\frac{\sigma^2\lambda^2}{2} + o_{a.s.}(\lambda^2)\tag1$$ (Where $o_{a.s.}$ is a small-o almost surely notation, and holds for $\lambda\to0$)

Now, dividing both sides of $(1)$ by $\lambda$ for $\lambda>0$ (respectively $<0$) and letting $\lambda\to0^+$ (respectively $0^-$), we can conclude that $\mathbb E[Y] \le 0$ (respectively $\ge 0$), from which it follows that $$\mathbb E[Y]=0$$ as desired.

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    $\begingroup$ I think that definition of sub-gaussianity is for mean zero random variables, and in general a random variable Y is subgaussian if E(Y) exists and Y-E(Y) satisfies your given subgaussian definition. The latter is what I'm trying to prove or disprove $\endgroup$
    – gwtw14
    Commented Mar 23, 2022 at 23:11
  • $\begingroup$ Hmm I see. After playing around with the definition for a bit I would say that $Y$ is not subgaussian, but I don't have a proof for now. Will update this answer if I find one. $\endgroup$ Commented Mar 24, 2022 at 12:29
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    $\begingroup$ There are two popular definitions of 'sub-Gaussian', one that implies zero mean and one that doesn't (basically, finite $\psi_2$-norm). $\endgroup$ Commented Feb 1 at 0:10

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