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How to prove the following known (Pinsker's) inequality?

For two strictly positive sequences $(p_i)^n_{i=l}$ and $(q_i)^n_{i=l}$ with $\sum_{i=1}^np_i=\sum_{i=1}^nq_i=1$ one has $$\sum_{i=1}^np_i\log\frac{p_i}{q_i}\ge \frac{1}{2}\left(\sum_{i=1}^n|p_i-q_i|\right)^2.$$

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    Retagged due to the relation to [Kullback-Leibler_divergence](http://en.wikipedia.org/wiki/Kullback%E2%80%93Leibler_divergence)comments may only be edited for 5 minutes(click on this box to dismiss)2012-04-04
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    If it is known, does it have a name? Where did you find it and was a reference to a proof missing?2012-04-04
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    @Aryabhata yes it is known. Pinsker's inequality: http://mathoverflow.net/questions/42667/two-reference-requests-pinskers-inequality-and-pontryagin-duality2012-04-04
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    @Kolmo: Please add an answer. Also, my point was OP knows it is known, and if OP knew the name, they could have done some research before posting it here.2012-04-04
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    Check also Beck & Teboulle 2003, "Mirror descent and nonlinear projected subgradient methods for convex optimization", Proposition 5.1 for elementary proof of a weaker inequality with symmetrized KL-divergence.2015-09-15

2 Answers 2

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See here. Pinsker's inequality

https://mathoverflow.net/questions/42667/two-reference-requests-pinskers-inequality-and-pontryagin-duality

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    Is there a simpler proof?2012-04-04
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    I find a simpler one from Borwein\& Levis, 2005, page 63.2012-04-10
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    @Kolmo can you please point out the condition for equality in Pinsker's inequality?2016-11-12
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Take a look at Tsybakov 2009 Introduction to Nonparametric Estimation, p.88, also available in the net. Good and simple proof.