# If $X$ is sub-Gaussian random variable with variance proxy $\sigma^2$, how to show that $E\{ \exp( t X^2) \} \leq (1 - 2 t \sigma^2)^{-1/2}$?

If $$X$$ is sub-Gaussian random variable with variance proxy $$\sigma^2$$, i.e., $$E(X) = 0$$ and $$E\{ \exp(s X) \} \leq \exp( \frac{\sigma^2 s^2}{2} )$$ for $$\forall s \in \mathbb{R}$$, then how to show that $$E\{ \exp( t X^2) \} \leq (1 - 2 t \sigma^2)^{-1/2}$$ for $$t < (2 \sigma^2)^{-1}$$?

My intuition is that sub-Gaussian is the random variable with tails similar or lighter than Gaussian distribution. So I guess that $$E\{ \exp( t X^2) \} \leq E [ \exp\{ t (\sigma Z)^2\} ]$$ should hold for $$Z \sim N(0,1)$$ and $$E [ \exp\{ t (\sigma Z)^2\} ] = (1 - 2 t \sigma^2)^{-1/2}$$ with $$t < (2 \sigma^2)^{-1}$$. However, I can not rigorously prove it.

You have just to use the integral representation $$e^{tX^2}=\int_{-\infty}^{\infty}e^{u\sqrt{2t}X-\frac{u^2}{2}}\frac{du}{\sqrt{2\pi}}$$ and Fubini.
Edit: Oh, I missed the point: what to do if $$t<0$$? Things are more difficult and the proof should involve complex variables for showing that $$E(e^{sX})\leq e^{s^2/2}$$ implies that $$f(s)=e^{-s^2/2}-E(\cos sX)\geq 0.$$ I will think about it...
• Do you know how to prove the result when $t < 0$? Commented Apr 16 at 15:41