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Let $X_1, \dots, X_n$ be random real-valued symmetric rank-one matrices, $$ X_i = x_i \otimes x_i, $$
where $x_i$ are such that the standard Euclidean norm satisfies $\|x_i\|^2 \leq a$ almost surely. Assume that $x_i$ are independently and identically distributed and let $M = \mathbb{E}[X_i] = \mathbb{E}[x_i \otimes x_i]$, denote their common mean.

Define their average $\overline{X}_{n} = n^{-1} S_n$ where $S_n = \sum_{i=1}^n X_i$.

Let $f(T) := \mathrm{trace}((T + I)^{-1})$.

Question: What is the smallest constant $C = C_d(a, n) \geq 1$ such that we have $$ \mathbb{E}[f(\overline{X}_n)] \leq C~f(M)? $$

It should be emphasized that the constant $C$ is universal: it is valid for any law of $x_i$, supported on the Euclidean ball of (squared) radius $a$. It should be dependent only on $a, n$ and the dimension $d$.

Comments:

  • Necessarily $C \geq 1$. Note that by Jensen's inequality, we have the following inequality, $ \mathbb{E}[f(\overline{X}_n)] \geq f(M), $ since $f$ is a convex function on the symmetric positive definite matrices.

  • In the case $d = 1$, I was able to solve this problem and calculate the extremal distribution. See this post.

Drew Brady
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    After some thought, one can check that the extremal distribution lies in the following family: $$ \pi = (1- \lambda) \delta_0 + \tfrac{\lambda}{2^d} \sum_{x \in {-1, 1}^d} \delta_{\sqrt{a} U \mathrm{diag}(\beta) x}, $$ where $U$ is a fixed orthogonal matrix, $\beta$ is a vector in the probability simplex, and $\lambda \in [0, 1]$. The support of this family is a set of size $2^{d} + 1$. In the case $d = 1$, then under $x \sim \pi$, the variable $x^2/a$ is Bernoulli $\lambda$, which is precisely the family which extremes in $d = 1$ as argued in the post above. – Drew Brady Feb 24 '23 at 07:32
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    I still don't see an easy way to compute the optimal constant, though. – Drew Brady Feb 24 '23 at 07:54
  • Sorry that I ask a question instead of providing an answer, but which version of (matrix-valued?) Jensen's inequality did you use in the first comment? Could you provide a reference? Thanks in advance! – IljaKlebanov Mar 01 '23 at 19:41
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    It follows from the fact that map $X \mapsto \mathrm{tr}(X^{-1})$ is convex on the space of (symmetric) positive definite matrices. You're welcome to use whatever favorite version of vector-valued Jensen you know applied to $\mathrm{vec}(X)$, if you like. – Drew Brady Mar 01 '23 at 23:44
  • Thanks! Somehow I got the wrong idea that you were comparing spd matrices in the Loewner order or similar, but yeah, of course, the trace makes everything real-valued.. – IljaKlebanov Mar 02 '23 at 09:45

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