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Selection and Bias in Prediction · Tutorial 793 of 989

Predictive Validity Is Not Causal Evidence

Learn why accurate out-of-sample prediction by a polygenic score does not, by itself, show that the score or its associated genetic variation causes the phenotype.

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What You'll Learn

  • Distinguish predictive validity from evidence about causal genetic effects
  • Derive the population least-squares slope under finite-moment conditions
  • Calculate how ancestry can produce a pooled score–phenotype association
  • Identify predictive association induced by correlated family environments
  • Work through how selection can create a score association absent in the source population
  • State what additional reasoning is needed before interpreting a score causally

Why Prediction and Causation Are Different Questions

A polygenic score can forecast a phenotype accurately in a particular population and still fail to identify a causal genetic effect. Prediction asks whether observed information helps estimate an outcome for people drawn from a specified population. Causal inference asks what would happen to the outcome under a defined change in a causal factor, under assumptions that make that comparison interpretable. These questions can have different answers even when they involve the same score and phenotype.

This distinction matters in statistical genetics because a score may track more than the genetic effects researchers hope to summarize. It can also track ancestry, family circumstances, or patterns of study participation. Any of these features may help forecast the phenotype in an evaluation sample. High predictive performance alone does not separate those explanations or establish that changing the score-associated genetic variation would change the outcome.

Definition: Predictive validity is the ability of a prediction rule to forecast outcomes in a specified target population, assessed using a stated measure of predictive performance. Causal evidence supports a claim about how outcomes would differ under a defined causal contrast, given the assumptions and design used to identify that contrast. Predictive validity is a property of a prediction rule in a population; it is not, by itself, identification of a causal effect.

For example, a score may be useful for forecasting a quantitative trait in the population where it was evaluated, while its association with the trait is partly due to population structure. Earlier material on population structure as predictive signal contamination distinguishes genetic signal from ancestry-related prediction. Here the emphasis is on the broader inference: even a score that predicts well does not tell us which causal process generated that predictive signal.

Prediction From a Score: What the Regression Slope Says

Let \(S\) denote a polygenic score and \(Y\) a quantitative phenotype, measured in a defined population. A simple population linear prediction rule has the form \(a+bS\). Its least-squares slope describes the best linear predictive association between score and phenotype in that population. It is not automatically the causal effect of changing the score, much less the causal effect of any particular allele.

The conditions for the standard slope formula matter. Suppose \(E[Y^2]<\infty\), \(E[S^2]<\infty\), and \(0<\operatorname{Var}(S)<\infty\). These conditions ensure that means, variances, and covariances exist and that the squared prediction error has finite expectation. The slope and intercept minimizing \(E[(Y-a-bS)^2]\) are then:

$$ b=\frac{\operatorname{Cov}(S,Y)}{\operatorname{Var}(S)}, \qquad a=E[Y]-bE[S]. $$

The slope has units of phenotype per score unit. It summarizes how the best linear prediction changes with observed score. It does not determine whether the score itself has a causal effect: the covariance in the numerator can arise through multiple causal and noncausal pathways.

Why the Least-Squares Formula Holds

A brief derivation clarifies both the interpretation and the needed assumptions. Define the population prediction risk \(R(a,b)=E[(Y-a-bS)^2]\). Under the finite-second-moment conditions above, this risk is finite for finite \(a\) and \(b\). For any fixed \(b\), the risk is minimized over \(a\) when the average residual is zero:

$$ E[Y-a-bS]=0 \quad\Longrightarrow\quad a=E[Y]-bE[S]. $$

Substitute this intercept into the residual. The prediction error becomes \((Y-E[Y])-b(S-E[S])\). Expanding its expected square gives:

$$ R(b)=\operatorname{Var}(Y)-2b\operatorname{Cov}(S,Y)+b^2\operatorname{Var}(S). $$

Because \(\operatorname{Var}(S)>0\), this quadratic has positive curvature. Differentiating and setting the derivative to zero gives \(-2\operatorname{Cov}(S,Y)+2b\operatorname{Var}(S)=0\), which yields the stated slope. The finite-moment conditions are not decorative: without them, expectations, covariance, or prediction risk might not exist, and this derivation would not be justified.

