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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.

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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...

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  • $\begingroup$ Thank you so much! $\endgroup$ Commented Apr 16, 2024 at 8:32
  • $\begingroup$ Do you know how to prove the result when $t < 0$? $\endgroup$ Commented Apr 16, 2024 at 15:41

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