1
$\begingroup$

Let $g = (g_1, \dots, g_d)$ be a sequence of independent standard Normal random variables, and suppose $\Sigma$ is a $d \times d$ (deterministic), real, symmetric, positive definite matrix. The Hanson-Wright inequality ensures that for $t > 0$,the upper and lower tails have the bound $$ \max\bigg(\mathbb{P}\Big\{g^T \Sigma^{-1} g - \mathrm{Tr}(\Sigma^{-1}) \geq t\Big\}, \mathbb{P}\Big\{g^T \Sigma^{-1} g - \mathrm{Tr}(\Sigma^{-1}) \leq - t\Big\}\bigg) \leq \exp\Big(-c \min\big\{\frac{t^2}{\|\Sigma^{-1}\|_F^2}, \frac{t}{\|\Sigma^{-1}\|_{\rm op}} \big\}\Big). $$ Above, $c > 0$ is a universal constant.

My question: do similar guarantees hold when $\Sigma$ is a random matrix? I am particularly interested in the case when we have $$ \Sigma = \frac{1}{n} \sum_{j=1}^n x_j \otimes x_j. $$ (We assume that $x_j \in \mathbb{R}^d$ are iid, and have law such that when $n > d$, $\Sigma$ is nonsingular almost surely.)

In this case, also assuming $x_j$ are mutually independent of $g$, I am wondering under what conditions we can expect to have exponential tail bounds for $$ \mathbb{P}_{x, g}\Big\{g^T \Sigma^{-1} g - \mathbb{E}_x[\mathrm{Tr}(\Sigma^{-1})] \geq t\Big\} \quad \mbox{or} \quad \mathbb{P}_{x,g}\Big\{g^T \Sigma^{-1} g - \mathbb{E}_x[\mathrm{Tr}(\Sigma^{-1})] \leq - t\Big\}, $$ where the probability is now over $g_1, \dots, g_d, x_1, \dots, x_n$. Any references or pointers to existing results are greatly appreciated.

$\endgroup$
4
  • $\begingroup$ What is the distribution of the $x_j$? If $P(x_j=0)>0$ then $E_x[trace\Sigma^{-1}]$ is $+\infty$. $\endgroup$
    – jlewk
    Commented Oct 20, 2021 at 1:27
  • $\begingroup$ But even somewhat mild conditions would prevent that from being true (e.g. $x_j$ has law absolutely continuous w.r.t. Lebesgue measure). This question makes sense, provided we assume that $\mathbb{E}_x[\mathrm{Tr}(\Sigma^{-1})]$ exists and is finite? $\endgroup$
    – Drew Brady
    Commented Oct 20, 2021 at 2:15
  • $\begingroup$ Absolute continuity is not suficient. If $d=1$ and $n\in\{1,2\}$ then $E[(\chi^2_n)^{-1}]=+\infty$. $\endgroup$
    – jlewk
    Commented Oct 20, 2021 at 3:38
  • $\begingroup$ The last part of the abstract of arxiv.org/abs/1912.10754 suggests the paper will have some answers to these issues, and possibly answers to your question. $\endgroup$
    – jlewk
    Commented Oct 20, 2021 at 3:42

0

You must log in to answer this question.