Slepian's inequality: equal variances and dominated increments imply $\mathbb P(\max_t X_t>\tau)\le\mathbb P(\max_t Y_t>\tau)$ for centered Gaussian vectors

Let $\iota$ be a finite index set. Let $X=(X_t)_{t\in\iota}$ be a random vector on a probability space $(\Omega,P)$ and $Y=(Y_t)_{t\in\iota}$ one on a probability space $(\Omega',Q)$. Assume both are centered and jointly Gaussian: 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 every $t$. Suppose that the variances agree and that the increments of $X$ are dominated by those of $Y$:

Then for every threshold $\tau\in\mathbb R$,

Equivalently, with equal variances the hypothesis says $\mathbb E X_sX_t\ge\mathbb E Y_sY_t$: the more strongly correlated vector $X$ has a stochastically smaller maximum.

This is Slepian's inequality, the comparison principle that Tropp and Webber use in the proof of Lemma B.1 of arXiv:2306.12418. There it is applied to $X_{u,v}=\langle S^{*}u,GTv\rangle+\|S^{*}u\|\|Tv\|\gamma$ and $Y_{u,v}=\|S^{*}u\|\langle h,Tv\rangle+\|Tv\|\langle g,S^{*}u\rangle$. The scalar Gaussian $\gamma$ is added to equalize the variances. This gives the sharp second-moment Chevet bound $\mathbb E\|SGT\|^2\le(\|S\|\|T\|_F+\|S\|_F\|T\|)^2$. Unlike the Sudakov–Fernique inequality, which needs no variance condition, it compares whole distributions of the maxima and not just their expectations. So it controls every moment $\mathbb E(\max X)_+^q$.

Formalization Note. Joint Gaussianity is expressed with Mathlib's HasGaussianLaw for the vector map $\omega\mapsto(t\mapsto X_t(\omega))$ into ι → ℝ. This also forces $P$ and $Q$ to be probability measures. Means, variances and increments are written as Bochner integrals; Gaussian coordinates are square-integrable, so these are the true moments. The maximum is ⨆ t, X t ω, a genuine maximum for finite nonempty $\iota$. For empty $\iota$ both maxima equal $0$ (the real convention $\sup\emptyset=0$), and the two sides coincide, since both are probability measures evaluated on $\{0>\tau\}$. The strict-inequality form stated here is equivalent to the form $P(\max X\ge\tau)\le Q(\max Y\ge\tau)$ found in textbooks, by monotone limits in $\tau$. The two vectors may live on different probability spaces; no coupling is assumed.