# Pinsker's inequality

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In information theory, Pinsker's inequality, named after its inventor Mark Semenovich Pinsker, is an inequality that bounds the total variation distance (or statistical distance) in terms of the Kullback–Leibler divergence. The inequality is tight up to constant factors.

## Formal statement

Pinsker's inequality states that, if $P$ and $Q$ are two probability distributions on a measurable space $(X,\Sigma )$ , then

$\delta (P,Q)\leq {\sqrt {{\frac {1}{2}}D_{\mathrm {KL} }(P\|Q)}},$ where

$\delta (P,Q)=\sup {\bigl \{}|P(A)-Q(A)|{\big |}A\in \Sigma {\text{ is a measurable event}}{\bigr \}}$ is the total variation distance (or statistical distance) between $P$ and $Q$ and

$D_{\mathrm {KL} }(P\|Q)=\operatorname {E} _{P}\left(\log {\frac {\mathrm {d} P}{\mathrm {d} Q}}\right)=\int _{X}\left(\log {\frac {\mathrm {d} P}{\mathrm {d} Q}}\right)\,\mathrm {d} P$ is the Kullback–Leibler divergence in nats. When the sample space $X$ is a finite set, the Kullback–Leibler divergence is given by

$D_{\mathrm {KL} }(P\|Q)=\sum _{i\in X}\left(\log {\frac {P(i)}{Q(i)}}\right)P(i)\!$ Note that in terms of the total variation norm $\|P-Q\|$ of the signed measure $P-Q$ , Pinsker's inequality differs from the one given above by a factor of two:

$\|P-Q\|\leq {\sqrt {2D_{\mathrm {KL} }(P\|Q)}}.$ A proof of Pinsker's inequality uses the partition inequality for f-divergences.

## History

Pinsker first proved the inequality with a worse constant. The inequality in the above form was proved independently by Kullback, Csiszár, and Kemperman.

## Inverse problem

A precise inverse of the inequality cannot hold: for every $\varepsilon >0$ , there are distributions $P_{\varepsilon },Q$ with $\delta (P_{\varepsilon },Q)\leq \varepsilon$ but $D_{\mathrm {KL} }(P_{\varepsilon }\|Q)=\infty$ . An easy example is given by the two-point space $\{0,1\}$ with $Q(0)=0,Q(1)=1$ and $P_{\varepsilon }(0)=\varepsilon ,P_{\varepsilon }(1)=1-\varepsilon$ . 

However, an inverse inequality holds on finite spaces $X$ with a constant depending on $Q$ . More specifically, it can be shown that with the definition $\alpha _{Q}:=\min _{x\in X:Q(x)>0}Q(x)$ we have for any measure $P$ which is absolutely continuous to $Q$ ${\frac {1}{2}}D_{\mathrm {KL} }(P\|Q)\leq {\frac {1}{\alpha _{Q}}}\delta (P,Q)^{2}.$ As a consequence, if $Q$ has full support (i.e. $Q(x)>0$ for all $x\in X$ ), then

$\delta (P,Q)^{2}\leq {\frac {1}{2}}D(P\|Q)\leq {\frac {1}{\alpha _{Q}}}\delta (P,Q)^{2}.$ 