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Blocking, pairs, and Latin squares — Examples

1 · Prerequisites

2 · Summary

These examples distinguish complete-block support from incomplete blocks, show why a Latin-square randomization restricts the assignments allowed by an RCBD, calculate a paired contrast with heterogeneous unit effects, and exhibit two interaction/error decompositions of the same two-by-two residual array. The paired counterexample shows why the observed contrast's design variance also depends on baseline outcomes.

3 · Logical flowchart

4 · Definitions, theorems and proofs

None yet.

5 · Examples, counterexamples and false statements

ExampleConstruction: AI-generatedVerification: AI-generatedjudge pass (gpt-6-luna)Open item page →

A missing treatment falls outside the complete-block assignment space

Example

There are three treatments $A, B, C$ and one block, but only two units receive treatments $A$ and $C$ . This is not a complete block: the assignment omits $B$ , so it is not an element of the RCBD assignment space.

Verification

Given: A block with treatments ${A, B, C}$ whose observed assignments are $A$ and $C$ only.

[F1] In an RCBD assignment, each block receives every treatment exactly once; equivalently its unit-to-treatment map is a permutation (Randomized complete-block design).

1.1F1given

The image of this block's assignment is ${A, C}$ , which omits $B$ , so the map is not onto ${A, B, C}$ and is not a permutation.

2.1F1step 1.1∎

Every assignment in the RCBD sample space is a permutation, while this assignment is not; therefore this block is incomplete and outside that sample space.

ExampleConstruction: AI-generatedVerification: AI-generatedjudge pass (gpt-6-luna)Open item page →

A Latin-square assignment is a strict subset of RCBD assignments

Example

With treatments $A, B, C$ , the cyclic array $\begin{array}{ccc} A & B & C \\ B & C & A \\ C & A & B \end{array}$ is a Latin square. The array with all three rows equal to $A, B, C$ is an allowed RCBD assignment of three blocks, but is not a Latin square. Under the uniform RCBD law that particular assignment has probability $1 / 216$ ; under the uniform Latin-square law it has probability zero.

Verification

Given: Three treatments, three blocks, and the two displayed assignments in the Example.

[F1] An RCBD assignment consists of an independently randomized treatment permutation in each block (Randomized complete-block design).

[F2] A Latin-square assignment uses every treatment once in every row and every column, and the specified law is uniform on all such squares (Randomized Latin-square design).

1.1F2given

In the cyclic array, each row and each column contains $A, B, C$ once, so it belongs to the Latin-square sample space.

1.2F1F2given

In the repeated-row array, each row contains $A, B, C$ once, so it is an RCBD assignment; each column repeats a single treatment, so it is not a Latin square.

2.1F1F2step 1.2algebra∎

There are $3! = 6$ permutations per RCBD block and the blocks are independently uniform, so the particular repeated-row assignment has probability $6^{- 3} = 1 / 216$ . It has probability zero under a law supported only on Latin squares.

ExampleConstruction: AI-generatedVerification: AI-generatedjudge pass (gpt-6-luna)Open item page →

A paired contrast with unequal unit-level effects

Example

Suppose $Y_{1} (0) = Y_{2} (0) = 0$ , $Y_{1} (1) = 4$ , and $Y_{2} (1) = 2$ . The unit effects are $δ_{1} = 4$ and $δ_{2} = 2$ . Under the two equally likely assignments, the treated-minus-control contrasts are $4$ and $2$ . Their mean is the pair average effect $3$ and their design variance is $1$ . The centered paired-contrast values are $+ 1$ and $- 1$ with equal weights, so their variance is $1$ .

Verification

Given: A two-unit, two-treatment pair with no interference, $Y_{1} (0) = Y_{2} (0) = 0$ , $Y_{1} (1) = 4$ , and $Y_{2} (1) = 2$ .

[F1] In the $t = 2, b = 1$ RCBD, the two treatment permutations are equally likely (Randomized complete-block design).

[F2] On a finite probability space, expectation is the weighted sum of the values (Expectation of a real random variable on a finite probability space).

[F3] Variance is the expectation of the squared centered value (Moments, variance, and covariance on a probability space).

1.1givenF1

If unit $1$ receives treatment, the contrast is $Y_{1} (1) - Y_{2} (0) = 4$ ; if unit $2$ receives treatment, it is $Y_{2} (1) - Y_{1} (0) = 2$ . They are equally likely by [F1].

