Sudakov–Fernique inequality: dominated increments imply $\mathbb E\max_t X_t\le\mathbb E\max_t Y_t$ for centered Gaussian vectors

Let $\iota$ be a finite index set, and let $X=(X_t)_{t\in\iota}$ on a probability space $(\Omega,P)$ and $Y=(Y_t)_{t\in\iota}$ on a probability space $(\Omega',Q)$ be centered, jointly Gaussian random vectors in $\mathbb R^\iota$. That is, the laws of $\omega\mapsto(X_t(\omega))_t$ and $\omega'\mapsto(Y_t(\omega'))_t$ are Gaussian measures on $\mathbb R^\iota$, and $\mathbb E X_t=\mathbb E Y_t=0$ for all $t$. Suppose the increments of $X$ are dominated by those of $Y$:

Then

This is the Gaussian comparison principle behind the sharp bound $\mathbb E\|G\|\le\sqrt N+\sqrt n$ for an $N\times n$ Gaussian matrix. On a finite net of $S^{n-1}\times S^{N-1}$, one compares $X_{u,v}=\langle Gu,v\rangle$ with $Y_{u,v}=\langle g,u\rangle+\langle h,v\rangle$. Unlike Slepian's lemma, no equality of variances is required.

Formalization Note. Joint Gaussianity is expressed with Mathlib's HasGaussianLaw for the vector-valued map $\omega\mapsto(t\mapsto X_t(\omega))$ into ι → ℝ; this hypothesis also forces $P$ and $Q$ to be probability measures. The maximum is written ⨆ t, X t ω, which is a genuine maximum for finite nonempty $\iota$. For empty $\iota$ both sides equal $0$ (the real convention $\sup\emptyset=0$), so the statement remains true. Both maxima are integrable: they are dominated by $\sum_t|X_t|$, and Gaussian coordinates are integrable. The two vectors live on possibly different probability spaces; no independence or coupling is assumed.