3
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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 $\sigma_T^2$.

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.