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Consider the following definition (from here).

A random variable $X$ with finite mean $\mu = \mathbb{E}[X]$ is said to be sub-Gaussian with parameter $\sigma^2$ if: $$\mathbb{E}[e^{\lambda(X - \mu)}] \le e^{\frac{\lambda^2\sigma^2}{2}}.$$

Given this definition, is X sub-Gaussian if it follows a half-normal distribution?

smz
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  • Yes it is sub-Gaussian. – Andrew Nov 19 '22 at 22:27
  • could you point me to a proof? – smz Nov 19 '22 at 22:56
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    See pg.24 of Vershynin's book – Andrew Nov 19 '22 at 23:05
  • Thank you for the reference. Would the following be correct? According to Proposition 2.5.2 in the book, since $\mathbb{P}[\vert X \vert \ge t] \le 2 e^{-t^2/2}$, then $\vert X \vert$ is sub-Gaussian. – smz Nov 19 '22 at 23:28
  • Looks good to me. Maybe you can be more careful, and recall the bound which holds for $t>0$ and $Z\sim\mathcal N(0,1)$. $$ \mathbb P(Z\geq t)\leq \frac{1}{t}\frac{1}{\sqrt{2\pi}}e^{-t^2/2} = \frac{1}{t}f_Z(t)$$ (And note that in fact $\mathbb P(Z\geq t) \sim t^{-1}f_Z(t)$) – Andrew Nov 20 '22 at 00:11
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    Be careful when applying Proposition 2.5.2. Equivalence of (i) and (v) relies on the assumption that $E[X]=0$. $P(|X - E[X]| \ge t) \le 2e^{-t^2/K^2}$ is the definition of sub-Gaussianity of $X$, but the analogue for $|X|$ would be $P(||X| - E[|X|]| \ge t) \le 2e^{-t^2/K^2}$. – angryavian Nov 20 '22 at 01:33
  • Thank you, @angryavian. I completely overlooked the condition for (i) to be equivalent to (v). – smz Nov 20 '22 at 01:55

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