# Comparing suprema of inner products of Gaussian variables

I'm given two i.i.d. standard normal vectors $$x, y \sim \mathcal{N}(0, I_n)$$, and vectors $$a \in \mathbb{S}^{n-1}$$, the unit sphere in $$n$$ dimensions. Additionally, given a set $$S \subseteq [n]$$, I want to prove (or disprove):

$$\mathbb{E}\left( \bigg| \sup_{a \in \mathbb{S}^{n-1}} \sum_{i = 1}^n \mathbf{1}\{i \in S\} a_i^2 x_iy_i \bigg| \right) \leq \mathbb{E}\left( \bigg| \sup_{a \in \mathbb{S}^{n-1}} \sum_{i = 1}^n a_i^2 x_iy_i \bigg| \right).$$

Are there any comparison inequalities or tools from probability that could potentially be useful here?

The rule you want is

$$\mathbf{z}$$ be any $$n$$-vector. The maximum value of $$\lambda\cdot \mathbf{z}$$ over all vectors $$\lambda$$ with non-negative components summing to unity is the maximum value of the components of $$\mathbf{z}$$.

Proof. This follows from the Hölder Inequality for $$p=1, q=\infty.$$ It has an elementary proof. Let $$i$$ be any index for which $$z_i$$ is largest. Then $$z_i - \lambda\cdot \mathbf{z} = (z_i - \lambda_1 z_1, z_i - \lambda_2 z_2, \ldots, z_i - \lambda_n z_n).$$ Since $$z_i\ge z_j$$ for all $$j$$ and $$0\le \lambda_j\le 1$$ for all $$j$$, each of the components on the right hand side is nonnegative. At least one will be zero when $$\lambda_i=1,$$ QED.

In the sequel, $$\lambda$$ will be the vector $$(a_1^2, a_2^2, \ldots, a_n^2)$$ where $$\mathbf a \in \mathbb{S}^{n-1}.$$

### Analysis

Let's simplify the expressions. Given vectors $$\mathbf{x}$$ and $$\mathbf y,$$ write $$z_i = x_i y_i.$$ Let $$P(\mathbf z)$$ be the set of coordinates $$i$$ for which $$z_i \gt 0.$$ The expression $$|\sum_{i=1}^n a_i^2 x_i y_i| = |\sum_{i=1}^n a_i^2 z_i|$$ is maximized by setting $$a_i=\pm 1$$ for some $$i\in P(\mathbf z)$$ at which $$z_i$$ is largest (and all $$a_j=0$$ for $$j\ne i$$). This yields

$$|\sup_{a\in\mathbb{S}^{n-1}} \sum_{i=1}^n a_i^2 x_i y_i| = |\max_{i\in P(\mathbf{z})} z_i| =\max_{i\in P(\mathbf{z})} z_i.\tag{1}$$

If $$P(\mathbf z)$$ is empty, this expression is maximized by setting $$a_i=1$$ for some $$i\in[n]$$ at which $$-x_iy_i$$ is smallest, producing

$$|\sup_{a\in\mathbb{S}^{n-1}} \sum_{i=1}^n a_i^2 x_i y_i| = \min_{i} (-z_i) .\tag{2}$$

Suppose $$S$$ is a proper subset of $$[n]$$ (if not, the equality of the two sides is trivially true) and is nonempty (it's easy to see the inequality holds in this case). When $$S\cap P(\mathbf z) \ne \emptyset,$$

$$\sup_{a\in\mathbb{S}^{n-1}} \sum_{i=1}^n I\{i\in S\} a_i^2 x_i y_i = \sup_{a\in\mathbb{S}^{n-1}} \sum_{i\in S}^n a_i^2 z_i\tag{3}$$

is maximized by setting $$a_i=\pm 1$$ for an $$i\in S\cap P(\mathbf z)$$ for which $$z_i$$ is largest, and it tells us

$$|\sup_{a\in\mathbb{S}^{n-1}} \sum_{i\in S}^n a_i^2 z_i| = \max_{i\in S\cap P(\mathbf z)} z_i.\tag{4}$$

Otherwise, $$z_i \lt 0$$ for all $$i\in S$$ and $$(3)$$ is maximized by setting $$a_i=0$$ for some $$i\in S$$ (and $$a_j=\pm 1$$ for some $$j\ne i$$). In this case

$$|\sup_{a\in\mathbb{S}^{n-1}} \sum_{i\in S}^n a_i^2 z_i| = 0.\tag{5}$$

When $$S \cap P(\mathbf Z)$$ is nonempty, $$(4)$$ does not exceed $$(1)$$ because it is the maximum over a subset of the values in $$(1)$$. In all other cases $$(4)$$ does not exceed $$(1)$$ and $$(5)$$ does not exceed either $$(1)$$ or $$(2).$$ Thus, no matter what the values $$\mathbf x$$ and $$\mathbf y$$ may be, the left hand side of the inequality never exceeds the right hand side. The inequality is maintained upon taking expectations.