Alphabeta Math
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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.

Initialization and signal propagation

1 · Prerequisites

2 · Logical flowchart

3 · Definitions, theorems and proofs

LemmaStatement: Literature-sourcedProof: AI-adaptedjudge pass (gpt-6-luna)Open item page →

Variance of a centered random weighted sum

Statement

Let n≥1. Let W1,…,Wn be mutually independent real random variables with finite second moments, means E[Wi]=0, and variances σi2. Let X=(X1,…,Xn) be a real random vector whose coordinates have finite second moments, and suppose the random elements W and X are independent. The coordinates of X need not be mutually independent. For S=∑i=1nWiXi, the variable S is square-integrable, E[S]=0, and Var⁡(S)=∑i=1nσi2E[Xi2].

Facts & Assumptions

Given: Mutually independent centered square-integrable real random variables W1,…,Wn and a square-integrable random vector X=(X1,…,Xn) independent of the vector W=(W1,…,Wn).

[F1]

Independence of random elements means independence of their generated sigma-algebras. (Independent random elements)

[F2]

The joint law of a finite family of independent random elements is the product of their marginal laws. (Independent random elements have product joint law)

[F3]

Measurable coordinatewise maps preserve independence. (Measurable coordinatewise functions preserve independence)

[F4]

Expectations of products of independent integrable random variables factor, and the product is integrable. (Expectations factor over finite products of independent random variables)

[F5]

Sums, products, and lattice operations of measurable real-valued functions used here are measurable. (Arithmetic and lattice operations preserve measurability whenever they are defined)

[F6]

If U,V∈L2(P), then E[∣UV∣]≤(E[U2])1/2(E[V2])1/2. (Cauchy-Schwarz for random variables)

[F7]

Expectation is integration against the probability measure. (Expectation of a nonnegative or integrable random variable)

[F8]

For a square-integrable real random variable Y, Var⁡(Y)=E[(Y−E[Y])2]. (Moments, variance, and covariance on a probability space)

[F9]

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

1.1givenF1F4F6F9

Each Wi and Xi is integrable by [F6] applied with the constant random variable 1. Independence of the two random elements gives independence of Wi and Xi by [F1], so [F4] yields E[WiXi]=E[Wi]E[Xi]=0. Linearity [F9] over the finite index set gives E[S]=0.

1.2givenF2F3F4F5F6

Fix i≠j. The coordinate-product maps (w,x)↦wiwj and (w,x)↦xixj are measurable by [F5]. Since the joint law of (W,X) is the product law by [F2], these maps, one applied to each independent factor, give independent random variables A=WiWj and B=XiXj by [F3]. Both are integrable by [F6]. Also E[A]=E[Wi]E[Wj]=0 by [F4] and mutual independence of the weights. Hence [F4] gives E[WiWjXiXj]=E[AB]=0.

1.3givenF2F3F4

For each i, Wi2 and Xi2 are independent nonnegative random variables by [F2] and [F3]. Applying the nonnegative case of [F4] gives E[Wi2Xi2]=E[Wi2]E[Xi2]=σi2E[Xi2]<∞.

2.1step 1.1step 1.2step 1.3F7F8F9algebra∎

Expanding the finite square S2 gives diagonal terms from step 1.3 and cross terms from step 1.2, so S2 is integrable. Linearity [F9] gives E[S2]=∑iσi2E[Xi2]. By step 1.1, E[S]=0; [F7] and [F8] therefore identify this second moment with Var⁡(S) and prove the claim.

LemmaStatement: AI-adaptedProof: AI-adaptedjudge pass (gpt-6-luna)Open item page →

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.

LemmaStatement: AI-adaptedProof: AI-generatedjudge pass (gpt-6-luna)Open item page →

An IID ReLU layer forces a zero atom in the next preactivation

Statement

Let n≥1 and let Z1,…,Zn be independent and identically distributed real random variables whose law is symmetric about zero and has no atom at zero. Put Xi=ReLU⁡(Zi). For any real random variables W1,…,Wn on the same probability space, define the next zero-bias preactivation T=∑i=1nWiXi. Then P(T=0)≥2−n>0. In particular, this next preactivation cannot be atomless at zero, regardless of the distribution or dependence of the weights. The same conclusion holds under the weaker assumption that the Xi are mutually independent and each Zi is symmetric with no atom at zero; independence of the Zi is not required in that formulation.

Facts & Assumptions

Given: The finite independent family Z1,…,Zn, its common symmetric atomless law, and the definitions in the Statement.

