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Average cluster size does not determine the unequal-size design effect

Statement refuted

For unequal cluster sizes, with independent PSUs, equal marginal variance, common within-cluster correlation, and negligible FPC, the design effect is still $1 + (\overset{ˉ}{m} - 1) ρ$ after substituting the arithmetic mean size $\overset{ˉ}{m}$ for the equal cluster size. In the example below, $\overset{ˉ}{m} = 2$ and the within-pair correlation is $1$ , so this proposed formula predicts $2$ , while the exact model-averaged design-effect ratio tends to $5 / 2$ as $M \to \infty$ with fixed $k$ .

The two designs are evaluated separately on each fixed realization of the model values: $S_{C}$ is a uniform SRSWOR of $k$ PSUs and $S_{E}$ is a uniform SRSWOR of $2 k$ SSUs. Define ${\hat{μ}}_{C} = \frac{M}{k N} \sum_{i \in S_{C}} T_{i}, {\hat{μ}}_{E} = \frac{1}{2 k} \sum_{(i, j) \in S_{E}} y_{i j},$ where $N = \sum_{i} n_{i} = 2 M$ . The design effect here is $E_{B} [{Var}_{d} ({\hat{μ}}_{C} ∣ B)] / E_{B} [{Var}_{d} ({\hat{μ}}_{E} ∣ B)]$ .

Facts & Assumptions

Given: An even number $M \geq 4$ of PSUs, half of size $1$ and half of size $3$ ; independent random variables $B_{i}$ with mean $0$ and variance $σ^{2} > 0$ ; fixed SSU values $y_{i j} = B_{i}$ conditional on $(B_{i})$ ; and a cluster SRSWOR of $k$ PSUs, $1 \leq k < M$ , compared with an element SRSWOR of $2 k$ SSUs from the same finite population. This matches the expected SSU count in the cluster sample. Define the design effect as the ratio of the two design variances, each averaged over the model for $(B_{i})$ .

[F1]

Cluster sampling units, cluster totals, and one-stage SRSWOR: A cluster design selects PSUs and, in one-stage sampling, observes all SSUs in selected PSUs; the cluster total is $T_{i} = \sum_{j} y_{i j}$ .

[F2]

Unbiasedness and variance of the SRSWOR cluster-total estimator: Under SRSWOR of $k$ from $M$ PSUs, the total estimator has variance $M^{2} (1 - k / M) S_{T}^{2} / k$ .

[F3]

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

[F4]

Covariance is symmetric and bilinear in finite linear combinations: Covariance is symmetric and bilinear in finite linear combinations of square-integrable variables.

[F5]

Variance of a finite sum as the sum of all variances and covariances: The variance of a finite sum of square-integrable real random variables is the sum of all pairwise covariances.

[A1]

Conditional on the model values, the finite population is fixed and each design is an ordinary uniform SRSWOR. The model variables $B_{i}$ are independent across PSUs and share the stated mean and variance.

Counterexample

1.1givenF1A1algebra

The cluster sizes are $1$ and $3$ in equal proportions, so $N = 2 M$ and $\overset{ˉ}{m} = 2$ ; in every size- $3$ cluster the values equal $B_{i}$ , giving correlation $ρ = 1$ for each within-cluster pair that exists (size- $1$ clusters have none), and every element has marginal variance $σ^{2}$ . The proposed average-size substitution therefore predicts $1 + (2 - 1) \cdot 1 = 2$ .

1.2givenA1F2F3F4F5algebra

For ${\overset{ˉ}{Y}}_{U} = N^{- 1} \sum_{i} n_{i} B_{i}$ , independence gives $E_{B} [{\overset{ˉ}{Y}}_{U}] = 0$ and ${Var}_{B} ({\overset{ˉ}{Y}}_{U}) = 5 σ^{2} / (4 M)$ ; each element has $E_{B} [y_{i j}^{2}] = σ^{2}$ , so the identity $\sum_{i, j} (y_{i j} - {\overset{ˉ}{Y}}_{U})^{2} = \sum_{i, j} y_{i j}^{2} - N {\overset{ˉ}{Y}}_{U}^{2}$ and $S_{U}^{2} = (N - 1)^{- 1} \sum_{i, j} (y_{i j} - {\overset{ˉ}{Y}}_{U})^{2}$ give $E_{B} [S_{U}^{2}] = (N σ^{2} - N {Var}_{B} ({\overset{ˉ}{Y}}_{U})) / (N - 1) = σ^{2} (2 M - 5 / 2) / (2 M - 1)$ . By [F2] on the singleton-PSU partition with sample size $2 k$ , the element mean estimator has variance $(1 - 2 k / N) S_{U}^{2} / (2 k) = (1 - k / M) S_{U}^{2} / (2 k)$ , so $E_{B} [{Var}_{d} ({\hat{μ}}_{E} ∣ B)] = [(1 - k / M) / (2 k)] σ^{2} (2 M - 5 / 2) / (2 M - 1)$ .

2.1givenA1F2F3F4F5step 1.1algebra

Let $n_{i} \in {1, 3}$ be the PSU sizes; then $T_{i} = n_{i} B_{i}$ , $\sum_{i} n_{i}^{2} = 5 M$ , and independence gives ${Var}_{B} (\overset{ˉ}{T}) = σ^{2} \sum_{i} n_{i}^{2} / M^{2}$ . Using $\sum_{i} (T_{i} - \overset{ˉ}{T})^{2} = \sum_{i} T_{i}^{2} - M {\overset{ˉ}{T}}^{2}$ and $E_{B} [T_{i}] = 0$ yields $E_{B} [\sum_{i} (T_{i} - \overset{ˉ}{T})^{2}] = \sum_{i} {Var}_{B} (T_{i}) - M {Var}_{B} (\overset{ˉ}{T}) = ((M - 1) / M) σ^{2} \sum_{i} n_{i}^{2}$ , hence $E_{B} [S_{T}^{2}] = (1 / (M - 1)) E_{B} [\sum_{i} (T_{i} - \overset{ˉ}{T})^{2}] = (σ^{2} / M) \sum_{i} n_{i}^{2} = 5 σ^{2}$ . By [F2], dividing the total-estimator variance by $N^{2} = (2 M)^{2}$ gives $E_{B} [{Var}_{d} ({\hat{μ}}_{C} ∣ B)] = (5 σ^{2} / (4 k)) (1 - k / M)$ .

3.1step 1.1step 1.2step 2.1algebra∎

The common positive factor $(1 - k / M) σ^{2} / (2 k)$ cancels, so $DEFF = (5 / 2) (2 M - 1) / (2 M - 5 / 2)$ ; at $M = 100$ this is $199 / 79$ , not $2$ , and it tends to $5 / 2$ as $M \to \infty$ , refuting the average-size substitution.

Remarks

This refutes only the arithmetic-mean-size substitution. It does not give a general design-effect formula for unequal cluster sizes, unequal weights, or nonexchangeable covariance structures; each requires its own stated estimator and variance calculation.

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Sources

Shalabh, Sampling Theory, Chapter 9: Cluster Sampling, PDF pp. 11–14, IIT Kanpur (standard reference, not scraped)
Sharon L. Lohr, Sampling: Design and Analysis, 2nd ed., §5.2.2, printed pp. 139–141 (standard reference, not scraped)