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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 — Examples

1 · Prerequisites

2 · Logical flowchart

3 · Definitions, theorems and proofs

None yet.

4 · Examples, counterexamples and false statements

ExampleConstruction: AI-generatedVerification: AI-generatedjudge pass (gpt-6-luna)Open item page →

Mean-zero weights or a nonzero bias do not supply the ReLU half-moment

Example

Compare two one-unit preactivations. In the first, the input is X=1, the bias is zero, and the weight W equals −1 with probability 2/3 and 2 with probability 1/3. In the second, the input X is uniform on [−2,2], the independent weight W equals −1 or 1 with equal probabilities, and the bias is 1. In each case compute E[ReLU⁡(Z)2] and compare it with 12Var⁡(Z), where Z=WX+b. The first case tests whether a zero-mean but nonsymmetric weight law suffices; the second tests the effect of a nonzero bias despite symmetric weights and input.

Facts & Assumptions

Given: The two one-unit probability models specified in the Example.

[F1]

If Z is symmetric about zero and has no atom at zero, then E[ReLU⁡(Z)2]=12E[Z2]. (Second moments of symmetric rectifiers)

[F2]

Variance is the centered second moment, Var⁡(Z)=E[(Z−E[Z])2]. (Moments, variance, and covariance on a probability space)

[F3]

Expectations of nonnegative random variables are integrals against their probability laws. (Expectation of a nonnegative or integrable random variable)

Verification

1.1givenF2algebra

In the first case, E[W]=(−1)(2/3)+2(1/3)=0 and Var⁡(Z)=Var⁡(W)=E[W2]=2. Since ReLU⁡(Z) is 2 with probability 1/3 and zero otherwise, E[ReLU⁡(Z)2]=4/3, whereas 12Var⁡(Z)=1.

1.2givenF1algebra

The first preactivation has P(Z>0)=1/3 and P(Z<0)=2/3, so its law is not symmetric despite its zero mean and absence of an atom at zero. The symmetric-preactivation hypothesis of [F1] fails; zero mean alone cannot replace it.

1.3givenF2F3algebra

In the second case, conditioning on either value of W shows that WX is uniform on [−2,2]. Thus Z=WX+1 is uniform on [−1,3], with mean 1 and variance (3−(−1))2/12=4/3; it has no atom at zero.

2.1givenF2F3algebra∎

The density of Z is 1/4 on [−1,3], so E[ReLU⁡(Z)2]=14∫03z2 dz=9/4, while 12Var⁡(Z)=2/3. The nonzero bias has shifted the symmetric preactivation and invalidated the half-moment identity.

ExampleConstruction: AI-adaptedVerification: AI-generatedOpen item page →

A symmetric atomless IID ReLU layer creates a zero atom next

Example

Let Z1,Z2 be independent and identically distributed symmetric real random variables with no atom at zero, set Xi=ReLU⁡(Zi), and form a second-layer zero-bias preactivation T=W1X1+W2X2 with arbitrary real weights. Although neither Zi has an atom at zero, the event that both rectified inputs vanish has probability 1/4 and forces T=0.

For a complete finite-network signal calculation, also take a two-layer width-one ReLU stack with X0=1, mutually independent W1,W2∈{−2,2} with equal sign probabilities, and an independent random sign R∈{−1,1}. Take objective J(X2)=RX2, so g2=R, and use zero-at-kink backpropagation. Then Var⁡(Z1)=Var⁡(Z2)=2 but P(Z2=0)=1/2 and Var⁡(g1)=1/2. The atomless half-factor coefficient would instead give 1. The general conditional recurrence allows zero atoms, but its weight/gated-signal independence hypothesis fails in this realized stack. Its half-factor special case also requires p2=0, whereas here p2=1/2.

Facts & Assumptions

Given: The two-coordinate zero-event model and the independent finite random-sign scalar stack in the Example.

[F1]

For n IID symmetric atomless preactivations followed by ReLU, any next zero-bias weighted sum has an atom at zero of probability at least 2−n. (An IID ReLU layer forces a zero atom in the next preactivation)

[F2]

Zero-at-kink backward signals satisfy h2=1Z2>0g2 and g1=W2h2; the atomless half-factor formula requires additional hypotheses. (ReLU backward gradient-variance recurrence)

Verification

1.1givenF1algebra

The example has n=2, so [F1] gives P(T=0)≥2−2=1/4, independently of the values of W1,W2.

1.2givenalgebra

In the scalar stack, Z1 is ±2 with equal probabilities. If W1<0 then X1=0 and Z2=0; if W1>0 then Z2 is 2 or −2 with equal probabilities. Thus the law of Z2 gives probabilities 1/4,1/2,1/4 to −2,0,2, respectively. Both preactivations have mean zero and variance 2, and the second has zero probability 1/2.

2.1step 1.1algebra

Therefore the two first-layer preactivations can be atomless while the next preactivation is not atomless. Requiring the no-atom condition at every layer is incompatible with this finite IID ReLU stack.

3.1givenF2step 1.2algebra∎

By [F2], g1=2R exactly when both weights are positive, an event of probability 1/4; it is zero otherwise. Independence and the centered sign R give mean zero and variance 2(1/4)=1/2. The atomless coefficient would be (1/2)(1)(2)E[R2]=1. Here h2=R on the same positive-weight event and zero otherwise: P(h2≠0∣W2<0)=0, whereas P(h2≠0)=1/4, so weight/gated-signal independence fails. The covariance identity and the specified kink convention remain valid; the atomless conditional recurrence is inapplicable.

