on GCI for asymmetric sets

Note

This post serves as a study note for ArXiv:1011.4166, which only uses elementary approaches. The Gaussian Correlation Inequality (GCI) was discussed in an earlier post on Gaussian Correlation Inequality. Related references are attached at the end[^1][^2]

🧩Gain & Loss

The original Gaussian Correlation Inequality assumes symmetry and Gaussian measure (can be extended to Gamma type on Gaussian Correlation Inequality). It turns out that the symmetry can be dropped on one set to gain some flexibility, while some more restrictions are required on the other set.

🌵Extension to star-shape sets

Let $f$ be a non-negative real-valued function on ${\mathbb{R}}^n$ with compact support for $|x|\to \infty$, and $B_t:=tB$ as a scaled ball centered at origin. The idea is to study the following inequality (yet simple):

$$\Phi (t)={\int }_{B_t}[f(x)-\mu (f)]d\mu (x),$$

where $\mu =\rho (|x|)dx$ is a radial probability measure and $\mu (f)={\mathbb{E}}_{\mu }(f)$. Of course $\Phi (0)=\Phi (\infty )=0$, we are concerned about the values in between. Moreover, around $t=0$, we have $f(tr(\theta )\theta )\approx f(0)$, thus if $f(0)>\mu (f)$, we can have ${\Phi }^{\prime }(t)>0$ near $t=0$. And for sufficiently large $t$, we find that $B_t\supset \mathrm{supp}f$,

$$\Phi (t)=\mu (f)-\mu (f)\mu (B_t)>0.$$

If we want to show the positivity of $\Phi$ on $(0,\infty )$, the naive idea is to ask for a single bump-shape graph for $\Phi$, although we cannot exclude the possibility for multi-bump shapes.

The derivative ${\Phi }^{\prime }(t)$ is easy to compute:

$${\Phi }^{\prime }(t)=t^{d-1}{\int }_{{\mathbb{S}}^{d-1}}[f(t\theta )-\mu (f)]\rho (t)d\theta$$

It appears that we only require $f(t\theta )<f(t^{\prime }\theta )$ for $t>t^{\prime }$, where $t^{\prime }$ is the first hitting time that ${\Phi }^{\prime }(t)=0$. This will be true if $f(tv)$ is non-increasing in all directions.

Suppose $A$ is a star-shape set, and define

$$f_n(x)=\{\begin{array}{ll}1,& x\in A\\ 1-n\min (n^{-1},\alpha (x)-1)& x\notin A\end{array}$$

where $\alpha (x)=\inf \{k>0\mid x\in kA\}$. This function is well-defined once $n$ is sufficiently large. We can see that $f_n$ is monotone decreasing. Therefore,

$$\mu (f_n{1}_B)\ge \mu (f_n)\mu (B).$$

We can take limit on both sides through monotone convergence theorem.

Theorem

Suppose $A$ is a star-shape set containing origin and $B$ is a ball centered at origin, then

$$\mu (A\cap B)\ge \mu (A)\mu (B)$$

for any radial measure $\mu (x)=\rho (|x|)dx$ (standard Gaussian included).

When $A$ is convex, there is an extremely short proof using the contraction property of Brenier map by Caffarelli, see the references(Cordero-Erausquin, 2002; Caffarelli, 2000) at the end (possibly another post to cover).

🗝️Sharpness

The assumption that $B$ is a ball is somewhat strong, but that is the price we pay to make $A$ quite flexible. It seems that without further assumption about $\rho$ or $A$, this is probably the best possible. In the following, we assume $\rho$ is standard Gaussian.

If $B$ is a perturbed ball with boundary described by $r(\theta )=1+ϵh(\theta )$ in polar coordinate, where $h$ has mean zero. Then

$${\Phi }^{\prime }(t)=t^{d-1}{\int }_{{\mathbb{S}}^{d-1}}[f(tr(\theta )\theta )-\mu (f)]r(\theta {)}^{d-1}\rho (tr(\theta ))d\theta$$

And we take derivative of $t^{1-d}{\Phi }^{\prime }(t)$, it becomes

$${\int }_{{\mathbb{S}}^{d-1}}(\theta \cdot \nabla f(tr(\theta )\theta ))r(\theta {)}^d\rho (tr(\theta ))d\theta +{\int }_{{\mathbb{S}}^{d-1}}[f(tr(\theta )\theta )-\mu (f)]r(\theta {)}^{d-1}{\rho }^{\prime }(tr(\theta ))r(\theta )d\theta$$

The first term is negative due to the property of $f$ (we can smooth out $f$ to make sure the gradient exists). The second term at $t=t^{\prime }$ becomes

$$\begin{array}{rl}& {\int }_{{\mathbb{S}}^{d-1}}[f(t^{\prime }r(\theta )\theta )-\mu (f)]r(\theta {)}^{d-1}(1+ϵh(\theta )){\rho }^{\prime }(t^{\prime }r(\theta ))d\theta \\ & =-C{t}^{\prime }{\int }_{{\mathbb{S}}^{d-1}}[f(t^{\prime }r(\theta )\theta )-\mu (f)]r(\theta {)}^{d-1}(2ϵh(\theta )+{ϵ}^2{h}^2(\theta ))\rho (t^{\prime }r(\theta ))d\theta .\end{array}$$

Since ${\rho }^{\prime }(t^{\prime }r(\theta ))\propto -\rho (t^{\prime }r(\theta ))t^{\prime }r(\theta )$. It implies that $h$ must satisfy (take $ϵ=0$)

$${\int }_{{\mathbb{S}}^{d-1}}[f(t^{\prime }\theta )-\mu (f)]h(\theta )\rho (t^{\prime })d\theta =0.$$

Otherwise, a flipping $f(v)\mapsto f(-v)$ will flip the sign. However, this means $f$ cannot be arbitrary.

References

🐻  Caffarelli, L.A. 2000. Monotonicity Properties of Optimal Transportation and the FKG and Related Inequalities. Communications in Mathematical Physics 214(3), 547–563.

🐻  Cordero-Erausquin, D. 2002. Some Applications of Mass Transport to Gaussian-type Inequalities. Archive for rational mechanics and analysis 161, 257–269.