Alphabeta Math
ExampleConstruction: AI-generatedVerification: AI-generatedPipeline-generated
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An identifiable nonlinear ANM with a cubic cause-to-effect function

Example

Assume AC. Let X,N be independent standard normal random variables and set Y=X3+N. The structural function is nonconstant and C3, and both noise densities are strictly positive and C3. For f(x)=x3 and standard normal input and noise, the triple satisfies Condition 19: at (x,y)=(1,0) its left side is 0 and its right side is 28. The bivariate direction-identifiability theorem therefore excludes every reverse additive-noise representation with an independent residual and a C3 structural function.

Facts & Assumptions

Given: Assume AC and take independent X,N∼N(0,1) with standard normal density ϕ(t)=(2π)−1/2e−t2/2.

[A1]

AC is the Axiom of Choice and implies countable choice (ACω), the additional axiom interface used by the standard-normal suppliers. The Axiom of Choice, The Axiom of Countable Choice (ACω), AC implies DC implies countable choice

[F1]

The standard normal law has density ϕ, and this density is positive and normalized. Standard normal and normal laws, The standard normal density has total mass one

[F2]

For a smooth bivariate ANM, Condition 19 fails when its displayed equation holds at every point of the nonzero domain. Condition 19 and restricted additive-noise models

[F3]

If Condition 19 holds, no reverse ANM in the stated smooth class represents the same observational law. Bivariate additive-noise direction identifiability under Condition 19

[F4]

The bivariate forward joint density is the product of the input density and shifted noise density. Positive joint density of a finite additive-noise model

Verification

Given: The independent standard normal variables X,N and Y=X3+N.

1.1givenA1F1F4construct

Under AC [A1], the equation is a bivariate additive-noise model with f(x)=x3, independent noise, strictly positive C3 input and noise densities, and nonconstant C3 function. By [F1] and [F4], its joint density is p(x,y)=ϕ(x)ϕ(y−x3).

1.2givenF1algebra

Put z=y−f(x). The standard normal log densities give ξ′′(x)=−1, ξ′′′(x)=0, ν′(z)=−z, ν′′(z)=−1, and ν′′′(z)=0. At (x,y)=(1,0), z=−1, f′=3, f′′=6, and f′′′=6, so ν′′(z)f′=−3≠0.

2.1F2F3step 1.2algebra

Substitution into Condition 19's right side gives (−1)(63)−2(−1)(6)(3)+(1)(6)+0−(1)623=−2+36+6−12=28, whereas ξ′′′(1)=0. Hence Condition 19 holds. The theorem [F3] now excludes a reverse ANM in its stated class, proving the Example's identifiability assertion.

3.1step 1.1step 1.2step 2.1∎

The input and noise densities, and the cubic function, satisfy the hypotheses in the Statement; the nonzero-domain witness in step 1.2 satisfies the Condition 19 inequality in step 2.1.

Depends on

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