ExampleConstruction: AI-generatedVerification: AI-generatedjudge pass (gpt-6-luna)Open item page →

Fan-in and fan-out target different signal directions

Example

Compare the source's forward and atomless conditional backward recurrence coefficients for fan-in 64 and fan-out 128. The respective coefficients are 64σ2/2 and 128σ2/2. Compute their factors at the two initializer choices σ2=2/64 and σ2=2/128. For a preceding preactivation variance V the forward result is a multiple of V; for a downstream second moment G the conditional backward result is a multiple of G (and of its variance when centered). These coefficient comparisons do not assert a nonzero actual ReLU gradient model satisfying the full backward independence assumptions: at a nondegenerate hidden ReLU layer those assumptions force G=0.

Diagram

The top row represents the forward linear map and ReLU; the lower row shows the reverse-mode maps with Dℓ=diag⁡(1Zℓ,j>0). Each box denotes the full vector of the stated dimension, and the arrow labelled Wℓ denotes the 128×64 matrix, not a single scalar edge. Thus the forward fan-in is 64 and the backward fan-out is 128. The diagram illustrates these dimensions; the verification below uses the recurrence formulas directly.

diagram failed to render:
\begin{tikzpicture}[>=stealth]
  \node[draw,rounded corners] (x) at (-4,1) {$X_{\ell-1}\in\mathbb R^{64}$};
  \node[draw,rounded corners] (z) at (0,1) {$Z_\ell\in\mathbb R^{128}$};
  \node[draw,rounded corners] (y) at (4,1) {$X_\ell\in\mathbb R^{128}$};
  \draw[->] (x) -- node[above] {$W_\ell$} (z);
  \draw[->] (z) -- node[above] {$\rho$} (y);
  \node[draw,rounded corners] (g0) at (-4,-1) {$g_{\ell-1}\in\mathbb R^{64}$};
  \node[draw,rounded corners] (h) at (0,-1) {$h_\ell\in\mathbb R^{128}$};
  \node[draw,rounded corners] (g1) at (4,-1) {$g_\ell\in\mathbb R^{128}$};
  \draw[<-] (g0) -- node[below] {$W_\ell^{\mathsf T}$} (h);
  \draw[<-] (h) -- node[below] {$D_\ell$} (g1);
\end{tikzpicture}

Facts & Assumptions

Given: The stated dimensions, source recurrence coefficients, and their conditional interpretations for V and G.

[F1]

Forward mode uses weight variance 2/nℓ, and backward mode uses 2/n^ℓ. (Rectifier variance-preserving initialization)

[F2]

The forward preactivation variance factor is nℓσℓ2/2. (ReLU forward preactivation-variance recurrence)

[F3]

Under its stated conditional assumptions, the backward gradient second-moment factor is n^ℓσℓ2/2. (ReLU backward gradient-variance recurrence)

Verification

1.1givenF1F2F3algebra

With fan-in variance σ2=2/64=1/32, the forward factor is 12(64)(1/32)=1, so V is preserved. The backward factor is 12(128)(1/32)=2, so the gradient variance becomes 2G.

1.2givenF1F2F3algebra

With fan-out variance σ2=2/128=1/64, the backward factor is 12(128)(1/64)=1, so G is preserved. The forward factor is 12(64)(1/64)=1/2, so the next preactivation variance is V/2.

2.1step 1.1step 1.2algebra∎

Since the two widths differ, the required variances 1/32 and 1/64 differ, and no shared weight variance makes both recurrence coefficients equal to one.

ExampleConstruction: AI-generatedVerification: AI-generatedjudge pass (gpt-6-luna)Open item page →

A PReLU slope changes both initialization scales

Example

For the PReLU source scale calculation, take negative-side slope a=1/2, fan-in n=5, and fan-out n^=10. Compare the exact forward recurrence coefficient (1+a2)nσ2/2 with the atomless conditional backward coefficient (1+a2)n^σ2/2. Compute the two unit-coefficient weight variances and determine whether they agree. The backward coefficient has the corollary's independence assumptions; it is not asserted to describe an arbitrary actual-network gradient, and a zero atom with a specified kink gate changes its factor.

Facts & Assumptions

Given: The dimensions, slope, and forward and atomless conditional backward coefficients stated in the Example.

[F1]

The PReLU forward and backward scale factors are (1+a2)/2 times fan-in or fan-out and the corresponding weight variances are 2/((1+a2)n) or 2/((1+a2)n^). (PReLU fan-in and fan-out variance scales)

Verification

1.1givenF1algebra

Since 1+a2=1+1/4=5/4, the forward-preserving variance is 2/((5/4)5)=8/25. The forward recurrence factor is (5/8)5(8/25)=1.

1.2givenF1algebra

The backward-preserving variance is 2/((5/4)10)=4/25. The backward recurrence factor is (5/8)10(4/25)=1.

2.1step 1.1step 1.2algebra∎

The two variances 8/25 and 4/25 differ because the fan-in and fan-out differ, so no shared weight variance makes both recurrence coefficients equal to one.

Sources