[F1]

For a symmetric atomless real random variable, each sign has probability 1/2. (Second moments of symmetric rectifiers)

[F2]

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

[F3]

Independence of random elements means independence of their generated sigma-algebras. (Independent random elements)

Proof

1.1givenF1

By [F1], P(Zi<0)=1/2 for each i. On the event E=⋂i=1n{Zi<0}, every Xi=ReLU⁡(Zi) is zero.

1.2givenF1F2F3

Since {Xi=0}={Zi≤0} and P(Zi=0)=0, the event A=⋂i{Xi=0} agrees with E up to a null set. Under the weaker hypothesis, independence of the Xi and [F2] give P(A)=∏iP(Xi=0)=2−n. Under the original hypothesis, each σ(Xi) is a sub-sigma-algebra of σ(Zi) because ReLU is measurable, so [F3] implies independence of the Xi and the same calculation applies.

2.1step 1.1step 1.2algebra∎

On E, T=∑iWiXi=0 for every value of the real weights. Hence P(T=0)≥P(E)=2−n>0, which proves the claim.

AlgorithmAlgorithm: Literature-sourcedProof: Not applicablejudge pass (gpt-6-luna)Open item page →

Rectifier variance-preserving initialization

Definition

Inputs. A finite layered perceptron with positive integer widths d0,…,dL, a real negative-side slope a, a mode m∈{forward,backward}, and a finite mutually independent family of real random variables ξℓji with a common symmetric law, mean zero, and variance one. The layer matrix has shape dℓ×dℓ−1, with row index j and column index i.

Output. Random weight matrices Wℓ and zero bias vectors bℓ.

Instructions. For each 1≤ℓ≤L, set the fan-in nℓ=dℓ−1 and fan-out n^ℓ=dℓ. Set vℓ=2(1+a2)nℓ(m=forward),vℓ=2(1+a2)n^ℓ(m=backward). For every matrix entry set Wℓ,ji=vℓ ξℓji, and set bℓ=0. The case a=0 is ReLU; the piecewise-linear activation for general a is t for t>0 and at for t≤0.

Remarks

The repeated forward recurrence concerns a layer whose inputs are outputs of a rectifier. A raw-input first affine layer has no preceding rectifier factor. If the input coordinates are square-integrable, centered, have common variance q, and the first-layer weights are independent of the input vector, then the finite weighted sum has mean zero. In its second moment, the diagonal terms each contribute v1q and the off-diagonal terms vanish because the weights are independent and centered. Thus Var⁡(Z1,j)=n1v1q, so v1=1/n1 preserves q. He et al. also use the rectifier scale at that first layer for simplicity; this changes one boundary factor, not the repeated hidden-layer factor. Fan-in mode and fan-out mode target different directions and need not preserve both when layer widths differ. The paper's depth and ImageNet accuracy comparisons are reported experiments, not consequences of this definition and not guarantees of training success or generalization.

The mode names specify the source's forward and conditional backward moment targets. They do not assert that finite-network gradients are independent of their weights or gates. A positive-probability zero preactivation requires a specified kink convention for backpropagation; it does not have an ordinary ReLU derivative. Consequently the backward scale alone supplies no nonzero actual-gradient preservation guarantee.

TheoremStatement: AI-adaptedProof: AI-adaptedjudge pass (gpt-6-luna)Open item page →

ReLU forward preactivation-variance recurrence

Statement

Let L≥2 and let d0,…,dL be positive integers. For each 1≤ℓ≤L, let Wℓ∈Rdℓ×dℓ−1 have mutually independent, identically distributed entries with a symmetric law, mean zero, and finite variance σℓ2, and suppose Wℓ is independent of the input activation vector Xℓ−1. Set the fan-in nℓ=dℓ−1, set every bias to zero, and define Zℓ=WℓXℓ−1 and Xℓ=ρ(Zℓ) coordinatewise, where ρ(t)=max⁡(t,0). Assume for every layer that the coordinates of Xℓ−1 have a common finite second moment; they need not be independent or identically distributed. Also assume that each preactivation coordinate Zℓ,j has a symmetric law; an atom at zero is allowed in this forward second-moment recurrence. Then, for every 2≤ℓ≤L and 1≤j≤dℓ, Var⁡(Zℓ,j)=12nℓσℓ2Var⁡(Zℓ−1,1). At the raw-input boundary, the exact base relation is Var⁡(Z1,j)=n1σ12E[X0,12]; if X0,1 is centered, its second moment here is its variance. In particular, writing Vℓ=Var⁡(Zℓ,1), the exact finite-depth formula is VL=V1∏ℓ=2Lnℓσℓ22. Thus σℓ2=2/nℓ for ℓ≥2 gives VL=V1, with zero atoms allowed at every depth. This is a statement about ensemble second moments, not about variance along every sampled network or training success.

