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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.
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These labels describe origin, not correctness: citations and verification chips remain separate evidence.

Continuous Neyman allocation may require constrained rounding

Example

Take two strata of sizes 2 and 10. In the first, let the values be (−1/2,1/2); in the second, take five copies each of −9/10 and 9/10. With total sample size six, the continuous Neyman allocation rounds to zero units in the first stratum under nearest- integer rounding, which violates the stratified design's positive sample-size requirement.

Facts & Assumptions

Given: The fixed strata above, independent within-stratum SRSWOR, total sample size six, and componentwise rounding to the nearest integer.

[F1]

Stratified sampling designs and the stratified total estimator: A stratified design requires integer sample sizes 1≤nh≤Nh; for a noncertainty stratum, Sh2=(Nh−1)−1∑i∈Uh(yi−yˉh)2.

[F2]

Continuous fixed-size Neyman allocation minimizes stratified variance: With positive noncertainty stratum variances and a fixed positive total sample size, the uncapped continuous optimum is proportional to NhSh; this relaxation imposes no integer, capacity, lower-bound, or minimum-sample constraints.

[F3]

Unbiasedness and variance of the stratified total estimator: For an admissible allocation, the stratified total's variance is the sum of the stratum terms Nh2(1−nh/Nh)Sh2/nh over noncertainty strata.

Verification

1.1givenchoosealgebra

The first stratum has mean zero and S12=((−1/2)2+(1/2)2)/(2−1)=1. The second has mean zero and S22=10(9/10)2/(10−1)=9, so S2=3.

2.1F1F2step 1.1algebra

The Neyman weights are N1S1=2 and N2S2=30. By [F2] and step 1.1, x1=6(2)/(2+30)=3/8 and x2=6(30)/(2+30)=45/8. Nearest-integer rounding gives (0,6), which violates [F1]'s n1≥1 condition.

3.1F1F3step 1.1step 2.1algebra∎

Allocation (n1,n2)=(1,5) is admissible. By [F3] and step 1.1 its variance is 22(1−1/2)1/1+102(1−5/10)9/5=2+90=92; this feasible correction is not claimed to solve every constrained integer problem.

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