Let $X \in R^n$ be a zero mean, random vector with sub-gaussian coordinates $X_i$.

prove that X is a sub-gaussian random vector no matter if coordinates are independent or dependent.

It is easy to prove the result in the case of independent coordinates.

When it comes to the case of dependent coordinates, I think of the definition of multivariate normal distribution but don't know if it works for sub-gaussian family.

Assume a random vector Z $\in R^n$ has independent zero mean, unit variance, sub-gaussian coordinates and denote $\Sigma_X$as the covariance matrix of X, we can find a Z such that $\Sigma_X^{1/2} Z$ has the same distribution as X.
Because for $\forall a \in R^n$ $a^{T}\Sigma_X^{1/2} \in R^n$, given the case of independent coordinates, we can say for $\forall a \in R^n, a^{T}\Sigma_X^{1/2}Z = a^{T}X$ is sub-gaussion distributed. So X is a sub-gaussian random vector.

I am not sure if the proof is right for the whole sub-gaussian family, because I am not sure we can find a Z that $\Sigma_X^{1/2} Z$ is distributed same as X.

Any suggestions and ideas?


1 Answer 1


Let $X=(X_1,...X_n) \in R^n$ be a random vector with sub-gaussians coordinates $X_i$. By definition it is sub-gaussian iff the random variable $\langle X,x\rangle$ is sub-gaussian for any $x \in \mathbb{R}^n$. Consider $x \in \mathbb{R}^n$ We will show that $Y:= \langle X,x\rangle$ satisfies $$\lVert Y\rVert_p \leq K_2 \sqrt{p} \enspace \forall p\geq 1$$ (one of the equivalent definitions for being sub-gaussian). Indeed, \begin{align*} \lVert Y\rVert_p= \lVert{\displaystyle \sum_{i=1}^nx_iX_i}\rVert_p \leq\sum_{i=1}^n|x_i|\lVert X_i \rVert_p\\ \leq (\sum_{i=1}^n|x_i|K_{2i})\sqrt{p} && (\lVert X_i \rVert_p \leq K_{2i}\sqrt{p}, \enspace \text{since $X_i$ are sub-gaussian}) \\ \implies\lVert Y\rVert_p \leq K\sqrt{p} \enspace \forall p\geq 1 && (\text{ where $K=\sum_{i=1}^n|x_i|K_{2i}$}) \end{align*} I used prop 2.5.2 and definition 3.4.1 from the following book of Roman Vershynin High-Dimensional Probability

  • $\begingroup$ Aren't you using independence assumption in one of the inequality?? @sakas $\endgroup$ May 11, 2022 at 13:03

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