Alphabeta Math
TheoremStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedprecheck passjudge pass (gpt-6-luna)
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.

Horvitz–Thompson variance and pair-inclusion estimate

Statement

Let (U,y) be a finite population with a sampling design as in Finite populations and sampling designs. Write Ii for the indicator that unit i is included, and assume πi>0 for every i∈U. The Horvitz–Thompson total is T^HT:=∑i∈UIiyiπi. Its design variance is Var⁡d(T^HT)=∑i,j∈UΔijyiyjπiπj, where Δij is defined in Inclusion-covariance difference. If, in addition, πij>0 for every ordered pair (i,j), the all-pairs inverse-inclusion expression V^HT:=∑i,j∈UIiIjΔijπijyiπiyjπj is defined and satisfies Ed[V^HT]=Var⁡d(T^HT). When a pair has zero inclusion probability, the corresponding all-pairs coefficient Δij/πij is undefined; omitting unsupported pairs is a different expression and does not by itself prove unbiasedness or non-estimability by some other method.

Facts & Assumptions

Given: A finite population (U,y), a finite sampling design, πi>0 for every unit, and additionally πij>0 for every ordered pair when proving the pair-inclusion identity.

[F1]

Finite populations and sampling designs: The population values are fixed; the finite design has support Sd and weights pd; Ii is the inclusion indicator; πi=Pd(Ii=1) and πij=Pd(Ii=Ij=1); also πii=πi.

[F2]

Inclusion-covariance difference: Δij=πij−πiπj.

[F3]

Moments, variance, and covariance on a probability space: Variance is the expectation of the squared centered variable, and covariance is the expectation of the product of centered variables.

[F4]

The expectation of an indicator is the probability of the event: The expectation of an event indicator equals the probability of its event.

[F5]

Variance of a finite sum as the sum of all variances and covariances: The variance of a finite sum of square-integrable variables on one probability space is the double sum of their pairwise covariances.

[F6]

Covariance is symmetric and bilinear in finite linear combinations: Covariance is bilinear for finite linear combinations of square-integrable variables on one probability space.

[F7]

Linearity, monotonicity, and the modulus bound for expectation: For integrable random variables X,Y and scalars a,b on one probability space, E[aX+bY]=aE[X]+bE[Y].

Proof

1.1givenF1choose

For each i∈U, set Xi=Iiyi/πi. The design support is finite by [F1], each Ii is bounded, and yi/πi is finite by the positivity assumption; hence every Xi is square-integrable and T^HT=∑i∈UXi.

2.1F1F5step 1.1

Apply [F5] to this finite family on the common design probability space to obtain Var⁡d(T^HT)=∑i,j∈UCov⁡d(Xi,Xj).

3.1F1F2F3F4F6step 2.1algebra

By [F6], Cov⁡d(Xi,Xj)=yiyjπiπjCov⁡d(Ii,Ij). For Ai={s:i∈s}, Ii=1Ai and IiIj=1Ai∩Aj, so [F4] and [F1] give Ed[Ii]=πi and Ed[IiIj]=πij; expanding [F3] gives Cov⁡d(Ii,Ij)=πij−πiπj=Δij by [F2], and substitution in step 2.1 proves the variance formula.

4.1givenF1F4F7step 3.1algebra

Under the additional pair-positivity assumption every coefficient in V^HT is defined and every term is integrable on the finite support. Linearity over the finite index set U×U follows by induction: the empty sum has expectation zero, and adjoining one term uses the two-variable rule [F7]. For each pair, IiIj is the indicator of joint inclusion by [F1], so [F4] gives Ed[IiIj]=πij; hence Ed[V^HT]=∑i,j∈UΔijπijyiyjπiπjEd[IiIj]=∑i,j∈UΔijyiyjπiπj=Var⁡d(T^HT) by step 3.1.

5.1givenstep 4.1algebra∎

If πij=0, the coefficient Δij/πij in the all-pairs expression is undefined; removing that unsupported term changes the expression, and step 4.1 no longer proves unbiasedness, without ruling out another design-specific estimator.

Depends on

Used by

Dependency tree · two levels

26 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources