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Measurement error and misclassification — Examples

1 · Prerequisites

2 · Summary

The worked example compares classical and Berkson models under their distinct assumptions. It then gives binary misclassification mechanisms whose observed associations reverse direction, one nondifferential relative to the named variables and one differential. A finite pair of latent-error decompositions has the same measured law but different variance splits; observing a validation variable can distinguish those two candidate laws.

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 →

Classical and Berkson slopes, misclassification, and validation

Example

Each construction below is a separate probability model.

Classical error. Let $(X, U)$ be uniform on ${- 1, 1}^{2}$ , put $W = X + U$ and $Y = 2 X$ . The latent-value slope is $2$ , while the measured-value slope is $1$ .
Berkson error. Let $(W, U)$ instead be uniform on ${- 1, 1}^{2}$ , put $X = W + U$ and $Y = 2 X$ . Under this different error direction and independence condition, the slope of $Y$ on $W$ is $2$ .
Nondifferential binary misclassification. Let $X$ be fair binary and $Y = X$ . Given either value of $X$ , set $W = X$ with probability $1 / 5$ and $W = 1 - X$ with probability $4 / 5$ . This mechanism is nondifferential relative to $Y$ given $X$ , but it reverses the sign of the observed mean contrast: the true contrast is $1$ and the contrast by $W$ is $- 3 / 5$ .
Differential binary misclassification. Give $(X, Y)$ probabilities $3 / 8, 1 / 8, 1 / 8, 3 / 8$ at $(0, 0), (0, 1), (1, 0), (1, 1)$ , respectively, and set $W = 1 - Y$ . The true contrast of $Y$ across $X$ is $1 / 2$ , while the contrast across $W$ is $- 1$ ; here $W$ is not conditionally independent of $Y$ given $X$ .
Unidentified variance split and validation. Let $A, B, C, D$ be the coordinate signs on the uniform space ${- 1, 1}^{4}$ . The decompositions $X_{0} = A + B, U_{0} = C + D$ and $X_{1} = A + B + C, U_{1} = D$ both give $W = A + B + C + D$ . Their latent-error variance pairs are $(2, 2)$ and $(3, 1)$ . The joint validation laws of $(X, W)$ differ: $P (X_{0} = 0) = 1 / 2$ but $P (X_{1} = 0) = 0$ . Thus observing $X$ adds information that separates these two candidate laws.

The fifth construction shows what additional validation information can do for this candidate pair; it does not assert a universal correction rule.

Facts & Assumptions

Given: The finite probability models and their cell probabilities specified in the Example.

[F1]

Classical and Berkson errors have the opposite additive equations and their own independence conditions (Additive measurement error and categorical misclassification models).

[F2]

Variance and covariance are centered expectations, and covariance is bilinear (Moments, variance, and covariance on a probability space, Covariance is symmetric and bilinear in finite linear combinations).

[F3]

Independence of real random variables is independence of their generated sigma-algebras (Independent random elements).

[F4]

Expectations of products of finite independent integrable families factor (Expectation factors over a finite product of mutually independent random variables).

[F5]

Under the stated classical hypotheses, the measured-value slope is the variance-ratio formula in the classical error theorem (Classical error attenuates the simple linear regression slope).

[F6]

Under the stated Berkson hypotheses, the measured-value slope equals the latent-value slope (Berkson error preserves the simple linear regression slope).

[F7]

Nondifferentiality is conditional independence relative to the named variables (Nondifferential measurement error relative to named variables).

[F8]

The general conditional-independence definition explicitly assumes the Axiom of Choice so that its conditional-expectation classes are available (Conditional independence given a sigma-algebra, The Axiom of Choice).

[F9]

The observed law need not identify the latent-error variance pair (The observed law need not identify the latent-error variance split).

[F10]

Conditional independence is the conditional-expectation factorization; it holds automatically when one side is measurable with respect to the conditioning sigma-algebra (Conditional independence given a sigma-algebra).

