Alphabeta Math
ExampleConstruction: AI-generatedVerification: AI-generatedPipeline-generatedjudge pass (gpt-6-luna)
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Polynomial-chaos coefficients group by variable support

Example

Let X1,X2 be independent symmetric two-point inputs on {−1,1}, and take the full degree-one tensor index set Λ={(0,0),(1,0),(0,1),(1,1)}. Its basis is 1,X1,X2,X1X2. For the exact finite expansion f=2+3X1+4X1X2, the coefficients are c00=2, c10=3, c01=0, and c11=4. The variance is 25, with S1=9/25, S2=0, ST1=1, and ST2=16/25.

Facts & Assumptions

Given: Independent symmetric two-point inputs, the full degree-one tensor index set, and the exact finite expansion in the Example.

[F1]

Finite product-law polynomial-chaos input data: finite independent real inputs use their product law and a finite admissible degree box; the degree condition alone asserts no polynomial completeness.

[F2]

Positive moment Gram matrices produce orthonormal polynomial families: for the symmetric two-point law with cutoff one, the orthonormal degree-one polynomials are 1 and x.

[F3]

Finite orthonormal polynomial tensor basis: a finite tensor basis consists of products of the univariate polynomials over a declared finite multi-index set.

[F4]

Product-coordinate averaging operators: PJ averages over coordinates outside J and lifts the result to the full product space.

[F5]

Independent product inputs have an orthogonal Hoeffding decomposition: coordinate averages define components that reconstruct f, are orthogonal, and contribute their squared norms to the variance.

[F6]

First-order and total Sobol indices: first and total indices divide variance-component sums by positive total variance.

[F7]

Moments, variance, and covariance on a probability space: variance is the expected square of the output after subtracting its mean.

Verification

Given: The finite product law, index set and exact expansion of the Example.

1.1F1F2F3givenalgebra

Each coordinate has mean zero and second moment one, so its degree-one monomial Gram matrix is the identity. The construction [F2] therefore gives 1 and xj, whose products in [F3] are the four displayed functions. On the four equiprobable outcomes (1,1),(1,−1),(−1,1),(−1,−1), their value vectors are (1,1,1,1), (1,1,−1,−1), (1,−1,1,−1), and (1,−1,−1,1). These vectors are pairwise orthogonal and each has squared length four, so under weights 1/4 the functions are orthonormal. Four independent functions on a four-point support form a basis of its four-dimensional L2 space.

2.1F7step 1.1givenalgebra

The functions in step 1.1 are linearly independent, so the exact expansion has unique coefficients: c00=2, c10=3, c01=0, and c11=4. The centered output is 3X1+4X1X2. Orthogonality gives V=E[(3X1+4X1X2)2]=9+16=25 by [F7]. The exact displayed finite expansion needs no infinite Parseval assertion.

2.2F1F4F5givenstep 1.1algebra

Product averaging [F4] and EXj=0 give P∅f=2, P{1}f=2+3X1, and P{2}f=2. The component definition in [F5] yields f{1}=3X1, f{2}=0, and f{1,2}=4X1X2. Their variance contributions are 9,0,16 because EXj2=1, matching the coefficient sums grouped by support.

3.1F6step 1.1step 2.1step 2.2algebra∎

Dividing these components by V=25 according to [F6] gives S1=9/25, S2=0, ST1=(9+16)/25=1, and ST2=16/25. The family is complete here by the finite four-point argument in step 1.1. Finite cutoff and positive moment Gram hypotheses alone do not imply completeness for a general input law.

Remarks

This example's full degree-one basis is complete because its input support has four points. In general the finite cutoff and positive moment Gram hypotheses of UQ1 do not imply completeness; an infinite coefficient formula needs a separate law-specific completeness hypothesis.

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