# Intuition behind Gaussian isoperimetric inequality

I was wondering whether or not there's an intuitive way of understanding the Gaussian isoperimetric inequality. I have been studying the Classical isoperimetric inequality and I finally understand it. I want to move to advance isoperimetric inequalities. I am interested in the Gaussian isoperimetric as it seems to have nice and practical applications in information theory.

I have no background in measure theory, but I understand that the concept of measure is a generalization of the notions of length, area and volume. I also understand that the Gaussian measure is a probability measure, meaning that it has the additional property of being normalized.

I've also looked at the definition of half spaces. I understand what a half space is. Most resources I've found do not explain the intuition behind the inequality, like the Wikipedia page , they simply provide the definition which is not easy not to understand.

How do you interpret the inequality?

The screenshots below come from the book Mathematical Foundations of Infinite-Dimensional Statistical Models.

The classical isoperimetric inequality is well-known: given a fixed perimeter, a circle achieves the largest area, or, given a fixed area, a circle achieves the smallest perimeter among other shapes. If we recall that perimeter can be seen as the derivative of the area, this is like saying

$$\mu(C+\epsilon O_2)\le\mu(A+\epsilon O_2)$$ where $$\mu$$ is Lebesgue measure, $$A$$ is a measurable set, $$C$$ is a circle with $$\mu(C)=\mu(A)$$, $$O_2$$ is the 2D unit disk, and the "$$+$$" is Minkowski addition. To see why, one can subtract $$\mu(C)=\mu(A)$$ from both two sides, divide them by $$\epsilon$$, and let $$\epsilon\to0^+$$, then he/she will get the usual form of isoperimetric inequality.

It turns out that this form of isoperimetric inequality is more convenient to generalize: we can allow higher dimensions, Riemannian manifolds equipped with some geodesic distances other than $$\mathbb{R}^2$$, or other measures. For example, we can have

where $$A_\epsilon\triangleq A+\epsilon O_n$$ and same for $$C_\epsilon$$.

We can interpret isoperimetric inequalities from the perspective of concentration inequality: it answers that if you perturb a set with some ϵ in distance/metric, how large its size will (at least) change. In the context of probabilistic measures, the change of "size" becomes a probability.

We need the following lemma to have a better understanding of or to prove Gaussian isoperimetric inequality:

where $$\gamma_n$$ is the standard Gaussian measure on $$\mathbb{R}^n$$. One can verify the claim via doing simulation for, say $$n=1$$, by projecting points uniformly distributed on $$\sqrt{m}S^{m+1}$$ onto $$\mathbb{R}^1$$. If in Python:

import numpy as np
import matplotlib.pyplot as plt

def runif_s(n_samples, n, m):
rnorm = np.random.randn(n_samples, n + m + 1)
return np.sqrt(m) * rnorm / np.linalg.norm(rnorm, axis=1)[..., np.newaxis]

def proj_hist(data, **kwargs):
n = 1
plt.figure()
plt.hist(data[:, :n], density=True, **kwargs)
plt.title('m = %d' % (data.shape[1] - n - 1))

if __name__ == '__main__':
n = 1
n_figures = 5
n_samples = 10**4
[proj_hist(runif_s(n_samples, n, int(m)), bins='auto') for m in np.logspace(0, 2, n_figures)]
plt.show()


And you may see the projected distribution is visually close to the normal distribution when $$m$$ grows large:

Let's return to Gaussian isoperimetric inequality. The inequality is a version of isoperimetric inequality w.r.t. Gaussian measure, by which we roughly mean, finding the counterpart of a "circle" under Gaussian measure. Recall that we already have an isoperimetrc inequality for $$S^{m+n}$$ w.r.t. Lebesgue measure (Theorem 2.2.1), and a relation between this measure and the standard Gaussian measure on $$\mathbb{R}^n$$ (Lemma 2.2.2), so all we have to do is to project the cap on $$S^{m+n}$$ back to $$\mathbb{R}^n$$, and let $$m\to\infty$$. For the convenience of doing projection, we choose the cap "perpendicular" to $$\mathbb{R}^n$$. If we look into the case of $$n=1$$ to gain intuition, the cap will be symmetric around $$\mathbb{R}^1$$ with the pole lying at $$-\sqrt{m}$$.

Let's say we now have a measurable set $$A$$ on $$\mathbb{R}^1$$, then we can find the cap $$C$$ on $$S^{m+1}$$ with $$\mu(C)=\gamma_1(A)$$. The projection of the cap onto $$\mathbb{R}^1$$ is the interval $$[-\sqrt{m},b(m)]$$ for some $$b(m)$$. By taking $$m\to\infty$$, the interval becomes $$(-\infty,b(\infty)]$$, and according to Lemma 2.2.2, we know that $$b(\infty)=\Phi^{-1}(\gamma_1(A))$$, giving the "circle" w.r.t. $$\gamma_1$$ being $$\{x:x\le \Phi^{-1}(\gamma_1(A))\}$$.

The proof of the general Gaussian isoperimetric inequality follows the same intuition, except that we need to replace the ray $$\{x:x\le \Phi^{-1}(\gamma_1(A))\}$$ with the hyperplane $$\{x:\langle x,u \rangle\le\Phi^{-1}(\gamma_n(A))\}$$, where $$u$$ is an arbitrary unit vector in $$\mathbb{R}^n$$. Finally we will have

and its countably-infinite-dimensional version

where $$\mathcal{C}$$ is the cylindrical $$\sigma$$-algebra on $$\mathbb{R}^\mathbb{N}$$.