# Is the half-normal distribution sub-Gaussian?

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.

• 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.)
– whuber
Aug 30, 2023 at 14:41

## 1 Answer

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.

• 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 Mar 23, 2022 at 23:11
• 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. Mar 24, 2022 at 12:29
• There are two popular definitions of 'sub-Gaussian', one that implies zero mean and one that doesn't (basically, finite $\psi_2$-norm). Feb 1 at 0:10