Why finite sub-gaussian norm indicates sub-gaussian distribution?

Question

The sub-gaussian norm of a random variable $\xi$ is defined as follows: $$\|\xi\| = \inf \left\{\lambda > 0 | \mathbb E\left(e^{\xi^2 / \lambda^2}\right) \leq 2\right\}.$$ When the moment generating function of a random variable $\xi$ satisfies $$\exists \sigma > 0 , \forall x > 0, \mathbb E\left(e^{x\xi}\right) \leq e^{\frac{\sigma^2 ~ x^2}{2}},$$ we say $\xi$ is sub-gaussian.

It is obvious that if a variable is sub-gaussian, then its sub-gaussian norm is finite. However, some resouces say the converse is also true, i.e. a random variable with finite sub-gaussian norm is sub-gaussian (e.g. in this note). I wonder how to prove that.

Answer 1

The notes you linked define a random variable to be sub-gaussian if the sub-gaussian norm is finite. But note that if $\xi$ has a finite sub-gaussian norm, then we have (by definition) $$ \mathbb{E}\exp(\xi^2 / \|\xi\|^2)\leq 2, $$ which is equivalent to $$ \mathbb{P}(|\xi| \geq t) \leq 2 \exp(-ct^2/\|\xi\|^2) $$ for all $t\geq 0$, where $c > 0 $ is a constant. This means $\xi$ is sub-gaussian. If in addition $\mathbb{E}(\xi) = 0$, then your definition is satisfied as well, but note that it is only under the zero mean condition. Otherwise, it does not necessarily define a sub-gaussian random variable.