ExampleConstruction: AI-generatedVerification: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-10

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.

Law of total variance

Example

Assume AC for conditional classes. On $Ω = {0, 1}^{2}$ with full sigma-algebra and each atom of mass $1 / 4$ , put $U (u, v) = u$ , $V (u, v) = v$ , $X = U + V$ and $G = σ (U)$ . Then $Var (X) = 1 / 2$ , $E [Var (X ∣ G)] = 1 / 4$ and $Var (E [X ∣ G]) = 1 / 4$ .

Facts & Assumptions

Given: The four-atom model, variables U,V,X and sigma-algebra G specified in the example; AC is the conditional-class convention.

[F1]

Finite weights summing to one define a probability measure. (Finite probability spaces are exactly finite full-power-set probability spaces)

[F2]

A known variable conditions to itself; an independent variable conditions to its mean. (Conditioning a known variable and an independent variable)

[F3]

Conditional expectation is linear and fixes constants. (Basic algebra and order properties of conditional expectation)

[F4]

Total variance is expected conditional variance plus the variance of the conditional mean. (Conditional variance decomposition)

[F5]

Conditional variance is the conditional mean of the squared residual. (Conditional variance)

Verification

technique · direct

1.1

The four masses are nonnegative and sum to one, so [F1] constructs the probability space. Each U and V marginal has mass one half at zero and at one, and $P (U = u, V = v) = 1 / 4 = P (U = u) P (V = v)$ for all four pairs. Summing over coordinate subsets gives independence for every Borel rectangle. Both variables are bounded and integrable, with $E U = E V = 1 / 2$ .

2.1

By [F2]–[F3], $E [X ∣ G] = U + 1 / 2$ . The residual is $V - 1 / 2$ , whose square equals 1/4 at every atom. Thus [F5] and the constant rule [F3] give $Var (X ∣ G) = 1 / 4$ and its expectation 1/4.

step 1.1F2F3F5

3.1

The four X values are (0,1,1,2), with mean 1, so $Var (X) = (1 + 0 + 0 + 1) / 4 = 1 / 2$ . The conditional mean takes values one half and three halves, each with probability one half; its mean is 1 and its variance is $((1 / 2)^{2} + (1 / 2)^{2}) / 2 = 1 / 4$ . Therefore the three computed quantities satisfy [F4] as $1 / 2 = 1 / 4 + 1 / 4$ .

step 2.1F4∎

Source notes

Durrett §4.1.2, printed pp.210–213, conditional identities; the explicit four-atom variance instance is locally calculated and has generated-example provenance.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

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Sources

Durrett, Probability: Theory and Examples, 5th ed. (standard reference, not scraped)