A homogenization principle for total variation

Abstract

A homogenization principle for total variation We prove an inequality comparing the variational distance between pairs of product probability measures to its homogenized counterpart. If $P_1,\dots ,P_n,Q_1,\dots ,Q_n$ are arbitrary probability measures on a measurable space and $\bar{P}:=\frac{1}{n}{\sum }_{i=1}^n{P}_i,\bar{Q}:=\frac{1}{n}{\sum }_{i=1}^n{Q}_i$, we show that $$TV\left(\underset{i=1}{\overset{n}{⨂}}{P}_i,\underset{i=1}{\overset{n}{⨂}}{Q}_i\right)\ge cTV({\bar{P}}^{\otimes n},{\bar{Q}}^{\otimes n}),$$ where $c>0$ is a universal constant. The proof is based on a one-dimensional representation of total variation between products. We embed pairs of probability distributions $P_i,Q_i$ into positive measures $\hat{I}{\cdot }_i$ on $\mathbb{R}$. We then define a functional $T$ over measures on $\mathbb{R}$ that realizes TV over products via convolution: $TV\left({⨂}_{i=1}^n{P}_i,{⨂}_{i=1}^n{Q}_i\right)=T(\hat{I}{\cdot }_1\ast \cdots \ast \hat{I}{\cdot }_n)$. Our main analytic discovery is that for the relevant class of positive measures $\hat{I}{\cdot }_i$, the convolution inequality $T(\hat{I}{\cdot }_1\ast \cdots \ast \hat{I}{\cdot }_n)\ge cT\left(\bar{\hat{I}}{\cdot }^{\ast n}\right)$ holds, where $\bar{\hat{I}}\cdot =\frac{1}{n}{\sum }_{i=1}^n\hat{I}{\cdot }_i$. Finally, a higher-dimensional lifting argument shows that $T\left(\bar{\hat{I}}{\cdot }^{\ast n}\right)\ge TV({\bar{P}}^{\otimes n},{\bar{Q}}^{\otimes n})$. To our knowledge, both the exact representation and the convolution inequality are new.