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ExampleConstruction: AI-generatedVerification: AI-generatedPipeline-generatedprecheck pass
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.

Exact variance against propagated variance for a nonlinear observation map

Example

Let Ω={ω0,ω1,ω2} carry the weights w(ω0)=1/2, w(ω1)=w(ω2)=1/4, let Y:Ω⟶R,Y(ω0)=0,Y(ω1)=1,Y(ω2)=3, and let h(y):=y2 on the open convex set U=(−1,4), which contains the three values of Y. Then h is C2 with constant Hessian Hh(x)=(2) and operator norm B=2, the mean is μ=E[Y]=1∈U, and E[Y2]=52,Var⁡(Y)=32,E[Y4]=412,Var⁡(h(Y))=412−(52)2=574. The linearization T(y)=h(1)+h′(1)(y−1)=1+2(y−1) has variance h′(1)2Var⁡(Y)=4⋅32=6=244≠574=Var⁡(h(Y)), so the propagated variance is not the exact variance of h(Y). The linearization error R=h(Y)−T(Y)=(Y−1)2 satisfies E[R]=32=B2E[(Y−1)2],E[R2]=92=B24E[(Y−1)4], so both displayed bounds of Mean, mean-square and covariance bounds for second-order linearization error are attained in this example. The example contains no measured quantity and asserts no physical model.

Facts & Assumptions

Given: The finite weighted set (Ω,w) and the random variable Y displayed in the Example; the function h(y)=y2 on U=(−1,4).

[L1]

The finite probability space (Ω,w) determines a probability measure P on (Ω,P(Ω)) with P(A)=∑ω∈Aw(ω), and expectations of finite-valued random variables computed by the finite weighted sum agree with their expectations as random variables on that probability space. (Finite probability spaces are exactly finite full-power-set probability spaces, Expectation of a real random variable on a finite probability space, Finite probability spaces, outcome weights, events, and event probabilities)

[L2]

If h:U→R is C2 on the open convex set U, ∥Hh(x)∥≤B on U, X takes values in U with E[X]=μ∈U and E∣X−μ∣2<+∞, then the linearization error R=h(X)−h(μ)−h′(μ)(X−μ) satisfies ∣E[R]∣≤E∣R∣≤B2E∣X−μ∣2, and E[R2]≤B24E∣X−μ∣4 when the fourth moment is finite. (Mean, mean-square and covariance bounds for second-order linearization error)

[L3]

For a square-integrable random variable X with mean μ, the covariance matrix of the one-dimensional random vector (X) is the 1×1 matrix (Var⁡(X)), and the linearization x↦h(μ)+h′(μ)(x−μ) has covariance matrix h′(μ)2Var⁡(X). (Mean vector and covariance matrix of a square-integrable random vector, The covariance of the linearization is AΣAT)

Verification

technique · direct computation on a finite probability space
1.1givenL1algebra

By [L1] the weights make (Ω,w) a probability space and the expectations below are the finite weighted sums of Expectation of a real random variable on a finite probability space: E[Y]=12⋅0+14⋅1+14⋅3=1,E[Y2]=14⋅1+14⋅9=52, E[Y4]=14⋅1+14⋅81=824=412,E[(Y−1)2]=12⋅1+14⋅0+14⋅4=32, and E[(Y−1)4]=12⋅1+14⋅0+14⋅16=12+4=92.

1.2givenL3L4algebra

The function h(y)=y2 is C2 on the open convex set U=(−1,4), with h′(y)=2y, so h(1)=1 and h′(1)=2; its Hessian is the constant matrix (2), of operator norm B=2 by [L3] and [L4], and the values 0,1,3 of Y and the mean μ=1 all lie in U.

2.1step 1.1L4algebra

By [L4], Var⁡(Y)=E[Y2]−E[Y]2=52−1=32 and Var⁡(h(Y))=E[Y4]−E[Y2]2=412−254=82−254=574.

2.2step 1.2algebra

The linearization error is R=h(Y)−1−2(Y−1)=Y2−2Y+1=(Y−1)2.

3.1step 1.2step 2.1L3algebra

By [L3] the propagated covariance is h′(1)2Var⁡(Y)=4⋅32=6=244, while step 2.1 gives Var⁡(h(Y))=574; these differ, so the propagated covariance is not the exact variance of h(Y) in this model.

4.1step 1.1step 2.2L2algebra∎

With B=2 and the moments of step 1.1, the bounds of [L2] read B2E[(Y−1)2]=1⋅32=32 and B24E[(Y−1)4]=1⋅92=92; since R2=(Y−1)4 by step 2.2, one has E[R]=32 and E[R2]=92, so both bounds are attained.

Depends on

Dependency tree · 2 levels

Sources