I am studying sub-gaussian random variables, and have come across the following two definitions of the sub-gaussian norm $\|X\|_{\psi_2}$ of a sub-gaussian random variable $X$:
- Vershynin's High-Dimensional Probability defines
$$
\|X\|_{\psi_2}=\inf\{t>0:\mathbb{E}\exp(X^2/t^2)\le 2\}.
$$ - Vershynin's Introduction to the Non-asymptotic Analysis of Random Matrices defines
$$
\|X\|_{\psi_2}=\sup_{p\ge 1}p^{-1/2}\left(\mathbb{E}|X|^p\right)^{1/p}.
$$
From Proposition 2.5.2 of Vershynin's High-Dimensional Probability, I know that the two definitions differ from each other by at most an absolute constant factor. My question is: do the two definitions give the exact same norm?
Best Answer
Define for any subgaussian variable $X$ $$\|X\|_{1}:=\inf\{t>0:\mathbb{E}\exp(X^2/t^2)\le 2\}$$ And $$\|X\|_{2}:=\sup_{p\ge 1}p^{-1/2}\left(\mathbb{E}|X|^p\right)^{1/p} $$
By the result you cite, we know that there exist two positive constants $c$ and $C$ such that $c\|X\|_{1} \le \|X\|_{2}\le C\|X\|_1 $. Thus I interpret your question as
The answer is no : Consider $X$ to be a Rademacher random variable, i.e. $\mathbb P(X=1) = \mathbb P(X = -1) = \frac 1 2$, such that $X^2 = |X|^p = 1$ almost surely (clearly, $X$ is subgaussian since it is bounded). It then is easy to check that $$\|X\|_1 = \sqrt{\frac{1}{\ln 2}} $$ But $$\|X\|_2 = 1 $$
Addendum : The OP asks
The answer remains the same : Let $X\sim \text{Bernoulli}(q)$ with $0<q<1$, i.e. $\mathbb P(X=1) = q = 1 - \mathbb P(X=0)$ (I name the parameter $q$ because there is already a $p$ in the definition of $\|\cdot\|_2$).
Now we have $X= X^2 = |X|^p$ almost surely for all $p\ge 1$, and it is not hard to check that $$\|X\|_1=\inf\left\{t>0:e^{1/t^2}\le \frac{e - (1 -q )}{q}\right\} = \sqrt{\frac{1}{\ln (e - (1-q)) - \ln(q)}}$$ And $$\|X\|_2= \sup_{p\ge 1}p^{-1/2}q^{1/p} $$
I'm pretty sure that the above supremum can be found in closed form by elementary calculus, but I'm feeling a bit lazy right now, so instead you can use your favourite computing tool to find out that for $q=1/2$ : $$\|X\|_1 > \|X\|_2 $$
You can actually see (or at least convince yourself) from this argument that changing the $2$ or $e$ in the definition of $\|\cdot\|_1$ to any other positive number will not change the conclusion : $\|\cdot\|_1 $ and $\|\cdot\|_2 $ are not equal.