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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?

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1 Answer 1

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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.

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