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Classical error attenuates the simple linear regression slope

Statement

Let $X, U, ϵ, W, Y$ be square-integrable real random variables on one probability space with $E [X] = E [U] = E [ϵ] = 0$ , $U$ independent of the random element $(X, ϵ)$ , and $ϵ$ independent of $X$ . Set $W = X + U, Y = α + β X + ϵ, α, β \in R,$ so $W$ has classical additive error. If $Var (X) + Var (U) > 0$ , the coefficient $γ$ of the best affine $L^{2}$ predictor of $Y$ from $W$ is $γ = \frac{Cov (Y, W)}{Var (W)} = β \frac{Var (X)}{Var (X) + Var (U)} .$ The attenuation factor lies in $[0, 1]$ . When both variances are positive, it is strictly between $0$ and $1$ , so for $β \neq 0$ the coefficient has the same sign as $β$ and smaller absolute value; when $β = 0$ , $γ = 0$ . The boundary cases are: if $Var (X) = 0$ , the factor is $0$ ; if $Var (U) = 0$ and $Var (X) > 0$ , it is $1$ .

Facts & Assumptions

Given: The square-integrable centered variables and independence hypotheses in the Statement.

[F1]

The classical model has $W = X + U$ , with centered $U$ independent of $X$ (Additive measurement error and categorical misclassification models).

[F2]

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

[F3]

Covariance is bilinear on finite linear combinations, including constants (Covariance is symmetric and bilinear in finite linear combinations).

[F4]

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

[F5]

For a finite independent family of integrable real random variables, the expectation of their product factors and the product is integrable (Expectation factors over a finite product of mutually independent random variables).

[F6]

The best affine $L^{2}$ predictor satisfies the covariance normal equations; with one predictor $W$ , its slope satisfies $γ Var (W) = Cov (Y, W)$ (Normal equations for best affine $L^{2}$ prediction).

[F7]

Square-integrable real variables are integrable on a probability space: apply Cauchy-Schwarz with the constant-one variable (Cauchy-Schwarz for random variables).

Proof

1.1givenF2F4F5F7

By [F7] the variables are integrable. Independence of $U$ from $(X, ϵ)$ implies independence of $U$ from each coordinate, because the coordinate sigma-algebras are contained in that of the pair; together with $ϵ$ independent of $X$ , [F4] and [F5] give $E [X U] = E [ϵ U] = E [X ϵ] = 0$ . Centering and [F2] therefore give $Cov (X, U) = Cov (ϵ, U) = Cov (X, ϵ) = 0$ .

2.1givenF1F2F3step 1.1algebra

By [F1], $W = X + U$ ; by bilinearity [F3] and the zero covariances in step 1.1, $Cov (Y, W) = Cov (α + β X + ϵ, X + U) = β Var (X) .$ The same expansion gives $Var (W) = Var (X) + 2 Cov (X, U) + Var (U) = Var (X) + Var (U) .$

3.1givenF6step 2.1algebra

The variance sum is positive by hypothesis, so [F6] determines the unique slope $γ = Cov (Y, W) / Var (W)$ . Substitution from step 2.1 gives the displayed formula.

4.1givenstep 3.1algebra∎

Nonnegative variances and a positive sum give $0 \leq Var (X) / (Var (X) + Var (U)) \leq 1$ . Equality at $0$ occurs exactly when $Var (X) = 0$ ; equality at $1$ occurs exactly when $Var (U) = 0$ . When both are positive the factor is strictly between the endpoints, so for $β \neq 0$ step 3.1 gives a coefficient of the same sign and smaller absolute value; when $β = 0$ , step 3.1 gives $γ = 0$ , so no strict reduction holds. This proves the stated attenuation and boundary cases.

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)