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I have a question about the definition of subgaussianity.

I have one version of the definition of sub-gaussianity on my textbook here, that is:

Suppose random variable X follows the inequality $\mathbb{E}[\exp(\lambda X)] \leq \exp(\frac{\lambda^2 \sigma^2}{2})$ for $\forall \lambda \in \mathbb{R}$, then we say that $X$ is $\sigma$-subgaussian.

However, this definition automatically results in the $X$ being centered. I also checked the wikipedia on the definition of sub-gaussianity, which seems to have slightly more relaxed condition called the Laplace condition: $$\exists B, b>0, \quad \forall \lambda \in \mathbb{R}, \quad \mathbb{E}[ e^{\lambda(X-\mathrm{E}[X])}] \leq B e^{\lambda^{2} b}$$

My question is: what's the relationship between these two condition? If we have the Laplace condition hold, what can we say about the sub-gaussian parameter of the random variable $X$? Is it automatically $\sqrt{2b}$-subgaussian?

Thanks for any help.

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Vershynin's book is a good reference for various equivalent definitions of sub-Gaussianity (and the proofs of their equivalence and keeping track of constants throughout).

Regarding the centering, there isn't an established convention.

  • Vershynin only states the $E[e^{\lambda X}] \le e^{\lambda^2 \sigma^2/2}$ definition in the case $E[X]=0$, but also provides other definitions of sub-Gaussianity that don't assume $E[X]=0$.
  • Rigollet makes $E[X]=0$ part of the definition of sub-Gaussianity.
  • Wikipedia and other texts (like Wainwright) define sub-Gaussianity by applying the MGF condition to the centered random variable $X-E[X]$ instead of $X$.

I'm not sure what textbook you are using, but the MGF condition $E[e^{\lambda X}] \le e^{\lambda^2 \sigma^2 /2}$ is the definition only when $E[X]=0$.

Also, the usage of "$\sigma$-subGaussian" (sometimes elsewhere as "$\sigma^2$-subGaussian" like in Rigollet's text) isn't standardized, and may differ from context to context.

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  • $\begingroup$ Thanks for your detailed answer and reference. I have looked at the book you suggest but I couldn't find any result addressing the constant $B$ before the exponential term. Once we derive that $\mathbb{E}\left[e^{\lambda(X-\mathrm{E}[X])}\right] \leq B e^{\lambda^{2} b}$, can we simply say that $X$ is $\sqrt{2b}$-subgaussian and ignore the disturbing $B$ constant before? $\endgroup$
    – Jamie Carr
    Commented Aug 28, 2021 at 8:27
  • $\begingroup$ @JamieCarr Oh I see your specific concern now. If you walk through the proof of "(v) $\implies$ (i)" in Vershynin's text with an extra factor of $B$, you will find that you still get the concentration in (i) but with a constant different than $2$ in the front of the bound. Note that Wikipedia's probability bound has a general $C$ instead of specifically $2$. So there is a subtle difference in the definitions, but I think in practice most would not care too much about this constant factor; it is the exponential decay that matters. $\endgroup$
    – angryavian
    Commented Aug 28, 2021 at 16:54
  • $\begingroup$ @JamieCarr Also note that Wikipedia only gives a definition for "sub-Gaussian" without any sort of parameterization by $\sigma$ or $\sigma^2$, so even disregarding the difference in my previous comment, you cannot use it to say something like "$\sqrt{2b}$-subGaussian." $\endgroup$
    – angryavian
    Commented Aug 28, 2021 at 16:55

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