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Let $T \subset \mathbb R^n$, and assume it's a finite set if that helps. Consider the symmetric Gaussian process $(X_t)_{t\in T}$ defined by $X_t = \langle G, t\rangle$, where $G$ is a standard Gaussian in $\mathbb{R}^n$ and $\langle\cdot, \cdot \rangle$ is the standard inner product. In other word, $\mathbb{E} X_t = 0$, and $\mathbb{E} (X_s - X_t)^2 = \|s-t\|_2^2$ for all $s,t\in T$

I want to know what happens to $\mathbb E \sup_{t\in T} X_t$ when we apply a contraction map to the index set $T$?

I.e. take a contraction map $f:\mathbb R^n \to \mathbb R^n$, $\|f(x)-f(y)\|_2 \leq \|x-y\|_2$ for all $x,y\in\mathbb R^n$, and define the process $(Y_t)_{t\in T}$ as $Y_t = \langle G, f(t)\rangle$. (Equivalently, $\mathbb{E} Y_t = 0$, and $\mathbb{E} (Y_s - Y_t)^2 = \|f(s)-f(t)\|_2^2$.) An application of Talagrand's majorizing measures theorem shows that for some absolute constant $L$,

$$ \mathbb E \sup_t Y_t \leq L\ \mathbb E \sup_t X_t. $$

Here are two questions:

  1. Is the bound true with $L=1$?

  2. Is there a direct proof that does not use Talagrand's theorem?

I am even interested in the simple map $f(t_1, \ldots, t_n) = (|t_1|, \ldots, |t_n|)$. Is there a simple direct proof in this case? Is the inequality true with constant 1 in this case?

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It is true, and it follows from the following fact, sometimes referred to as Sudakov-Fernique inequality, sometimes as Slepian-Fernique lemma:

Assume that $(X_t)$ and $(Y_t)$ are two centered Gaussian processes. If $\|Y_s - Y_t\|_2 \leqslant \|X_s - X_t\|_2$ then $\mathbb{E} \sup Y_t \leqslant \mathbb{E} \sup X_t$.

In this inequality the index set $T$ is assumed to be countable so that we do not have any problems with measurability.

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  • $\begingroup$ Thank you Mateusz! I now realize this is something I should know, for example as a step in proving Sudakov's minoration. $\endgroup$ – Sasho Nikolov Apr 24 '15 at 17:00

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