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The observed law need not identify the latent-error variance split

Statement

Within the class of centered, square-integrable classical additive error models with $U$ independent of $X$ , the observed law of $W = X + U$ need not identify the target $ψ (θ) = ({Var}_{θ} (X_{θ}), {Var}_{θ} (U_{θ})) \in [0, + \infty)^{2} .$ In particular, there is a two-point parametric model whose two parameter values give the same law of $W$ but different latent-error variance pairs.

Facts & Assumptions

Given: A classical additive error model has $W = X + U$ , and variances and independence have their probability-space meanings.

[F1]

Classical additive error is the equation $W = X + U$ with centered $U$ independent of $X$ (Additive measurement error and categorical misclassification models).

[F2]

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

[F3]

For independent integrable random variables, products are integrable and their expectations factor (Expectation factors over a finite product of mutually independent random variables).

[F4]

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

[F5]

A parametric statistical model is a parameter-indexed family of sampling laws on a fixed measurable sample space (Parametric statistical models).

[F6]

An estimand is a function from the parameter set to a target space (Estimands on a statistical model).

[F7]

An estimand is identifiable only if equal sampling laws always give equal target values (Identifiability of an estimand).

Proof

1.1givenF2F3F4algebra

Let $Ω = {- 1, 1}^{4}$ with its full power set and uniform probability, and label its coordinate variables $A, B, C, D$ . For any subsets $S_{i} \subseteq {- 1, 1}$ on $k$ distinct coordinates, counting gives $P (A_{i} \in S_{i} for all selected i) = \prod_{i} \frac{∣ S_{i} ∣}{2},$ where $(A_{1}, A_{2}, A_{3}, A_{4}) = (A, B, C, D)$ . Thus the coordinate sigma-algebras are mutually independent in the sense of [F2]. Each coordinate has mean $0$ and variance $1$ ; [F3] gives zero expectation for products of distinct coordinates, so [F4] gives their zero covariances.

2.1givenF1F2step 1.1

Define $(X_{0}, U_{0}, W_{0}) = (A + B, C + D, A + B + C + D)$ and $(X_{1}, U_{1}, W_{1}) = (A + B + C, D, A + B + C + D)$ . Thus $W_{0} = W_{1}$ pointwise. For any Borel $B_{0}, B_{1} \subseteq R$ , the event $X_{0} \in B_{0}$ is determined by $(A, B)$ and $U_{0} \in B_{1}$ by $(C, D)$ ; their preimage sets $E, F \subseteq {- 1, 1}^{2}$ satisfy $P ((A, B) \in E, (C, D) \in F) = ∣ E ∣ ∣ F ∣ / 16 = (∣ E ∣ / 4) (∣ F ∣ / 4)$ . The same count for the three-coordinate block and the last coordinate proves $X_{i}$ independent of $U_{i}$ for $i = 0, 1$ . Both errors are centered.

2.2F3F4step 1.1algebra

Bilinearity and the zero covariances from step 1.1 give $Var (X_{0}) = Var (A + B) = 1 + 1 + 2 (0) = 2$ , $Var (U_{0}) = Var (C + D) = 1 + 1 + 2 (0) = 2$ , $Var (X_{1}) = Var (A + B + C) = 1 + 1 + 1 + 2 (0 + 0 + 0) = 3$ , and $Var (U_{1}) = Var (D) = 1$ . The variance pairs $(2, 2)$ and $(3, 1)$ are different.

3.1F5F6F7step 2.1step 2.2algebra∎

Let $Θ = {0, 1}$ , take sample space $(R, B (R))$ , and define $P_{i}$ to be the law of $W_{i}$ . Then $P : i \mapsto P_{i}$ is a parametric model [F5], and $ψ (i) = (Var (X_{i}), Var (U_{i}))$ is an estimand [F6]. Since $W_{0} = W_{1}$ , $P_{0} = P_{1}$ , but step 2.2 gives $ψ (0) \neq ψ (1)$ . This violates the defining implication for identifiability [F7], so the split is not identifiable from the observed law of $W$ alone.

Depends on

Used by

Classical and Berkson slopes, misclassification, and validation Example

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Sources

Hernán and Robins, Causal Inference: What If, Chapter 9 (standard reference, not scraped)
Keogh et al., STRATOS guidance on measurement error and misclassification, Part 1 (standard reference, not scraped)