91 Communication Complexity — Pinsker
91.1 Overview
This module develops a self-contained proof of Pinsker’s inequality, which bounds the total variation distance between two probability measures by their KL divergence. The argument proceeds via the Radon-Nikodym density and a variational characterisation of total variation, culminating in \(2 \cdot \mathrm{TV}(\mu ,\nu )^2 \le \mathrm{KL}(\mu \| \nu )\).
91.2 Declarations
For two probability measures \(\mu \) and \(\nu \) on a measurable space \(\Omega \), the Radon-Nikodym density \(\mathrm{rnDensity}(\mu ,\nu )(x) \in \mathbb {R}\) is the real part of the Radon-Nikodym derivative \(\frac{d\mu }{d\nu }(x)\), obtained by converting the extended non-negative real value \(\mu .\mathrm{rnDeriv}\, \nu \, x\) to a real number via ENNReal.toReal.
For all \(x \in \Omega \), \(\mathrm{rnDensity}(\mu ,\nu )(x) \ge 0\).
The function \(x \mapsto \mathrm{rnDensity}(\mu ,\nu )(x)\) is measurable.
The function \(\mathrm{rnDensity}(\mu ,\nu )\) is integrable with respect to \(\nu \).
If \(\mu \ll \nu \), then \(\int \mathrm{rnDensity}(\mu ,\nu )(x)\, d\nu = 1\).
The function \(x \mapsto \mathrm{rnDensity}(\mu ,\nu )(x) - 1\) is integrable with respect to \(\nu \).
The function \(x \mapsto |\mathrm{rnDensity}(\mu ,\nu )(x) - 1|\) is integrable with respect to \(\nu \).
If \(\mu \ll \nu \), then \(\int (\mathrm{rnDensity}(\mu ,\nu )(x) - 1)\, d\nu = 0\).
If \(\mu \ll \nu \) and \(S \subseteq \Omega \) is measurable, then
Let \(\mu \) be a finite measure and \(g : \Omega \to \mathbb {R}\) measurable, integrable, and mean-zero (\(\int g\, d\mu = 0\)). Then
Let \(\mu \) be a finite measure and \(g\) measurable, integrable, and mean-zero. Then
The quantity \(\mathrm{densityAbsIntegral}(\mu ,\nu ) := \int |\mathrm{rnDensity}(\mu ,\nu )(x) - 1|\, d\nu \) is the \(L^1(\nu )\)-distance of the Radon-Nikodym density from \(1\).
The density positive set is \(\{ x \in \Omega : \mathrm{rnDensity}(\mu ,\nu )(x) \ge 1\} \), the region where \(\mu \) locally dominates \(\nu \).
The density positive set \(\{ x : \mathrm{rnDensity}(\mu ,\nu )(x) \ge 1\} \) is a measurable subset of \(\Omega \).
The density positive integral is \(\mathrm{densityPositiveIntegral}(\mu ,\nu ) := \int _{\mathrm{densityPositiveSet}} (\mathrm{rnDensity}(\mu ,\nu )(x)-1)\, d\nu \), the excess mass of \(\mu \) over \(\nu \) on the region where \(\mu \) dominates.
If \(\mu \ll \nu \), then \(\mathrm{tvDistanceSup}(\mu ,\nu ) = \mathrm{densityPositiveIntegral}(\mu ,\nu )\).
If \(\mu \ll \nu \), then \(\mathrm{TV}(\mu ,\nu ) = \mathrm{densityPositiveIntegral}(\mu ,\nu )\).
If \(\mu \ll \nu \), then \(\mathrm{densityPositiveIntegral}(\mu ,\nu ) = \tfrac {1}{2}\, \mathrm{densityAbsIntegral}(\mu ,\nu )\).
For real numbers \(u \ge 0\) and \(y\), one has \(u \cdot y \le \mathrm{klFun}(u) + e^y - 1\), where \(\mathrm{klFun}(u) = u \ln u - u + 1\).
Let \(\mu \) be a probability measure, \(X : \Omega \to \mathbb {R}\) and \(t \in \mathbb {R}\). If \(e^{tX}\) is \(\mu \)-integrable, then \(\int e^{t X(x) - \Lambda _\mu (t)}\, d\mu = 1\), where \(\Lambda _\mu (t) = \log \int e^{tX}\, d\mu \) is the cumulant generating function.
If \(e^{tX}\) is \(\mu \)-integrable, then so is \(x \mapsto e^{t X(x) - \Lambda _\mu (t)}\).
Let \(f, X : \Omega \to \mathbb {R}\) with \(f \ge 0\) and \(\int f\, d\mu = 1\). Under suitable integrability conditions on \(f\), \(X\), and \(e^{tX}\),
Let \(f : \Omega \to \mathbb {R}\) with \(f \ge 0\), \(\int f\, d\mu = 1\), \(f\) integrable, and \(\mathrm{klFun}(f)\) integrable. Then
If \(\mu \ll \nu \), then \(\mathrm{TV}(\mu ,\nu ) = \tfrac {1}{2}\, \mathrm{densityAbsIntegral}(\mu ,\nu )\).
If \(\mu \ll \nu \) and the log-likelihood ratio \(\mathrm{llr}(\mu ,\nu )\) is \(\mu \)-integrable, then
If \(\mu \ll \nu \) and \(\mathrm{llr}(\mu ,\nu )\) is \(\mu \)-integrable, then
If \(\mu \ll \nu \) and \(\mathrm{llr}(\mu ,\nu )\) is \(\mu \)-integrable, then
If \(\mu \ll \nu \) and \(\mathrm{llr}(\mu ,\nu )\) is \(\mu \)-integrable, then
If \(\mu \ll \nu \) and \(\mathrm{llr}(\mu ,\nu )\) is \(\mu \)-integrable, then
in \(\mathbb {R}_{\ge 0}^\infty \).
If \(\mu \ll \nu \) (without any integrability assumption), then \(\mathrm{ofReal}(2\, \mathrm{TV}(\mu ,\nu )^2) \le \mathrm{KL}(\mu \| \nu )\) in \(\mathbb {R}_{\ge 0}^\infty \).
For any two probability measures \(\mu \) and \(\nu \) on \(\Omega \),
where the inequality is in \(\mathbb {R}_{\ge 0}^\infty \) and KL divergence uses natural logarithms.
If \(\mathrm{KL}(\mu \| \nu ) \ne \infty \), then
providing a real-valued form of Pinsker’s inequality convenient when finiteness of the KL divergence is already known.