As per this lecture notes, any random variable with $\mathbf{E}[X] = 0$ and $Var(X) \leq \sigma^2$ is a sub-gaussian RV.
The proof given there uses Fubini's theorem to deduce that
$ \sum_\limits{n \geq 0} t^n\frac{\mathbf{E}(X^n)}{n!} \leq \exp({\frac{t^2\sigma^2}{2}})$
I expanded the series on the LHS, but could not establish the inequality as LHS contains terms involving higher order moments of X. Any help on it ?
Best Answer
As mentioned in the comments above, the statement as you give it cannot hold. Here is a counter example.
Let $X$ be a real random variable with pdf equal to $$f(x) = \begin{cases}\tfrac{C}{x^4} & \text{if } |x|\geq 1\\ 0 & \text{otherwise}\end{cases}$$ where $C$ is chosen so that $\int_\mathbb{R} f = 1$. Then by symmetry $\mathbb{E} X = 0$. Moreover, $$\sigma^2 = \mathbb{E}X^2 = \int_{\mathbb{R} \setminus [-1,1]}\frac{Cx^2}{x^4} = 2C\int_1^\infty \frac{1}{x^2} = 2C.$$
On the other hand, the fourth moment $$\mathbb{E}X^4 = \int_{\mathbb{R}\setminus[-1,1]} \frac{x^4}{x^4} = 2\int_1^\infty 1 = \infty.$$ This is impossible if $X$ was $2C$-sub Gaussian since if it was, symmetry ensures that $$\frac{1}{4!}\mathbb{E} X^4\leq \sum_{n=0}^{\infty}\frac{\mathbb{E} X^{2n}}{(2n)!} = \sum_{n=0}^{\infty}\frac{\mathbb{E} X^n}{n!} = \mathbb{E} e^{X} \leq e^{\tfrac{4C^2}{2}} = e^{2C^2},$$ or in other words that $$\infty = \mathbb{E} X^4 \leq 24 e^{2C^2},$$ a contradiction.