Boundedness of Gaussian Process for Borell-TIS inequality

Question

I have a question regarding the Borell-TIS Theorem which presents a result for a Gaussian process. An assumption is that the Gaussian Process $\{f_t\}_{t\in T} $ is almost surely finite, i.e. $P(\sup_{t\in T} |f_t|<\infty)=1$, where $T$ denotes a topological space.

What are the conditions for the covariance function of the Gaussian Process such that this assumption is fulfilled? Boundedness? Lipschitz Continuity?

Thanks!

Answer 1

A sufficient condition is given in Anderson and Dobric (1987), "The Central Limit Theorem for Stochastic Processes." Ann. Probab. 15 (1) 164 - 177, January, 1987.

Let $X(t)$, $t\in T$ be a Gaussian process indexed by a set $T$. For $1\leq p<\infty$, define the pseudometric $\varrho_{p}$ on $T$ by $\varrho_{p}(s,t):=\bigl(\mathbb{E}[|X(t)-X(s)|^{p}]\bigr)^{1/p}$, $s,t\in T$. Then $X$ is a tight random element in $\ell^{\infty}(T)$ if and only if $(T,\varrho_{p})$ is totally bounded and $X$ has sample paths almost surely unifromly $\varrho_{p}$-continuous.

That being said, if $(T,\varrho_{p})$ is totally bounded and $X$ has sample paths almost surely uniformly $\varrho_{p}$-continuous, then $X$ is almost surely bounded. Here $p$ can be any $1\leq p<\infty$, with a typical choice $p=2$, this becomes a joint condition about the set $T$ and covariance kernel, as you wish.

However, this is necessary if we also require $X$ to be tight. I'm not sure if there exists a weaker condition that guarantees $X\in\ell^{\infty}(T)$.

Answer 2

If $E(\sup_{t\in T} f_t) < \infty$, then clearly $P(\sup_{t\in T}f_t<\infty)= 1$. In fact, the converse also holds true (apply Borel-Tis inequality to $E(\sup_{t\in T}f_t ) \leq \int_0^\infty P(\sup_{t\in T}f_t > u) du)$. Therefore, a neccesary and sufficient condition for almost sure sample path boundedness of a centered Gaussian random field is $E(\sup_{t\in T} f_t) < \infty$.

The quantity $E(\sup_{t\in T} f_t)$ is of central importance in the study of Gaussian random field. Assume that $T$ is compact. Then we have the celebrated result that gives sufficient conditions under which this expectation is finite: $$E(\sup_{t\in T} f_t ) \leq \int_0^{diam(T)/2}\sqrt{\log N(T, d, \epsilon)}d\epsilon,$$ where $N(T, d, \epsilon)$ stands for the smallest number of $\epsilon$-balls (induced by the canonical metric $d(s,t):= \sqrt{E(X_s-X_t)^2}$) that is needed to cover $T$. For stationary fields, $E(\sup_{t\in T}f_t)$ is finite iff the right hand side integral is finite. Explicit constraints on the covariance function can then be derived.

A good reference for this is Random Field and Geometry by Robert J. Adler.