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Method agreement — Examples

1 · Prerequisites

2 · Summary

The examples show a constant method offset hidden by perfect correlation and a finite repeated-measures model whose shared random intercept makes paired differences dependent. Both calculations keep the population/model assumptions visible and do not claim endpoint coverage or empirical interchangeability.

3 · Logical flowchart

4 · Definitions, theorems and proofs

None yet.

5 · Examples, counterexamples and false statements

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

Perfect correlation can coexist with a method offset

Example

Let $Ω = {0, 1}$ have probability $1 / 2$ at each point, and let two methods on the same target report $A (ω) = ω$ and $B (ω) = ω + c$ for a fixed $c > 0$ . Their correlation is $1$ , but their difference is the nonzero constant $- c$ and their conventional limits of agreement both equal $- c$ .

Facts & Assumptions

Given: $Ω = {0, 1}$ with weights $1 / 2, 1 / 2$ , $A (ω) = ω$ , $B (ω) = ω + c$ , and $c > 0$ . For this example define $Corr (A, B) : = \frac{Cov (A, B)}{\sqrt{Var (A) Var (B)}} .$

[F1]

For $D = A - B$ , the bias is $E [D]$ , the spread is $\sqrt{Var (D)}$ , and conventional limits are their mean plus or minus $1.96$ times that spread. Method agreement and limits of agreement

[F2]

Variance and covariance are expectations of the centered square and centered product. Moments, variance, and covariance on a probability space

[F3]

On a finite probability space, expectation is the weighted sum of values over its atoms. Expectation agrees with the published finite weighted sum

[F4]

Covariance is symmetric and bilinear on finite linear combinations, including constant summands. Covariance is symmetric and bilinear in finite linear combinations

Verification

Given: The two-point probability model and method values in the Example. Verify the correlation and limits.

1.1F2F3F4givenalgebra

By [F3], $E [A] = 1 / 2$ and $E [B] = 1 / 2 + c$ . Directly summing the two centered squares gives $Var (A) = Var (B) = 1 / 4$ . By [F4], $Cov (A, B) = Cov (A, A + c) = Var (A) + Cov (A, c) = 1 / 4$ , since covariance with a constant is zero. Thus the denominator in the definition of correlation is $1 / 4$ and $Corr (A, B) = 1$ .

1.2F1F2F3algebra

At both points $D = A - B = - c$ , so by [F2] and [F3], $E [D] = - c$ and $Var (D) = 0$ . Hence its conventional limits are $- c \pm 1.96 \cdot 0 = (- c, - c)$ , while its bias is nonzero.

2.1step 1.1step 1.2given∎

Therefore perfect correlation does not imply zero bias or interchangeable measurements; any practical tolerance must be supplied for the intended use.

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

Repeated method differences can share a random intercept

Example

Let $Ω = {- 1, 1}^{3}$ have uniform probability, and let $U, ϵ_{1}, ϵ_{2}$ be its coordinate variables. For two repeated paired measurements on one target, suppose the differences follow the random-intercept model $D_{j} = U + ϵ_{j}$ , $j = 1, 2$ . Then each difference has mean $0$ and variance $2$ , but $Cov (D_{1}, D_{2}) = 1$ . In fact, $P (D_{1} = 2, D_{2} = 2) = \frac{1}{8} \neq \frac{1}{16} = P (D_{1} = 2) P (D_{2} = 2) .$ The repeated differences are dependent and hence are not an iid sample.

Facts & Assumptions

Given: The eight equally likely triples in $Ω$ , coordinate functions $U, ϵ_{1}, ϵ_{2}$ , and differences $D_{j} = U + ϵ_{j}$ .

[F1]

Independent random elements are those whose generated sigma-algebras are independent. Independent random elements

[F2]

An iid family is independent and has a common law. Identical distribution and IID families

[F3]

Finite-space expectations are weighted sums over atoms. Expectation agrees with the published finite weighted sum

[F4]

Variance and covariance are the expectations of centered squares and centered products. Moments, variance, and covariance on a probability space

[F5]

Covariance is bilinear on finite linear combinations. Covariance is symmetric and bilinear in finite linear combinations

[F6]

Independent square-integrable real random variables have zero covariance. Independence forces covariance to vanish

Verification

Given: $Ω = {- 1, 1}^{3}$ with uniform weights and $D_{j} = U + ϵ_{j}$ . Verify the marginal moments and the within-target dependence.

1.1F1F3F4givenalgebra

Each triple has mass $1 / 8 = (1 / 2)^{3}$ , so for arbitrary subsets $S_{1}, S_{2}, S_{3} \subseteq {- 1, 1}$ , $P (U \in S_{1}, ϵ_{1} \in S_{2}, ϵ_{2} \in S_{3}) = \prod_{j = 1}^{3} ∣ S_{j} ∣ / 2$ . Thus [F1] makes the coordinates independent. Each coordinate has mean $0$ and variance $1$ by the finite weighted sum [F3] and centered-square definition [F4].

2.1F3F4F5F6step 1.1algebra

By [F5] and [F6], $E [D_{j}] = 0$ , $Var (D_{j}) = Var (U) + Var (ϵ_{j}) + 2 Cov (U, ϵ_{j}) = 2$ , and $Cov (D_{1}, D_{2}) = Var (U) = 1$ ; all cross terms vanish because they are covariances of distinct independent coordinates.

3.1F1F2F3givenalgebra∎

The event $D_{1} = D_{2} = 2$ requires the single triple $(1, 1, 1)$ , so its probability is $1 / 8$ . Each event $D_{j} = 2$ requires $U = 1$ and $ϵ_{j} = 1$ and has probability $1 / 4$ , making the product of marginal probabilities $1 / 16$ . The factorization required for independence therefore fails; by [F2] the two differences are not iid.