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Second moments of symmetric rectifiers

Statement

Let Z be a real random variable whose law is symmetric about zero, meaning that Z and −Z have the same distribution. Write p0=P(Z=0). Define the ReLU function ρ(t)=max⁡(t,0) and, for a∈R, define ϕa(t)=t for t>0 and ϕa(t)=at for t≤0. Then P(Z>0)=P(Z<0)=(1−p0)/2. If E[Z2]<∞, the following ReLU moment identities hold: E[ρ(Z)]=12E[∣Z∣],E[ρ(Z)2]=12E[Z2]. The corresponding PReLU identities are E[ϕa(Z)]=1−a2E[∣Z∣],E[ϕa(Z)2]=1+a22E[Z2]. In particular, Var⁡(ρ(Z))=12E[Z2]−14(E[∣Z∣])2. For any specified value c∈R at the kink, put ηa,c(t)=1t>0+a1t<0+c1t=0. Then, without a moment hypothesis on Z, E[ηa,c(Z)2]=1+a22(1−p0)+c2p0. The zero-at-kink ReLU backpropagation gate is η0,0, with second moment (1−p0)/2. It agrees almost surely with the ordinary ReLU derivative if p0=0. A specified value at zero is a backpropagation convention, and is not an ordinary derivative there unless a=1 and c=1.

Facts & Assumptions

Given: A symmetric real random variable Z; finite second moment for the activation-moment conclusions; and the specified finite gate values a,c.

[F1]

Expectations of measurable functions of a random element are integrals against its law. (Change of variables for expectation)

[F2]

Expectation is the Lebesgue integral on the probability space. (Expectation of a nonnegative or integrable random variable)

[F3]

Variance is Var⁡(Y)=E[(Y−E[Y])2] for square-integrable Y. (Moments, variance, and covariance on a probability space)

[F4]

A square-integrable random variable has integrable products by Cauchy–Schwarz; in particular E[∣Z∣]≤(E[Z2])1/2. (Cauchy-Schwarz for random variables)

[F5]

Expectation is linear in scalar multiples and binary sums of integrable random variables; repeated use gives the finite-sum rule. (Linearity, monotonicity, and the modulus bound for expectation)

Proof

1.1givenF1F2

Symmetry gives P(Z>0)=P(Z<0), and these probabilities sum to 1−p0, so each is (1−p0)/2. The gate takes the values 1,a,c on the positive, negative, and zero events, respectively, so its squared expectation is ((1+a2)/2)(1−p0)+c2p0. For the remaining moment claims assume E[Z2]<∞. Applying [F1] and [F2] to t↦t21t>0 and t↦t21t<0 shows that the positive-side and negative-side second moments are equal. Their sum is E[Z2], since the contribution at zero is zero, so each is 12E[Z2].

1.2givenF1F4F5algebra

By [F1], symmetry gives E[Z]=E[−Z], while scalar linearity [F5] gives E[−Z]=−E[Z], hence E[Z]=0. The expectation E[∣Z∣] is finite by [F4]. The positive and negative parts of Z have equal expectations by symmetry and sum to ∣Z∣, so linearity [F5] gives each expectation as 12E[∣Z∣]. Therefore E[ρ(Z)]=12E[∣Z∣] and E[ϕa(Z)]=E[Z1Z>0]+aE[Z1Z<0]=1−a2E[∣Z∣], using [F5] for the sum and scalar factor.

2.1step 1.1algebra

The identities ρ(Z)2=Z21Z>0 and ϕa(Z)2=Z21Z>0+a2Z21Z<0 now give the two asserted second moments from step 1.1.

3.1step 2.1step 1.2F3F5algebra∎

Expanding the variance definition [F3] using expectation linearity [F5] gives Var⁡(ρ(Z))=E[ρ(Z)2]−(E[ρ(Z)])2. Substitution of steps 2.1 and 1.2 proves the stated centered-variance identity and completes all formulas.

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