# Concentration inequality for max component of a multivariate Gaussian in the general case

I am looking to bound the variance of the maximum component of a vector distributed multivariate Gaussian in the general case where the Gaussian distribution has arbitrary mean and full covariance matrix.

Specifically, let $$X \sim N(\mathbf{\mu}, \Sigma)$$ where $$\mu \in \mathbb{R}^d$$ and $$\Sigma$$ is an arbitrary PSD matrix. I am hoping to bound the distance between $$\|x\|_\infty$$ and $$\mathbb{E}\,\|X\|_\infty$$, in other words$$\mathbb{E}_{x \sim N(\mu,\Sigma)}\,\left|\left(\max_{i=1, \ldots, d} x_i\right) - \left(\mathbb{E}_{x \sim N(\mu,\Sigma)}\,[\max_{i=1,\ldots,d} X_i]\right) \right|^2.$$

All references I've come across seem to handle the special case of $$X$$ being centered (zero mean) and usually with spherical covariance matrix:

Any references or suggestions are gladly welcomed.

• The standard proof of Borell's inequality works for this case as well. For example, see page 13 of wisdom.weizmann.ac.il/~zeitouni/notesGauss.pdf – ofer zeitouni May 26 at 18:41
• Thanks Ofer! Some questions: the first paragraph of section 2 states that the GP is centered, which seems it may not help me. Also, does $\sup_{t\in T} X_t \le \infty$ where $T$ compact imply the sampled Gaussian RVs are assumed to never exceed some max value? If so, this is bad since Gaussian RVs have full support over the real line and therefore have no max value. Lastly, prop 4 does in fact seem helpful since $f$ can rescale and translate the input Gaussian RVs $Y$ s.t. they are no longer centered with have unit variance. Is this the right intuition? Thanks again. – ted May 27 at 6:54
• I didn't mean that the statement is there - just that the proof goes over to the non-centered case. About the rest of your question, I think you are confusing two things: the range of Gaussian variables (which is R), and the index set (which is $T$). In your case, $T=\{1,\ldots,d\}$ and thus $T$ is finite and it is always the case that $E\sup_{t\in T} X_t<\infty$. To summarize - just go over the proof and check that it carries over to your case. – ofer zeitouni May 27 at 14:51
• Many thanks. So if $f(x) = R^{1/2} x + \mu$, then we have $|f(x) - f(y)| \le \max_i R^{1/2}_{ii} \|(x - \mu) - (y - \mu)\|$ and the means cancel. Btw, why then is Borell's inequality stated for the case of centered Gaussians? (See for example, en.wikipedia.org/wiki/Borell%E2%80%93TIS_inequality) Does something break down for non-centered Gaussians when $T$ is infinite? – ted May 28 at 21:42
• Not really, although of course you need to make sure that (say) the maximum of the mean is bounded. – ofer zeitouni May 29 at 4:10