2.1F1F2F3step 1.1algebra∎

By [F2] the contrast mean is $(4 + 2) / 2 = 3$ , the pair average effect. By [F3] its variance is ${(4 - 3)^{2} + (2 - 3)^{2}} / 2 = 1$ . Thus the example has unequal unit effects and directly computes the observed contrast moments.

ExampleConstruction: AI-generatedVerification: AI-generatedjudge pass (gpt-6-luna)Open item page →

A two-by-two interaction and error share one residual array

Example

The residual array $R = (\begin{array}{cc} 1 & - 1 \\ - 1 & 1 \end{array})$ has zero row and column sums. It can be represented as an interaction $γ = R$ plus projected error $e = 0$ , or as interaction $γ = 0$ plus projected error $e = R$ . Both decompositions give the same observed additive residual.

Verification

Given: The $2 \times 2$ array $R$ in the Example and the residual-space conclusion of One cell observation confounds unrestricted interaction with residual error.

[F1] For one observation per cell, a double-centered interaction and the additive-model residual both lie in the same residual subspace (One cell observation confounds unrestricted interaction with residual error).

1.1givenF1algebra

Each row and each column of $R$ sums to zero, so $R$ is double-centered and belongs to that residual subspace.

2.1F1step 1.1algebra∎

Both pairs $(γ, e) = (R, 0)$ and $(γ, e) = (0, R)$ consist of vectors in the residual subspace and sum to $R$ ; hence they yield the identical observed residual with different interaction/error allocations.

CounterexampleConstruction: AI-generatedVerification: AI-generatedjudge pass (gpt-6-luna)Open item page →

Observed paired-contrast variance is not determined by effect heterogeneity

Statement refuted

Define $δ_{i} = Y_{i} (1) - Y_{i} (0)$ , and let $D$ be the observed treated-minus-control contrast. The claimed formula is that in every two-unit paired randomization, ${Var}_{d} (D) = \frac{(δ_{1} - δ_{2})^{2}}{4} .$

Facts & Assumptions

Given: Two units with no-interference binary potential outcomes $Y_{1} (0) = 0$ , $Y_{1} (1) = 1$ , $Y_{2} (0) = 10$ , and $Y_{2} (1) = 10$ , and a uniform assignment of exactly one unit to treatment $1$ .

[F1]

The observed paired contrast has the design expectation and variance proved in the local corrected theorem (Design expectation and variance of a paired contrast).

[F2]

The effect of the unit selected for treatment $1$ has variance $(δ_{1} - δ_{2})^{2} / 4$ (Moments of the effect for the randomly selected treated unit).

Counterexample

1.1givenF1algebra

The fixed effects are $δ_{1} = 1$ and $δ_{2} = 0$ . If unit $1$ is treated, the observed contrast is $D_{1} = Y_{1} (1) - Y_{2} (0) = - 9$ ; if unit $2$ is treated, it is $D_{2} = Y_{2} (1) - Y_{1} (0) = 10$ . The two assignments each have weight $1 / 2$ .

2.1F1step 1.1algebra

The observed contrast has mean $(- 9 + 10) / 2 = 1 / 2$ and variance ${(- 9 - 1 / 2)^{2} + (10 - 1 / 2)^{2}} / 2 = 90.25$ , also given by [F1]. But the proposed expression is $(δ_{1} - δ_{2})^{2} / 4 = 1 / 4$ . Hence it is not the variance of the observed paired contrast.

3.1F2step 2.1algebra∎

The variance $1 / 4$ is instead the variance of the randomly selected unit effect $S \in {1, 0}$ from [F2]. This distinguishes that latent schedule quantity from the observed contrast, and completes the counterexample.

Blocking, pairs, and Latin squares — Examples

1 · Prerequisites

2 · Summary

3 · Logical flowchart

4 · Definitions, theorems and proofs

5 · Examples, counterexamples and false statements

A missing treatment falls outside the complete-block assignment space

Example

Verification

A Latin-square assignment is a strict subset of RCBD assignments

Example

Verification

A paired contrast with unequal unit-level effects

Example

Verification

A two-by-two interaction and error share one residual array

Example

Verification

Observed paired-contrast variance is not determined by effect heterogeneity

Statement refuted

Facts & Assumptions

Counterexample

Sources