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I'm trying to understand sub-gaussian RVs to see if they could be relevant to my work.

The common definition of a sub-gaussian RV is the following. X is $\sigma$ sub-gaussian if its laplace transform / moment generating function is smaller than that of a Gaussian RV of standard deviation $\sigma$

$$ E(\exp(tX)) \leq \exp(\sigma^2 t^2 / 2) $$

Note that this characterizes

Another characterization of sub-gaussian variables is:

$$ \exists a, E(\exp(a X^2)) \leq 2 $$

And it seems to me that (almost) all random variables check that condition. Indeed if we look at the function (which is a form of moment generating function):

$$ a \rightarrow f(a) = E(\exp(a X^2)) $$

then we know the value at 0: $f(0)=1$ and, if f is continuous, then we can find a value of $a$ that checks the condition.

Does that mean that pretty much everyone (unless $f$ is absurdly miss-behaved) is sub-gaussian ?

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  • $\begingroup$ A random variable with Cauchy distribution is not sub-gaussian, although this might fit the 'absurdly miss-behaved' part. Same for slash distribution. $\endgroup$
    – Budenn
    Commented Jul 30, 2015 at 13:34
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    $\begingroup$ I don't know what "most" random variables are, but you can take anything whose tail is merely exponential, like the exponential distribution, and $E(\exp(aX^2)$ won't even exist for $a>0$. Ans this is not a research-level question. $\endgroup$
    – Brendan McKay
    Commented Jul 30, 2015 at 13:43
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    $\begingroup$ As Brendan points out, but to put a more fine point on it, the condition that $\mathbb{E} e^{a X^2}$ exists for some $a > 0$ is a quite strong condition that the tails of $X$ be "light". To put it another way, let $Y = \exp(a X^2)$; the condition is that the expectation of $Y$ exists. So conversely, if we have a $Y$ whose expectation exists and we want to guarantee to get a subgaussian $X$, we need to first take its logarithm and then its square root, which really "shrinks" the tails. $\endgroup$
    – usul
    Commented Jul 30, 2015 at 13:57
  • $\begingroup$ Thank you very much for the insight usul. I now realize what I got wrong. $\endgroup$
    – Guillaume Dehaene
    Commented Jul 30, 2015 at 14:24

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Note that most distributions which arise in nature are actually finitely supported, so are trivially sub-gaussian.

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  • $\begingroup$ Doesn't the Poisson distribution count as arising in nature? $\endgroup$
    – Yemon Choi
    Commented Jul 31, 2015 at 2:45
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    $\begingroup$ It is an idealization - for practical purposes, the truncated version is no less useful... $\endgroup$
    – Igor Rivin
    Commented Jul 31, 2015 at 4:06

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