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Finite KL nonnegativity for subprobability references

Statement

Proves ln⁡t≤t−1 for t>0 and, for finite probability weights p and positive-on-support reference weights q with ∑iqi≤1, proves ∑i:pi>0pilog⁡2(pi/qi)≥0 by an explicit finite argument without choice.

Facts & Assumptions

Given: A finite probability vector p=(pi) and nonnegative reference weights qi with qi>0 whenever pi>0 and ∑iqi≤1.

[F1]

Finite-block information and entropy in bits: log⁡2t=ln⁡t/ln⁡2 and ln⁡2>0.

[F2]

The integral logarithm L(x):=∫1xdtt for x>0: L(t)=∫1tdu/u, with oriented limits.

[F4]

If f≤g on [a,b] and both are integrable then ∫abf≤∫abg; and m(b−a)≤∫abf≤M(b−a): if integrable functions obey a pointwise order on an interval with increasing endpoints, their integrals obey the same order.

[F5]

The natural logarithm as the inverse of the exponential function: log⁡ is the natural logarithm, the inverse of exp⁡.

Proof

Given: The finite weights and hypotheses above.

1.1F2F3F4F5F6F7algebra

The integral logarithm is the natural logarithm by [F3] and [F5]. If t>1, then 1/u≤1 on [1,t], so [F2]–[F4] give ln⁡t=∫1tdu/u≤t−1. If 0<t<1, then 1/u≥1 on [t,1], so ∫t1du/u≥1−t; [F3] and [F7] give ln⁡t=−∫t1du/u≤t−1. At t=1, [F6] gives equality. Therefore ln⁡t≤t−1 for all t>0.

2.1F1step 1.1givenalgebra∎

Apply step 1.1 to t=qi/pi for every i with pi>0. Then ∑i:pi>0piln⁡(pi/qi)=−∑i:pi>0piln⁡(qi/pi)≥∑i:pi>0(pi−qi)=1−∑i:pi>0qi≥0. Dividing by ln⁡2>0 gives the same nonnegativity for the bit-log sum. Thus finite KL nonnegativity against a subprobability reference, as well as the logarithm inequality used to prove it, follows from finite sums and elementary integral bounds; no choice axiom enters.

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