# Sub Gaussian concentration of correlated random variables

Let $$X_1, X_2, \dots, X_n$$ be $$n$$ zero mean random variables. It is given that $$X_i$$ is sub-Gaussian with parameter $$\sigma_i^2$$ (assumed known) for $$i = 1,2, \dots, n$$. Consider the random variable $$Y = \sum_{i = 1}^n X_i$$. It is known that $$Y$$ is a sub-Gaussian random variable. We denote the corresponding sub-Gaussian parameter as $$\sigma_Y^2$$.

If $$X_i$$'s are given to be independent, then it is well-known that $$\sigma_Y^2 = \sum_{i = 1}^n \sigma_i^2$$. Similarly, a well known upper bound for the general case is $$\sigma_Y^2 = \left( \sum_{i = 1}^n \sigma_{i}\right)^2$$. However, when the random variables $$X_i$$'s are correlated such that the covariance matrix is known, can this bound on $$\sigma_Y$$ be improved using this information? In particular, if we know that the random variables are negatively correlated for all $$i$$, I would expect the bound on $$\sigma_Y$$ to be closer to that obtained in the independent case than the one obtained in the general case.

I was wondering if there was any way to prove the same or any standard results that account for correlation between random variables. I was also interested in the case where the random variables are sub-Exponential. Any leads or references will be appreciated. Thanks!

If $$X\sim \mathcal{N}(0,\sigma^2)$$, its sub-gaussian norm is $$||X||_{\psi_2}=\sigma$$. If $$(X_1,\dots,X_n)$$ is jointly Gaussian-distributed, then $$\sum_{i=1}^n X_i$$ is also Gaussian. So, you can determine its sub-gaussian norm by analyzing the variance of $$\sum_{i=1}^n X_i$$. Here, the covariance matrix will indeed show up.
For example, if $$(X,Y)$$ are jointly Gaussian distributed with mean zero, then: $$||X+Y||_{\psi_2}=\sqrt{Var[X+Y]}=\sqrt{E[X^2]+E[Y^2]+2E[XY]}=\sqrt{\sigma^2_X+\sigma^2_Y+\rho_{XY}\sigma_X\sigma_Y}$$ Note that for $$\rho_{XY}=0$$, we have $$||X+Y||_{\psi_2}^2=||X||_{\psi_2}^2+||Y||_{\psi_2}^2$$ and for $$\rho_{XY}=1$$, $$||X+Y||_{\psi_2}^2=(||X||_{\psi_2}+||Y||_{\psi_2})^2$$. So, for the Gaussian case, we can indeed nicely interpolate between the two extreme cases you describe. As stated in the beginning, I doubt that the general case can be this terrific.
• Is the sub-gaussian norm of $X \sim \mathcal{N}(0, \sigma^2)$ not $C\sigma$ for some absolute constant $C$? How do you justify it being $\sigma$ exactly? Aug 1, 2023 at 11:42
• @LSK21 There are various definitions of the sub-gaussian norm, see e.g. Vershynin's book for a list of definitions that are equivalent up to a constant. If you use the definition with moment-generating functions, the sub-gaussian norm of a normal is $\sigma$, see e.g. Wainwright's book. If you prefer, multiply all the statements in my answer with $C$. Aug 2, 2023 at 7:50