Let $(G_t)_{t\in T}$ be a centered Gaussian process (with $T = [0,1]$). Can we say anything about the distribution of $$\Vert G\Vert := \sup_{t\in T}\vert G_t\vert?$$

For a multivariate normal (i.e., $T = \{1,2,...,n\}$), we know that the distribution function is given by $$F(x) = \int_{[- x, x]^n}\phi_\Sigma(u)\,\mathrm du,$$ where $\phi_\Sigma$ is the multivariate normal density of a centered $n$-variate normal distribution with covariance matrix $\Sigma$ (c.f., Distribution of the maximum of absolute value of multivariate Gaussian).

Still assuming $T = \{1,2,...,n\}$, it's also known that the expectation of $\Vert G\Vert$ can be bounded by $\sqrt{2\log(n)}$, which diverges as $n$ approaches infinity (despite a Gaussian process is known to be bounded almost surely?!).

I hope someone can help me sort my thoughts and resolve my confusion.

  • $\begingroup$ The answer would depend on the specific process. Take for example, $G_t$ to be 1) Brownian bridge process and 2) the standard Brownian motion. You get different results. $\endgroup$
    – Raghav
    Commented Sep 14, 2023 at 2:32
  • 2
    $\begingroup$ In general, the supreme of Gaussian processes is a very hard to characterize, and has been the subject of intense study for the past few years. You might want to look into Talagrand's majorizing measure theorem tcsmath.wordpress.com/2010/07/18/…. $\endgroup$
    – abacaba
    Commented Sep 14, 2023 at 3:21
  • $\begingroup$ @raghav can you show how the distribution of $\Vert G\Vert$ looks like in both cases? $\endgroup$
    – lmaosome
    Commented Sep 15, 2023 at 3:34


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