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.
[1] Pinsker's inequality states that, if
are two probability distributions on a measurable space
, then where is the total variation distance (or statistical distance) between
is a finite set, the Kullback–Leibler divergence is given by Note that in terms of the total variation norm
of the signed measure
, Pinsker's inequality differs from the one given above by a factor of two: A proof of Pinsker's inequality uses the partition inequality for f-divergences.
Note that the expression of Pinsker inequality depends on what basis of logarithm is used in the definition of KL-divergence.
(logarithm in base
is typically defined with
(logarithm in base 2).
Then, Given the above comments, there is an alternative statement of Pinsker's inequality in some literature that relates information divergence to variation distance: i.e., in which is the (non-normalized) variation distance between two probability density functions
[2] This form of Pinsker's inequality shows that "convergence in divergence" is a stronger notion than "convergence in variation distance".
A simple proof by John Pollard is shown by letting
: Here Titu's lemma is also known as Sedrakyan's inequality.
Note that the lower bound from Pinsker's inequality is vacuous for any distributions where
, since the total variation distance is at most
For such distributions, an alternative bound can be used, due to Bretagnolle and Huber[3] (see, also, Tsybakov[4]): Pinsker first proved the inequality with a greater constant.
The inequality in the above form was proved independently by Kullback, Csiszár, and Kemperman.
[5] A precise inverse of the inequality cannot hold: for every
An easy example is given by the two-point space
[6] However, an inverse inequality holds on finite spaces
[7] More specifically, it can be shown that with the definition
which is absolutely continuous to
has full support (i.e.