Alphabeta Math
TheoremStatement: AI-adaptedProof: AI-adaptedPipeline-generatedjudge pass (gpt-6-luna)
How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

  • Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
  • AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
  • AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

ReLU backward gradient-variance recurrence

Statement

Consider a finite rectified stack in a fully connected layered perceptron, with widths d0,…,dL, zero biases, Zℓ=WℓXℓ−1, and Xℓ=ρ(Zℓ), where ρ(t)=max⁡(t,0). For an objective J differentiable as a function of the last activation, set gL=∇XLJ(XL) and define backward signals by the zero-at-kink convention γℓ,j=1Zℓ,j>0, hℓ=γℓ⊙gℓ, and gℓ−1=WℓThℓ. At realizations where all relevant preactivations are nonzero, these are ordinary chain-rule gradients. At zeros they are specified backpropagation signals; no ordinary derivative of ReLU at zero is asserted.

Without any independence assumptions, fix a layer ℓ and suppose every Tji=Wℓ,jiγℓ,jgℓ,j is square-integrable. Then E[gℓ−1,i]=∑jE[Tji],Var⁡(gℓ−1,i)=∑jVar⁡(Tji)+2∑j<kCov⁡(Tji,Tki). This is an exact atom-aware identity for the realized network signals. For any specified bounded measurable random gate vector qℓ in place of γℓ, define Tji=Wℓ,jiqℓ,jgℓ,j. If every redefined Tji is square-integrable, the same mean and covariance formulas hold.

For the conditional recurrence, additionally assume at the layers in question that the entries of Wℓ are independent with a common symmetric centered law of finite variance σℓ2, that Zℓ,j is symmetric with common zero probability pℓ, and that the coordinates of gℓ have common finite second moment Gℓ. Assume the gate vector γℓ is independent of gℓ and Wℓ is independent of hℓ. Write the fan-out as n^ℓ=dℓ. Then E[gℓ−1,i]=0,Gℓ−1=Var⁡(gℓ−1,i)=1−pℓ2n^ℓσℓ2Gℓ. If these hypotheses hold for every r≤ℓ≤L, then Gr−1=GL∏ℓ=rL1−pℓ2n^ℓσℓ2. In particular, the source's atomless conditional case pℓ=0 gives Var⁡(gℓ−1,i)=12n^ℓσℓ2E[gℓ,12] and the corresponding product of half factors. If the downstream coordinate is centered, Gℓ is its variance. The conditional fan-out choice σℓ2=2/n^ℓ makes each atomless product factor one.

These conditional identities do not establish that a nonzero actual-network gradient satisfies their hypotheses. Indeed, if ℓ≥2, Xℓ−1 is a ReLU output and σℓ2>0, the assumption Wℓ⊥hℓ already forces hℓ=0 almost surely. If also pℓ<1 and gate independence holds, it forces Gℓ=0. Thus the nonzero atomless backward preservation model with these full independence assumptions is infeasible for such an actual ReLU layer; the conditional implication and product remain valid, and the general covariance identity above supplies the unconditional finite-network formula.

Facts & Assumptions

Given: The finite stack, an objective differentiable as a function of XL with gL=∇XLJ(XL), and the backward-signal convention; square-integrable summands for the general identity; and, only for the conditional formulas, the stated symmetry, moments, and independence assumptions.

[F1]

A fully connected layered perceptron evaluates each layer as a finite weighted sum followed by its specified activation. (Finite feedforward computation graphs and layered MLPs)

[F2]

At totally differentiable points, derivatives of a composition multiply in layer order. (The chain rule for total derivatives: D(g∘f)(a)=Dg(f(a))∘Df(a))

[F3]

A symmetric preactivation with zero probability p has positive sign probability (1−p)/2. (Second moments of symmetric rectifiers)

[F4]

Independent centered weights, independent of the input vector, give weighted-sum mean zero and variance equal to the sum of weight variances times input second moments. (Variance of a centered random weighted sum)

[F5]

Products of independent integrable variables have factored expectations. (Expectations factor over finite products of independent random variables)

[F6]

Variance and covariance are centered second moments. (Moments, variance, and covariance on a probability space)

[F7]

Products of square-integrable variables are integrable by Cauchy–Schwarz. (Cauchy-Schwarz for random variables)

[F8]

A finite independent family has product joint law. (Independent random elements have product joint law)

[F9]

Expectation is linear on integrable random variables, and respects their almost-sure order. (Linearity, monotonicity, and the modulus bound for expectation)

Proof

1.1givenF1F2

At a realization with nonzero preactivations, the derivative of ReLU is the diagonal matrix with entries γℓ,j, and the affine derivative is Wℓ. The chain rule [F2] therefore gives the two backward identities in the Statement. The specified convention defines the same identities at zeros without using [F2] there.

1.2givenF6F7F9algebra

For fixed i, gℓ−1,i=∑jTji. A finite sum of square-integrable variables is square-integrable: (∑jTji)2≤dℓ∑jTji2. By [F7] each pairwise product is integrable. Expanding the mean and centered square, then applying [F6] and expectation linearity [F9], gives exactly the asserted mean and covariance sums, without factoring any dependent variables.

1.3givenF3F4F5

Under the conditional assumptions, [F3] and gate independence give E[hℓ,j2]=E[γℓ,j]E[gℓ,j2]=(1−pℓ)Gℓ/2 by [F5]. For fixed i, the column of weights is centered, independent, and independent of hℓ. Applying [F4] yields mean zero and variance σℓ2∑jE[hℓ,j2]=((1−pℓ)/2)dℓσℓ2Gℓ. This also equals the upstream second moment, proving the one-step formula.

2.1step 1.3algebra

Repeat step 1.3 for the finitely many layers L,L−1,…,r. All generated upstream coordinates are centered and have the common second moment in step 1.3, so substitution gives the stated product, including a possibly uncentered terminal signal. Setting pℓ=0 gives the original half-factor formulas and setting σℓ2=2/dℓ makes those factors one.

3.1givenF3F5F8step 1.3algebra∎

To prove the feasibility obstruction, let q=P(Wℓ,11<0). Symmetry and positive variance imply q>0: if q=0, symmetry gives P(Wℓ,11>0)=0, forcing variance zero. By [F8], the event E that all dℓdℓ−1 matrix entries are strictly negative has probability qdℓdℓ−1>0. Since Xℓ−1 is coordinatewise nonnegative, all coordinates of Zℓ are nonpositive on E, so γℓ=0 and hℓ=0 there, including when the input vector is zero. Independence of Wℓ and hℓ now gives 0=P(E∩{hℓ≠0})=P(E)P(hℓ≠0), hence hℓ=0 almost surely. If pℓ<1, step 1.3's gate calculation gives 0=E[hℓ,j2]=(1−pℓ)Gℓ/2, hence Gℓ=0. This proves the obstruction without declaring the valid conditional identities false.

Remarks

He et al. §2.2, Eqs. (14)–(16), multiply the atomless half factors under the stated weight/incoming-gradient and gate/downstream-gradient independence assumptions. Those are a conditional moment calculation, not a proof of their realization by a nonzero finite ReLU gradient. Zero atoms also invalidate the ordinary derivative at those realizations; choosing the zero-at-kink convention addresses the algorithmic signal but does not establish the independence assumptions.

Depends on

Used by

Dependency tree · two levels

35 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources