# Can anyone give a reference to the proof of this concentration inequality?

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.

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