Verification

1.1givenF1F2F3F4F5algebra

In the first construction, each of the four sign pairs has probability $1 / 4$ and each coordinate sign has probability $1 / 2$ , so $X$ and $U$ are independent. Each has $E [X] = 0$ and $Var (X) = 1$ , and likewise for $U$ . Their covariance is $0$ by [F2]–[F4]. The error is classical, $ϵ = 0$ , and the response coefficient is $β = 2$ . Applying [F5] gives measured slope $2 (1) / (1 + 1) = 1$ , while $Y = 2 X$ gives latent slope $2$ .

1.2givenF1F2F3F4F6

In the second construction, $W$ and $U$ are independent centered signs with variance $1$ , $X = W + U$ , and $ϵ = 0$ . Thus the distinct Berkson hypotheses hold, and [F6] gives measured slope $2$ .

1.3givenF7F8F10algebra

In the third construction $Y = X$ . On each positive-probability stratum $X = x$ , the variable $Y$ is constant. The conditional-independence definition [F10], applied under the explicit Axiom of Choice premise [F8], therefore gives for all bounded measurable $f, g$ , $E [f (W) g (Y) ∣ σ (X)] = g (X) E [f (W) ∣ σ (X)] = E [f (W) ∣ σ (X)] E [g (Y) ∣ σ (X)] .$ By [F7], the error is nondifferential relative to $Y$ given $X$ .

1.4givenF7F8F10algebra

In the fourth construction, $E [Y ∣ X = 1] - E [Y ∣ X = 0] = 3 / 4 - 1 / 4 = 1 / 2$ . Since $W = 1 - Y$ , $E [Y ∣ W = 1] - E [Y ∣ W = 0] = 0 - 1 = - 1$ . On the positive stratum $X = 1$ , the conditional expectation of $1_{{W = 1}} 1_{{Y = 1}}$ is $0$ , whereas the product of the two conditional expectations is $(1 / 4) (3 / 4) = 3 / 16$ . The factorization in [F10] fails, so the error is differential relative to $Y$ given $X$ .

1.5givenF1F3F4algebra

On the uniform four-sign space, every specified pattern of $k$ distinct signs has probability $2^{- k}$ ; thus the coordinates are mutually independent, centered, and have variance $1$ . For sign-pattern sets $E, F$ on disjoint blocks of sizes $2, 2$ or $3, 1$ , the joint probability is $∣ E ∣ ∣ F ∣ / 16$ , the product of the two block probabilities; hence $X_{i}$ is independent of $U_{i}$ for $i = 0, 1$ . Their distinct-coordinate covariances are zero by [F4]. They satisfy $W_{0} = W_{1} = A + B + C + D$ pointwise.

2.1givenF2step 1.3algebra

The true contrast is $E [Y ∣ X = 1] - E [Y ∣ X = 0] = 1$ . The table gives $P (W = 1) = 1 / 2$ , $E [Y ∣ W = 1] = 1 / 5$ , and $E [Y ∣ W = 0] = 4 / 5$ , so the measured contrast is $1 / 5 - 4 / 5 = - 3 / 5$ . Thus this nondifferential mechanism reverses the contrast direction.

2.2F2F3F4F9step 1.5algebra

Bilinearity [F2] and the zero coordinate covariances from step 1.5 give $Var (X_{0}) = 1 + 1 = 2$ , $Var (U_{0}) = 1 + 1 = 2$ , $Var (X_{1}) = 1 + 1 + 1 = 3$ , and $Var (U_{1}) = 1$ . Thus the same observed $W$ law is compatible with distinct variance pairs, as asserted in [F9].

3.1givenstep 1.5step 2.2algebra∎

The event $X_{0} = 0$ requires opposite signs among $A, B$ , so has probability $2 / 4 = 1 / 2$ . Since $X_{1} = A + B + C$ is a sum of three signs, it takes values only in ${- 3, - 1, 1, 3}$ , and therefore $P (X_{1} = 0) = 0$ . These unequal marginals show that the joint validation laws of $(X, W)$ differ, so observing $X$ adds information that separates these two candidate laws.