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HT variance and SRSWOR — Examples

1 · Prerequisites

2 · Summary

These examples enumerate a three-unit SRSWOR, expose what zero pair support does and does not permit, check the one-unit and census boundaries, and show how a valid unbiased Sen–Yates–Grundy estimate can be negative for one sample even while the true design variance remains nonnegative.

3 · Logical flowchart

4 · Definitions, theorems and proofs

None yet.

5 · Examples, counterexamples and false statements

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A three-unit SRSWOR variance calculation

Example

Let $U = {1, 2, 3}$ with fixed values $y = (1, 2, 4)$ , and select an SRSWOR sample of size $n = 2$ . The three samples each have probability $1 / 3$ . Their Horvitz–Thompson totals are $9 / 2$ , $15 / 2$ , and $9$ , respectively, so the true design variance of the total is $7 / 2$ . The pair-inclusion and Sen–Yates–Grundy sample estimates are $3 / 4$ , $27 / 4$ , and $3$ ; their mean is also $7 / 2$ . The sample means are $3 / 2$ , $5 / 2$ , and $3$ , with variance $7 / 18$ . The finite support of three values is not an exact normal law.

Sample	$p_{d} (s)$	${\hat{T}}_{H T}$	${\hat{V}}_{H T}$	${\hat{V}}_{S Y G}$	${\overset{ˉ}{y}}_{s}$
${1, 2}$	$1 / 3$	$9 / 2$	$3 / 4$	$3 / 4$	$3 / 2$
${1, 3}$	$1 / 3$	$15 / 2$	$27 / 4$	$27 / 4$	$5 / 2$
${2, 3}$	$1 / 3$	$9$	$3$	$3$	$3$

Facts & Assumptions

Given: $U = {1, 2, 3}$ , $y_{1} = 1, y_{2} = 2, y_{3} = 4$ , and SRSWOR of size two.

[F1]

Simple random sampling without replacement: The support consists of all two-element subsets, each with probability $1 / (\binom{3}{2}) = 1 / 3$ .

[F2]

Horvitz–Thompson variance and pair-inclusion estimate: If $π_{i} > 0$ , the total variance is $\sum_{i, j} Δ_{i j} y_{i} y_{j} / (π_{i} π_{j})$ ; if all $π_{i j} > 0$ , the pair-inclusion estimate has expectation equal to that variance.

[F3]

Sen–Yates–Grundy variance identity: For a fixed-size design, ${Var}_{d} ({\hat{T}}_{H T}) = - \frac{1}{2} \sum_{i, j} Δ_{i j} (y_{i} / π_{i} - y_{j} / π_{j})^{2}$ , and the sample pair-inclusion version is unbiased when every $π_{i j} > 0$ .

[F4]

SRSWOR inclusion covariance and sample-mean variance: For $N > 1$ and $1 \leq n \leq N$ , ${Var}_{d} ({\overset{ˉ}{y}}_{S}) = (1 - n / N) S_{U}^{2} / n$ .

Verification

1.1givenF1choosealgebra

By [F1], the support is ${{1, 2}, {1, 3}, {2, 3}}$ with equal probabilities. Counting occurrences gives $π_{i} = 2 / 3$ , $π_{i j} = 1 / 3$ for $i \neq j$ , $Δ_{i i} = 2 / 9$ , and $Δ_{i j} = - 1 / 9$ for $i \neq j$ ; the population mean is $7 / 3$ and $S_{U}^{2} = ((1 - 7 / 3)^{2} + (2 - 7 / 3)^{2} + (4 - 7 / 3)^{2}) / 2 = 7 / 3$ .

2.1givenF1F2step 1.1algebra

Each observed total is $(3 / 2)$ times the sum of the two observed values, so the HT totals are $9 / 2, 15 / 2, 9$ and have mean $7$ ; their design variance is $\frac{1}{3} ((9 / 2 - 7)^{2} + (15 / 2 - 7)^{2} + (9 - 7)^{2}) = 7 / 2$ .

2.2F1F2F3step 1.1algebra

Put $x_{i} = y_{i} / π_{i}$ , so $(x_{1}, x_{2}, x_{3}) = (3 / 2, 3, 6)$ . For any sampled pair ${i, j}$ , the HT pair estimate and the SYG estimate both equal $\frac{1}{3} (x_{i} - x_{j})^{2}$ , since $Δ_{i i} / π_{i} = 1 / 3$ and $Δ_{i j} / π_{i j} = - 1 / 3$ ; thus they are $3 / 4, 27 / 4, 3$ and their expectation is $\frac{1}{3} (3 / 4 + 27 / 4 + 3) = 7 / 2$ by [F2] and [F3].

3.1F1F4step 1.1algebra∎

The sample means are $3 / 2, 5 / 2, 3$ , with average $7 / 3$ and design variance $\frac{1}{3} ((3 / 2 - 7 / 3)^{2} + (5 / 2 - 7 / 3)^{2} + (3 - 7 / 3)^{2}) = 7 / 18$ ; [F4] gives $(1 - 2 / 3) (7 / 3) / 2 = 7 / 18$ , and the three supported means form a finite set rather than an exact normal calibration.

