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Sen–Yates–Grundy variance identity

Statement

Let (U,y) be a finite population with a fixed-size sampling design as in Finite populations and sampling designs, so every supported sample has the same size n. Assume πi>0 for every i∈U and put xi:=yi/πi. Then ∑j∈UΔij=0(i∈U), and the Horvitz–Thompson variance satisfies the Sen–Yates–Grundy identity Var⁡d(T^HT)=−12∑i,j∈UΔij(xi−xj)2. If additionally πij>0 for every ordered pair, then the sample expression V^SYG:=−12∑i,j∈UIiIjΔijπij(xi−xj)2 is unbiased for Var⁡d(T^HT). If Δij≤0 for every distinct pair, this condition guarantees that V^SYG is nonnegative for every supported sample. It is a sufficient sign condition for the sample expression; the true variance in the identity is nonnegative for every valid design.

Facts & Assumptions

Given: A fixed finite population and design with sample size n on every supported outcome, πi>0 for every unit, and additionally πij>0 for every pair when proving the sample-estimate claim.

[F1]

Finite populations and sampling designs: Ii indicates inclusion of unit i, πi=Pd(Ii=1), πij=Pd(Ii=Ij=1), and πij=πji.

[F2]

Inclusion-covariance difference: Δij=πij−πiπj.

[F3]

Horvitz–Thompson variance and pair-inclusion estimate: For πi>0, Var⁡d(T^HT)=∑i,j∈UΔijxixj.

[F4]

The expectation of an indicator is the probability of the event: An event indicator has expectation equal to the probability of its event.

[F5]

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].

[F6]

Moments, variance, and covariance on a probability space: Variance is the expectation of a square and therefore is nonnegative.

Proof

1.1F1F2F4F5algebra

For each i,j, IiIj indicates joint inclusion and Ii indicates inclusion by [F1], so [F4] gives their expectations as πij and πi. Since ∑jIj=n on every sample, finite linearity by induction using [F5] gives ∑jπij=Ed[Ii∑jIj]=nπi and ∑jπiπj=πiEd[∑jIj]=nπi; subtracting by [F2] gives ∑jΔij=0.

2.1F1F2F3F6step 1.1algebra

By [F1] and [F2], Δij=Δji and the column sums also vanish; expansion gives −12∑i,jΔij(xi−xj)2=−12(∑ixi2∑jΔij+∑jxj2∑iΔij−2∑i,jΔijxixj)=∑i,jΔijxixj, which is Var⁡d(T^HT) by [F3] and is nonnegative by [F6].

3.1givenF1F3F4F5step 2.1algebra

If every πij>0, each coefficient in V^SYG is defined and every term is integrable on the finite support. Finite-sum linearity follows by induction from [F5]; by [F1], each IiIj indicates joint inclusion, so [F4] gives Ed[IiIj]=πij and Ed[V^SYG]=−12∑i,jΔijπijEd[IiIj](xi−xj)2=−12∑i,jΔij(xi−xj)2=Var⁡d(T^HT) by step 2.1.

4.1givenF1step 3.1algebra∎

On a supported sample, diagonal terms vanish and symmetry gives V^SYG(s)=−∑i<j, i,j∈sΔijπij(xi−xj)2; if every distinct-pair Δij≤0, every summand is nonnegative, proving the sufficient condition.

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