Let $X$ be a $m$-dimensional Gaussian vector, and $Y$ a $q$-dimensional Gaussian vector, for some $m,q\geq 1$. Assume that the $X\_i$ and $Y\_j$ are centred and have unit variance. Assume that $E X\_i Y\_j \leq \mu$ for any $i,j$, for some $\mu< 1$.


Is there a constant $C\_{\mu,m,q}$ such that for every Borel $A,B$,
$$P(X\in A,Y\in B)\leq C\_{\mu,m,q} P(X\in A)P(Y\in B)?$$


It would imply $E \varphi(X)\psi(Y)\leq C\_{\mu,m,q} E\varphi(X)E\psi(Y)$ for (say) nonnegtive $\varphi,\psi$.


The inequality "feels" right, it works if $X$ and $Y$ are both uni-dimensional.


it looks like some of the measure concentration inequalities verified by the Gaussian, but I don't manage to find a proof and actually such a proof should be simple given the simplicity of the inequality (it might actually mean that it is completely wrong). Anyone could share ideas on this?



reverse FKG type inequality for slightly correlated Gaussian vectors
In this formulation, it still isn't true.
Take $\{X\_i\}\_{i=1}^n$i.i.d standard Gaussian. Take $\{Y\_i\}\_{i=1}^{n-1}$ i.i.d. standard Gaussian. Take $Y\_n=\sum\_{i=1}^n X\_i/\sqrt{n}$. Then $|EX\_i Y\_j|\leq 1/\sqrt{n}$ for any $i,j$.
Fix $D\subset R$ measurable.
Take $A=\{\sum X\_i/\sqrt{n}\in D\}$ and $B=\{Y\_n\in D\}$. Then the left side of your inequality is simply $P(Y\_n\in D)$ while the right side is $C\_{\mu,n} P(Y\_n\in D)^2$.
Now take a sequence $D\_m$ so that $P(Y\_n\in D\_m)\to\_{m\to\infty} 0$.


