Sub-Gaussian distribution

Remarks

Up to changing $K$ by a constant factor, the condition is equivalent to each of the following: $E[e^{X^2/K^2}] \le 2$; and $(E[\abs{X}^p])^{1/p} \le K \sqrt{p}$ for all $p \ge 1$. If $E[X] = 0$, it is also equivalent to a bound on the moment-generating function: there is $\sigma > 0$ with $E[e^{\lambda X}] \le e^{\lambda^2 \sigma^2 / 2}$ for all real $\lambda$. In this last form one says that $X$ is sub-Gaussian with variance proxy $\sigma^2$. Example: if $X$ is normal with mean $0$ and variance $\sigma^2$, then $E[e^{\lambda X}] = e^{\lambda^2 \sigma^2 / 2}$, so $X$ is sub-Gaussian with variance proxy $\sigma^2$. Example: by Hoeffding's lemma, if $a \le X \le b$ and $E[X] = 0$, then $E[e^{\lambda X}] \le e^{\lambda^2 (b - a)^2 / 8}$, so $X$ is sub-Gaussian with variance proxy $(b - a)^2 / 4$. Non-example: an exponential variable with rate $1$ has $P(X \ge t) = e^{-t}$, and $e^{-t} \le 2 e^{-t^2/K^2}$ fails for large $t$, whatever $K$ is. If $X_1, \dots, X_n$ are independent and centred with variance proxies $\sigma_i^2$, then $X_1 + \dots + X_n$ has variance proxy $\sigma_1^2 + \dots + \sigma_n^2$.