The following concentration inequality for the supremum of a Gaussian process indexed by a separable metric space appears here: http://math.iisc.ac.in/~manju/GP/6-Concentration%20and%20comparison%20again.pdf (this, to the best of my knowledge, is one of several lecture notes prepared by Prof. Manjunath Krishnapur, IISC Bangalore), page-22 (or page 1 in the pdf), exercise-2.

Let $X$ be a centered, continuous Gaussian process on a separable metric space $T$ and suppose that $X^* := \sup_{t \in T} X_t$ is finite with probability $1$. Then, show that: $$\lim_{x\rightarrow +\infty}\frac{1}{x^2} \log \mathbb{P}\left(X^* \geq x\right) = -\frac{1}{2\sigma_T^2}~,$$ where $\sigma_T^2 := \sup_{t \in T} \mathbb{E} (X_t^2)$.

I need to use this result in one of my research works, so I need a proper reference, where this, or anything similar is proved. I basically want a reference, where an exponential concentration of the supremum of a continuous Gaussian process in a separable metric space (for me, Euclidean spaces suffice) in terms of its maximum variance, is proved. Can anyone help me? Thanks in advance.


This follows immediately from the Borell-TIS inequality.

Indeed, for any $x>EX^*$ this inequality yields $$\frac1{x^2}\,\ln P(X^*\ge x)\le-\frac{(x-EX^*)^2}{2\sigma_T^2 x^2}\to-\frac1{2\sigma_T^2} $$ as $x\to\infty$.

On the other hand, for any $t\in T$ $$\frac1{x^2}\,\ln P(X^*\ge x)\ge\frac1{x^2}\,\ln P(X_t\ge x)\to-\frac1{2\sigma_t^2} $$ as $x\to\infty$. It remains to choose $t\in T$ so that $\sigma_t^2$ be arbitrarily close to $\sigma_T^2$.

  • $\begingroup$ Thanks a lot @losif. Now I know that $x$ must be larger than $E X^*$ for the inequality to hold. $\endgroup$ – Somabha Jul 25 at 2:20

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