Index: The Book of Statistical Proofs ▷ Statistical Models ▷ Univariate normal data ▷ Simple linear regression ▷ Population regression under Berkson measurement error

Theorem: Let $W$ be an assigned or recorded exposure with $0<\mathrm{Var}(W)<\infty$. Let $U$ be the difference between true and assigned exposure, and let $\varepsilon$ be outcome noise. Assume $W,U,\varepsilon$ are mutually independent and

\[\label{eq:berkson-slope-assumptions} \begin{gathered} \mathrm{E}(U)=\mathrm{E}(\varepsilon)=0,\\ \mathrm{Var}(U)=\tau^2<\infty,\qquad \mathrm{Var}(\varepsilon)=\sigma^2<\infty \; . \end{gathered}\]

For fixed real coefficients $\alpha$ and $\beta$, define true exposure $X$ and outcome $Y$ by

\[\label{eq:berkson-slope-model} X=W+U,\qquad Y=\alpha+\beta X+\varepsilon \; .\]

Then

\[\label{eq:berkson-slope-conditional} \mathrm{E}(Y\mid W)=\alpha+\beta W,\qquad \mathrm{Var}(Y\mid W)=\beta^2\tau^2+\sigma^2 \; .\]

The population least-squares regression of $Y$ on an intercept and $W$ has intercept $\alpha$ and slope $\beta$.

The variables $X,W,U$ have the same exposure units. The variables $Y,\alpha,\varepsilon$ have outcome units, and $\beta$ has outcome units per exposure unit. Thus $\beta^2\tau^2$ and $\sigma^2$ both have squared outcome units. No normality assumption is required.

Proof: Substitute the true-exposure equation into the outcome equation:

\[\label{eq:berkson-slope-substitute} Y=\alpha+\beta W+\beta U+\varepsilon \; .\]

Because $U$ and $\varepsilon$ are independent of $W$ and have mean zero,

\[\label{eq:berkson-slope-mean} \mathrm{E}(Y\mid W) =\alpha+\beta W+\beta\mathrm{E}(U\mid W) +\mathrm{E}(\varepsilon\mid W) =\alpha+\beta W \; .\]

Mutual independence also makes the conditional joint distribution of $(U,\varepsilon)$ given $W$ equal to its independent marginal distribution. The random part after conditioning on $W$ is therefore $\beta U+\varepsilon$, with variance

\[\label{eq:berkson-slope-variance} \begin{split} \mathrm{Var}(Y\mid W) &=\beta^2\mathrm{Var}(U)+\mathrm{Var}(\varepsilon)\\ &\quad{}+2\beta\mathrm{Cov}(U,\varepsilon)\\ &=\beta^2\tau^2+\sigma^2 \; . \end{split}\]

Population least squares chooses an intercept and slope to minimize expected squared prediction error. Let $a$ and $b$ denote trial values of those coefficients. Conditional centering, or expansion of the square, gives

\[\label{eq:berkson-slope-least-squares} \begin{split} \mathrm{E}[(Y-a-bW)^2] &=\mathrm{E}[(\alpha-a+(\beta-b)W)^2]\\ &\quad{}+\beta^2\tau^2+\sigma^2\\ &=[\alpha-a+(\beta-b)\mathrm{E}(W)]^2\\ &\quad{}+(\beta-b)^2\mathrm{Var}(W) +\beta^2\tau^2+\sigma^2 \; . \end{split}\]

The first two terms are nonnegative. Since $\mathrm{Var}(W)>0$, they are both zero only when $b=\beta$ and $a=\alpha$. This proves the population regression claim. Equivalently, its slope is $\mathrm{Cov}(W,Y)/\mathrm{Var}(W)=\beta$.

This model specifies $X=W+U$: true exposure varies around assigned exposure. Classical measurement error specifies the reverse direction, $W=X+U$, together with different independence assumptions. It generally gives an attenuated slope. These are different joint models.

∎
Sources:
  • Emmanuel Lesaffre, Andrew B. Lawson (2012): "Measurement error"; in: Bayesian Biostatistics, Section 13.1.2, pp. 376-378.

Metadata: ID: P564 | shortcut: berkson-slope | author: kjpetrykowski | date: 2026-10-20, 20:00.