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Zero pair support can prevent universal unbiased variance estimation

Example

Let $U = {1, 2}$ have fixed values $y_{1} = a, y_{2} = b$ , and sample one unit by SRSWOR. Then $π_{1} = π_{2} = 1 / 2$ but $π_{12} = 0$ . The HT total is $2 a$ or $2 b$ , each with probability $1 / 2$ , so its variance is $(a - b)^{2}$ . The all-pairs inverse-inclusion formula cannot use the unsupported pair $(1, 2)$ . If one keeps only supported pairs, its expectation is $a^{2} + b^{2}$ , which is not the variance for all $a, b$ . In fact, no function of the observed singleton label and value is unbiased for this variance for every finite population.

Facts & Assumptions

Given: $U = {1, 2}$ , fixed real values $(a, b)$ , and a uniformly sampled singleton.

[F1]

Finite populations and sampling designs: The observed record contains the selected unit labels and their observed values.

[F2]

Simple random sampling without replacement: Each singleton has probability $1 / (\binom{2}{1}) = 1 / 2$ .

[F3]

Horvitz–Thompson variance and pair-inclusion estimate: The HT total uses weights $1 / π_{i}$ ; its variance has a pair term $Δ_{i j} y_{i} y_{j} / (π_{i} π_{j})$ , and the all-pairs inverse estimate is established under positive $π_{i j}$ for every ordered pair.

[F4]

Statistics and estimators: An estimator is a statistic, hence a function of the observed data record, with the target as its specified estimand.

Verification

1.1givenF2F3algebra

The two possible samples are ${1}$ and ${2}$ , each with probability $1 / 2$ by [F2]. Thus $π_{1} = π_{2} = 1 / 2$ and $π_{12} = 0$ ; the HT total is $2 a$ or $2 b$ , with mean $a + b$ and variance $\frac{1}{2} (2 a - a - b)^{2} + \frac{1}{2} (2 b - a - b)^{2} = (a - b)^{2}$ .

2.1givenF1F2F3step 1.1algebra

The coefficient $Δ_{12} / π_{12}$ is undefined. Restricting the sample-pair expression to supported ordered pairs leaves only diagonal terms: on ${1}$ the value is $\frac{Δ_{11}}{π_{11}} (a / π_{1})^{2} = (1 / 4) / (1 / 2) (2 a)^{2} = 2 a^{2}$ , and on ${2}$ it is $2 b^{2}$ . Its expectation is $a^{2} + b^{2}$ , which at $a = b = 1$ is $2$ while the variance in step 1.1 is $0$ ; omitting the unsupported cross pair is not universally unbiased.

3.1givenF1F2F4step 1.1algebracontradiction∎

Any estimator $G$ of the observed record has values $g_{1} (a)$ on ${1}$ and $g_{2} (b)$ on ${2}$ by [F1] and [F4], so universal unbiasedness would require $(g_{1} (a) + g_{2} (b)) / 2 = (a - b)^{2}$ for every $a, b$ . At $(0, 0), (0, 1), (1, 0), (1, 1)$ this gives respectively $g_{1} (0) + g_{2} (0) = 0$ , $g_{1} (0) + g_{2} (1) = 2$ , $g_{1} (1) + g_{2} (0) = 2$ , and $g_{1} (1) + g_{2} (1) = 0$ ; the first and fourth left sides sum to the same value as the second and third, but the right sides are $0$ and $4$ , a contradiction.

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Census and singleton SRSWOR boundaries

Example

For $U = {1, 2, 3}$ with values $(1, 2, 4)$ , the population mean is $7 / 3$ and $S_{U}^{2} = 7 / 3$ . At a census ( $n = 3$ ), the sample mean is fixed and its design variance is zero. At $n = 1$ , the sample mean takes values $1, 2, 4$ equally often, with variance $14 / 9$ , agreeing with the SRSWOR formula. The usual sample variance $s^{2} = (n - 1)^{- 1} \sum_{i \in S} (y_{i} - {\overset{ˉ}{y}}_{S})^{2}$ is undefined when $n = 1$ . A population of size $N = 1$ is outside the theorem's $S_{U}^{2}$ convention because its denominator $N - 1$ is zero.

Facts & Assumptions

Given: $U = {1, 2, 3}$ with fixed values $(1, 2, 4)$ .

[F1]

Simple random sampling without replacement: At each admissible sample size, every subset of that size is equally likely.

[F2]

SRSWOR inclusion covariance and sample-mean variance: For $N > 1$ and $1 \leq n \leq N$ , ${Var}_{d} ({\overset{ˉ}{y}}_{S}) = (1 - n / N) S_{U}^{2} / n$ .

