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The Gaussian activation normalizer is finite and positive

Statement

Assume AC. Let Z have the standard normal law on R. Suppose σ:R→R is analytic, nonpolynomial, satisfies ∣σ(0)∣≤B for some finite B, and is L-Lipschitz for a finite L. Then 0<E[σ(Z)2]<∞. Consequently the normalization constant

cσ:=(E[σ(Z)2])−1

is a well-defined positive finite real number.

Facts & Assumptions

Given: AC, the activation σ, its constants B,L, and a standard normal random variable Z.

[A1]

AC is assumed by the standard-normal-law and Gaussian-moment suppliers. The Axiom of Choice

[F1]

Proof

technique · bound the activation by an affine function of $|Z|$, then use continuity and nonpolynomiality for strict positivity
1.1A1F1givenalgebra

Lipschitz continuity gives ∣σ(z)∣≤B+L∣z∣, hence σ(z)2≤2B2+2L2z2. By [F1], Eσ(Z)2≤2B2+2L2<∞.

1.2A1F1givenchoose

A nonpolynomial analytic function is not identically zero, so choose z0 with σ(z0)≠0. Continuity gives an open interval around z0 on which ∣σ∣≥∣σ(z0)∣/2. The standard normal density is strictly positive on that interval, so its integral of σ2 is positive.

2.1step 1.1step 1.2algebra∎

The expectation is both finite and strictly positive by steps 1.1 and 1.2, so its reciprocal exists and is positive and finite.

Depends on

Used by

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