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KL divergence between equal-variance Gaussian location laws

Statement

Assume AC and let a,b∈R and τ>0. For the normal location laws with common variance τ2, D(N(a,τ2)∥N(b,τ2))=(a−b)22τ2.

Facts & Assumptions

Given: AC, real means a,b, and a positive scale τ.

[A1]

The Axiom of Choice: AC supplies the normal-law construction and the Radon–Nikodym assumptions used by the KL definition.

[F1]

Standard normal and normal laws: N(a,τ2) is the law of a+τZ when Z has standard normal law, whose density is ϕ(z)=(2π)−1/2e−z2/2.

[F2]

The standard normal density has total mass one: ϕ is a probability density on R.

[F3]

AC supplies countable selections and prescribed serial paths: AC supplies countable choice, the axiom required by the change-of-variables supplier.

[F4]

A C^1 diffeomorphism satisfies the change-of-variables formula for nonnegative Lebesgue measurable functions: for a C1 diffeomorphism T and nonnegative measurable g, ∫g(y) dy=∫g(T(x))∣T′(x)∣ dx.

[F5]

Integrating against a density agrees with integrating the product: ∫g d(f dμ)=∫gf dμ for nonnegative measurable f,g.

[F6]

The natural logarithm as the inverse of the exponential function: the natural logarithm is the inverse of the real exponential, so log⁡(exp⁡y)=y. For positive u,v, log⁡(u/v)=log⁡u−log⁡v (Order, continuity, range, and the product, quotient, and reciprocal laws for the natural logarithm).

[F7]

Multivariate normal law, including singular covariance: a normal law with mean a has expectation a; in dimension one this is the scalar law N(a,τ2).

[F8]

Kullback–Leibler divergence: for P≪Q, D(P∥Q)=∫log⁡(dP/dQ) dP.

Proof

1.1A1F1F2F3F4F5givenconstruct

Set pa(x)=1τ2πexp⁡(−(x−a)22τ2). Apply [F4] to gpa with T(z)=a+τz; since τpa(a+τz)=ϕ(z) by [F1], for every nonnegative Borel g, ∫g(x)pa(x) dx=∫g(a+τz)ϕ(z) dz. Thus N(a,τ2) has Lebesgue density pa; the same calculation gives pb for N(b,τ2). Both are strictly positive. Let r=pa/pb. By [F5], ∫Er dN(b,τ2)=∫Epa dx=N(a,τ2)(E) for every Borel set E, so r=dN(a,τ2)/dN(b,τ2).

1.2F1F6F7algebra

For every x, cancel the common positive prefactor of pa and pb and apply the quotient law and inverse identity in [F6] to obtain log⁡pa(x)pb(x)=(x−b)2−(x−a)22τ2=(a−b)(2x−a−b)2τ2. This affine function is integrable under N(a,τ2), whose mean is a by [F7].

2.1A1F8step 1.2algebra∎

By [F8] and step 1.2, D(N(a,τ2)∥N(b,τ2))=∫(a−b)(2x−a−b)2τ2 N(a,τ2)(dx)=(a−b)(2a−a−b)2τ2=(a−b)22τ2. The identity also covers a=b, when both sides are zero.

Depends on

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