# Multivariate Sub-Gaussian Deviation Inequality

2022-11-24

Abstract

We derive a deviation inequality for multivariate sub-Gaussians using mixture supermartingales and Ville’s inequality.

# Introduction

Consider a random process $$(X_t)_t$$ taking values $$X_t \in \mathbb R^d$$. Let us assume that $$X_t - \mu_t$$ is $$\Sigma_t$$ sub-Gaussian given the past. That is, at every $$t$$ we have for every $$\eta \in \mathbb R^d$$ that \begin{equation}\label{eq:subg} \operatorname*{\mathbb E}\left[ e^{ \left\langle\eta, X_t-\mu_t\right\rangle } \middle| \mathcal F_{t-1} \right] ~\le~ e^{\frac{1}{2} \left\|\eta\right\|^2_{\Sigma_t}} . \end{equation} Let us denote the total deviation vector from the mean by $$S_t = \sum_{s=1}^t (X_s-\mu_s)$$ and the total covariance matrix by $$V_t = \sum_{s=1}^t \Sigma_s$$. By $$\eqref{eq:subg}$$, we can make a test super-martingale with value $M_t^\eta ~=~ e^{ \left\langle\eta, S_t\right\rangle - \frac{1}{2} \left\|\eta\right\|^2_{V_t} } .$ Finally, by mixing over $$\eta$$ (here we do the normal prior case) we can make a test super-martingale with value $M_t ~=~ \operatorname*{\mathbb E}_{\eta \sim \mathcal N(0, \sigma^2 I)} \left[ M_t^\eta \right] ~=~ \int (2 \pi \sigma^2)^{-d/2} e^{ - \frac{1}{2 \sigma^2} \|\eta\|^2 + \left\langle\eta, S_t\right\rangle - \frac{1}{2} \left\|\eta\right\|^2_{V_t} } \,\textrm{d}\eta ,$ which results in $\ln M_t ~=~ \frac{1}{2} \left\|S_t\right\|^2_{\left(V_t + \sigma^{-2} I\right)^{-1}} - \frac{1}{2} \ln \det\left( I + \sigma^2 V_t \right) .$ So all in all, using Ville’s inequality, we find $\operatorname*{\mathbb P}\left\{ \exists t : \frac{1}{2} \left\|S_t\right\|^2_{\left(V_t + \sigma^{-2} I\right)^{-1}} > \ln \frac{1}{\delta} + \frac{1}{2} \ln \det\left( I + \sigma^2 V_t \right) \right\} \le \delta .$ This is a pretty nice result obtained in a clean way. We may go many ways from here. Trying to learn $$\sigma^2$$. Or trying to choose $$\sigma^2$$ of order $$1/V_t$$ so that the log-determinant disappears. Going beyond sub-Gaussian. Or making things more interpretable. Next we will look at one way to simplify the result.

# Special case of bounded deviations

In particular, if all $$X_t - \mu_t$$ are bounded in a Euclidean norm ball of radius $$B$$, then sub-Gaussianity holds with $$\Sigma_t = B^2 I$$. So with probability at least $$1-\delta$$, for all $$t$$ $\frac{1}{2 \left(t B^2 + \sigma^{-2}\right)} \left\|S_t\right\|^2 \le \ln \frac{1}{\delta} + \frac{d}{2} \ln \left( 1 + \sigma^2 B^2 t \right) .$ Setting $$\sigma = 1/B$$ then gives $\frac{1}{2 \left(t + 1\right)} \left\|S_t/B\right\|^2 \le \ln \frac{1}{\delta} + \frac{d}{2} \ln \left( t + 1 \right) .$