Facts & Assumptions

Given: A finite fully connected layered perceptron with zero biases, symmetric centered independent weights, layer inputs independent of their weights, symmetric preactivations, and the stated layerwise moment assumptions.

[F1]

A fully connected layered perceptron has a real weight on every edge between consecutive layers, so each preactivation is the corresponding finite weighted sum. (Finite feedforward computation graphs and layered MLPs)

[F2]

For independent centered weights Wi independent of a square-integrable input vector X, Var⁡(∑iWiXi)=∑iVar⁡(Wi)E[Xi2]. (Variance of a centered random weighted sum)

[F3]

If Z is symmetric about zero, E[ρ(Z)2]=12E[Z2], including when Z has an atom at zero. (Second moments of symmetric rectifiers)

Proof

1.1givenF1F2

In the fully connected layer from Xℓ−1 to Zℓ, each coordinate is Zℓ,j=∑i=1nℓWℓ,jiXℓ−1,i. Apply [F2] to this row of weights and the input vector. Since its entries have common variance σℓ2 and the input coordinates have common second moment, this gives E[Zℓ,j]=0 and Var⁡(Zℓ,j)=nℓσℓ2E[Xℓ−1,12]. For ℓ=1 this is the stated raw-input base relation.

2.1givenF3step 1.1

For ℓ≥2, Xℓ−1,1=ρ(Zℓ−1,1). By [F3], E[Xℓ−1,12]=12E[Zℓ−1,12]. Symmetry makes E[Zℓ−1,1]=0, so this is 12Var⁡(Zℓ−1,1). Substitution into step 1.1 proves the claimed recurrence.

3.1step 2.1algebra∎

Apply step 2.1 successively for ℓ=2,…,L. The product of the finitely many coefficients is the displayed expression for VL; setting each coefficient to 1 proves preservation. No division by a variance occurs, so the identity also holds when V1=0.

Remarks

The equal-second-moment assumption allows the dependence between coordinates that naturally arises after averaging over a random previous layer. For a concrete finite-depth model satisfying all these hypotheses, take every width to be one, X0=1, and mutually independent weights Wℓ=±vℓ with equal probabilities, where vℓ>0. At each layer the fresh weight is independent of the preceding activation and its random sign makes the preactivation symmetric. The event W1<0 already forces all subsequent preactivations to vanish, yet V1=v1 and VL=v1∏ℓ=2L(vℓ/2) hold exactly. Thus atomlessness is unnecessary for finite-depth forward propagation.

TheoremStatement: AI-adaptedProof: AI-adaptedjudge pass (gpt-6-luna)Open item page →

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.

CorollaryStatement: AI-adaptedProof: AI-adaptedjudge pass (gpt-6-luna)Open item page →

PReLU fan-in and fan-out variance scales

Statement

Let a∈R and consider a finite rectified stack with zero biases, widths d0,…,dL, Zℓ=WℓXℓ−1, and Xℓ=ϕa(Zℓ), where ϕa(t)=t for t>0 and ϕa(t)=at for t≤0. Put nℓ=dℓ−1 and n^ℓ=dℓ. Suppose each Wℓ has independent entries with a common symmetric centered law of finite variance σℓ2 and is independent of Xℓ−1. Suppose each layer input has coordinates with a common finite second moment, and all preactivation coordinates are symmetric about zero. Input coordinates may be dependent, and zero atoms are allowed. Then, writing Vℓ=Var⁡(Zℓ,1), V1=n1σ12E[X0,12],Vℓ=1+a22nℓσℓ2Vℓ−1(2≤ℓ≤L),VL=V1∏ℓ=2L1+a22nℓσℓ2. Thus σℓ2=2/((1+a2)nℓ) preserves the forward preactivation variance across the hidden layers.

For backward signals, take an objective J differentiable as a function of the last activation and set gL=∇XLJ(XL). Specify a finite kink value cℓ and put ηℓ,j=1Zℓ,j>0+a1Zℓ,j<0+cℓ1Zℓ,j=0, hℓ=ηℓ⊙gℓ, and gℓ−1=WℓThℓ. They are ordinary chain-rule gradients when all relevant preactivations are nonzero; when a=1 and cℓ=1 the activation is linear and this also holds at zeros. Otherwise a specified kink value is a backpropagation convention. If the coordinates of Zℓ have common zero probability pℓ, the coordinates of gℓ have common finite second moment Gℓ, ηℓ is independent of gℓ, and Wℓ is independent of hℓ, then Gℓ−1=Var⁡(gℓ−1,i)=n^ℓσℓ2κℓGℓ,κℓ=1+a22(1−pℓ)+cℓ2pℓ. The upstream coordinates have mean zero. If these conditional hypotheses hold for every r≤ℓ≤L, then Gr−1=GL∏ℓ=rLn^ℓσℓ2κℓ. In the source's atomless case pℓ=0, this retains the original backward factor (1+a2)n^ℓσℓ2/2 and the fan-out choice σℓ2=2/((1+a2)n^ℓ). For a noncentered terminal signal GL is its second moment; for centered signals it is their variance. The source's forward and atomless backward weight variances agree exactly when nℓ=n^ℓ. With atoms the conditional backward-preserving choice is 1/(n^ℓκℓ) when κℓ>0; if κℓ=0 the upstream signal has zero variance for every finite weight variance. These are conditional backward statements, and no independence assumption or nonzero actual-gradient realization is inferred from initialization.