Verification

1.1givenF1F2algebra

The population mean is $(1 + 2 + 4) / 3 = 7 / 3$ and $S_{U}^{2} = ((1 - 7 / 3)^{2} + (2 - 7 / 3)^{2} + (4 - 7 / 3)^{2}) / 2 = 7 / 3$ . At $n = 3$ , [F1] gives the single sample $U$ , so ${\overset{ˉ}{y}}_{S} = 7 / 3$ is constant with variance zero; [F2] gives $(1 - 3 / 3) (7 / 3) / 3 = 0$ .

2.1givenF1F2step 1.1algebra∎

At $n = 1$ , [F1] gives sample means $1, 2, 4$ , each with probability $1 / 3$ ; their expectation is $7 / 3$ and variance is $\frac{1}{3} ((1 - 7 / 3)^{2} + (2 - 7 / 3)^{2} + (4 - 7 / 3)^{2}) = 14 / 9$ . Formula [F2] gives $(1 - 1 / 3) (7 / 3) = 14 / 9$ . The denominator $n - 1$ in $s^{2}$ is zero, while $N = 1$ would make the population-variance denominator zero; neither is substituted into an undefined expression.

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Positive pair covariances and a negative SYG estimate

Example

Let $U = {1, 2, 3, 4}$ and use the fixed-size design $p_{d} ({1, 2}) = p_{d} ({3, 4}) = \frac{2}{5}, p_{d} ({1, 3}) = p_{d} ({1, 4}) = p_{d} ({2, 3}) = p_{d} ({2, 4}) = \frac{1}{20} .$ All units have $π_{i} = 1 / 2$ . The pair differences are $Δ_{12} = Δ_{34} = 3 / 20 > 0$ and $Δ_{i j} = - 1 / 5$ for the four cross pairs. Thus a positive pair difference does not force a negative Sen–Yates–Grundy (SYG) estimate for every fixed population: for $y = (0, 0, 1, 1)$ all six sample estimates are nonnegative. But for $y = (0, 1, 0, 0)$ the supported sample ${1, 2}$ produces the negative estimate $- 3 / 2$ , even though the true design variance is $1$ . In both schedules the SYG estimates average to the corresponding true variance.

Facts & Assumptions

Given: The six two-unit samples and their probabilities are as stated, with fixed population values chosen separately for each schedule.

[F1]

Finite populations and sampling designs: A sampling design is a finite support of subsets with positive weights summing to one; $π_{i}$ and $π_{i j}$ are the first- and second-order inclusion probabilities.

[F2]

Inclusion-covariance difference: $Δ_{i j} = π_{i j} - π_{i} π_{j}$ .

[F3]

Sen–Yates–Grundy variance identity: For fixed sample size and positive first-order inclusion probabilities, the SYG sample estimate is unbiased when all $π_{i j} > 0$ ; pairwise nonpositive distinct-unit $Δ_{i j}$ is a sufficient guarantee of nonnegative sample estimates.

Verification

1.1givenF1F2algebra

The six weights sum to $2 (2 / 5) + 4 (1 / 20) = 1$ and every sample has size two; each unit's incident weights sum to $1 / 2$ , so $π_{i} = 1 / 2$ . Pair probabilities are $π_{12} = π_{34} = 2 / 5$ and $1 / 20$ for cross pairs; by [F2], $Δ_{12} = Δ_{34} = 2 / 5 - 1 / 4 = 3 / 20$ and $Δ_{i j} = 1 / 20 - 1 / 4 = - 1 / 5$ for cross pairs. All $π_{i j}$ are positive.

2.1F1F3step 1.1algebra

For $y^{A} = (0, 0, 1, 1)$ , $x = (0, 0, 2, 2)$ . Samples ${1, 2}$ and ${3, 4}$ give SYG estimate zero; each cross sample gives $- Δ_{i j} (x_{i} - x_{j})^{2} / π_{i j} = 16$ . The six estimates are $0, 0, 16, 16, 16, 16$ and have expectation $4 (1 / 20) (16) = 16 / 5$ . HT totals are $0, 4, 2$ with probabilities $2 / 5, 2 / 5, 1 / 5$ , so their mean is $2$ and variance is $(2 / 5) 4 + (2 / 5) 4 = 16 / 5$ .

3.1F1F2F3step 1.1step 2.1algebra∎

For $y^{B} = (0, 1, 0, 0)$ , $x = (0, 2, 0, 0)$ . The HT total is $2$ or $0$ with probability $1 / 2$ each, so its variance is $1$ . Sample ${1, 2}$ has probability $2 / 5$ and SYG estimate $- Δ_{12} (x_{1} - x_{2})^{2} / π_{12} = - (3 / 20) (4) / (2 / 5) = - 3 / 2$ ; samples ${2, 3}$ and ${2, 4}$ have estimate $16$ and the other three have zero. The expectation is $(2 / 5) (- 3 / 2) + 2 (1 / 20) (16) = 1$ , equal to the variance despite the negative sample value; thus the nonpositive-Delta condition is sufficient but not necessary for all supported estimates to be nonnegative for a particular fixed population, as schedule $y^{A}$ in step 2.1 shows.