Facts & Assumptions

Given: The finite stack and its forward moment assumptions, an objective differentiable as a function of XL with gL=∇XLJ(XL), the specified backward-signal convention, and the additional layerwise conditional hypotheses only for backward preservation.

[F1]

The source initializer uses forward variance 2/((1+a2)nℓ) or backward variance 2/((1+a2)n^ℓ). (Rectifier variance-preserving initialization)

[F2]

Symmetry gives PReLU second moment (1+a2)E[Z2]/2, and the squared gate moment is ((1+a2)/2)(1−p0)+c2p0. (Second moments of symmetric rectifiers)

[F3]

Independent centered weights independent of an input vector give weighted-sum mean zero and variance ∑iVar⁡(Wi)E[Xi2]. (Variance of a centered random weighted sum)

[F4]

Totally differentiable compositions obey the total-derivative chain rule. (The chain rule for total derivatives: D(g∘f)(a)=Dg(f(a))∘Df(a))

[F5]

The ReLU forward recurrence allows zero atoms and has factor nℓσℓ2/2 and the corresponding finite-depth product. (ReLU forward preactivation-variance recurrence)

[F6]

Exact backward signals obey a covariance-sum identity; the ReLU half-factor product is conditional, and its full atomless independence assumptions force zero downstream second moment at a nondegenerate actual ReLU hidden layer. (ReLU backward gradient-variance recurrence)

[F7]

Expectations of products of independent integrable variables factor. (Expectations factor over finite products of independent random variables)

Proof

1.1givenF1F2F3algebra

Apply [F3] to a row of Wℓ. The preactivation is centered and has variance nℓσℓ2E[Xℓ−1,12], which proves the raw-input relation. For ℓ≥2, [F2] and centered symmetry give E[Xℓ−1,12]=(1+a2)Vℓ−1/2, proving the forward recurrence without an atomlessness or coordinate-independence assumption. Successive substitution gives the finite product; the forward choice [F1] makes every hidden-layer factor one.

1.2givenF2F4F7

At nonzero preactivations the derivative of PReLU is 1 on the positive half-line and a on the negative half-line, so [F4] gives the stated backward recursion for ordinary gradients there. At zeros the recursion is instead the specified convention, except in the linear case a=1,cℓ=1, where it is the ordinary derivative everywhere. By [F2] and independence of the gate and gℓ, [F7] gives E[hℓ,j2]=κℓGℓ.

2.1givenF1F3F6step 1.2algebra

Apply [F3] to each column of Wℓ and hℓ, using the assumed matrix/signal independence. The upstream signal is centered and has variance σℓ2∑jE[hℓ,j2]=n^ℓσℓ2κℓGℓ. Iteration over the finite set of layers proves the backward product, just as for [F6]. If pℓ=0 then κℓ=(1+a2)/2 and the source's backward choice [F1] gives unit coefficient. For positive κℓ, solving the unit-coefficient equation gives 1/(n^ℓκℓ); for zero κℓ the coefficient vanishes for every finite weight variance.

3.1F1F5F6step 1.1step 2.1algebra∎

The positive source scales 2/((1+a2)nℓ) and 2/((1+a2)n^ℓ) agree exactly when nℓ=n^ℓ. At a=0,cℓ=0, the forward and backward formulas specialize to [F5] and [F6], including the atom correction and feasibility caveat; at a=1,cℓ=1, κℓ=1 for every zero probability, as required for the linear derivative. These substitutions also show why an arbitrary kink convention must not be silently identified with an ordinary derivative.

Remarks

The atomless half-factor product is retained as the precise conditional calculation in He et al. §2.2; the forward product is valid for actual finite stacks satisfying the forward assumptions, with dependent coordinates and zero atoms allowed. The backward independence hypotheses require separate verification for an actual network. In particular, the ReLU specialization has the explicit nonzero-feasibility obstruction proved in the backward theorem, even at one hidden layer. No mean-field limit or empirical training claim supplies that gap.

4 · Examples, counterexamples and false statements

None yet.

Sources