Worked Example: Ancestry Produces a Pooled Prediction Signal

Consider a hypothetical evaluation population with two equally common ancestry groups. Let \(A=0\) or \(1\) indicate group, so \(P(A=1)=0.5\). Suppose \(S=2A+U\), where \(U\) has mean 0 and variance 1, and \(Y=10+4A+\varepsilon\), where \(\varepsilon\) has mean 0 and variance 4. Assume \(U\), \(\varepsilon\), and \(A\) are mutually independent. In this construction, there is no causal effect of \(S\) on \(Y\); both variables share the group indicator \(A\).

The mean score is \(E[S]=2(0.5)=1\), and the mean phenotype is \(E[Y]=10+4(0.5)=12\). Since \(\operatorname{Var}(A)=0.5(0.5)=0.25\), the score variance is \(\operatorname{Var}(S)=4(0.25)+1=2\). The phenotype variance is \(\operatorname{Var}(Y)=16(0.25)+4=8\).

The covariance comes from the shared group indicator: \(\operatorname{Cov}(S,Y)=\operatorname{Cov}(2A+U,10+4A+\varepsilon)=8\operatorname{Var}(A)=2\). Thus the pooled regression slope is \(b=2/2=1\) phenotype unit per score unit, and the intercept is \(a=12-1(1)=11\). The fitted linear prediction is \(11+S\).

The score is positively predictive in this pooled population. Its squared correlation with the phenotype is \(\operatorname{Cov}(S,Y)^2/[\operatorname{Var}(S)\operatorname{Var}(Y)]=4/(2\cdot8)=0.25\). Yet within either ancestry group, \(S\) varies through \(U\) and \(Y\) varies through \(\varepsilon\), which are independent. The pooled signal reflects group differences, not a causal score effect. The units of the slope are phenotype units per score unit.

Prediction Can Track Family and Environmental Factors

An observed genetic score may be correlated with family-level conditions even when the score has no causal effect on the phenotype under study. Inherited genetic variation and family circumstances can be statistically related through population structure, assortative mating, or other social and familial processes. If family circumstances also predict the phenotype, a score can carry useful predictive information about them.

A predictive association under these conditions is not meaningless: it may genuinely improve forecasts for a target population that shares the same relationships. The limitation is interpretive. That performance does not reveal how much is due to causal genetic effects, how much is due to correlated environments, or whether the same prediction will transfer to a population with different family and social patterns. Within-family benchmarks can provide a different comparison, but a between-person prediction result alone does not substitute for that analysis.

Worked Example: A Score Predicts Through a Shared Family Environment

In a hypothetical population, let \(F\) be a centered family-environment measure with variance 1. Let \(U\) and \(\varepsilon\) be mutually independent of \(F\), each with mean 0, with variances 1 and 4 respectively. Suppose the observed score and phenotype follow \(S=F+U\) and \(Y=5+3F+\varepsilon\). The score has no causal effect on \(Y\) in this model; both depend on \(F\).

The means are \(E[S]=0\) and \(E[Y]=5\). The score variance is \(1+1=2\), and the covariance is \(\operatorname{Cov}(S,Y)=\operatorname{Cov}(F+U,5+3F+\varepsilon)=3\operatorname{Var}(F)=3\). Therefore the least-squares slope is \(b=3/2=1.5\) phenotype units per score unit, and the intercept is \(a=5\). The fitted prediction is \(5+1.5S\).

The phenotype variance is \(9+4=13\), so the squared correlation is \(3^2/(2\cdot13)=9/26\), approximately 0.346. This is a real predictive association under the specified population distribution. But the score’s association arises because it tracks \(F\); the model contains no causal path from \(S\) to \(Y\). The numerical slope therefore cannot be read as a causal effect of increasing the score by one unit.

Selection Can Change the Association Being Predicted

Predictive performance is specific not only to a population, but also to how that population was observed. If inclusion in a study depends on variables related to both score and phenotype, conditioning on participation can create or alter an association. A score may then predict within the selected sample even when it has no predictive association in the source population. Conversely, selected-sample prediction may fail to describe the target population.

This possibility is related to cohort participation as a collider and to selective ascertainment. The key point here is not a particular correction method. It is that a predictive association measured after selection does not, by itself, identify the causal process that produced it. A study’s participation mechanism and the intended target population must be considered before interpreting its predictive results.

Worked Example: Selection Creates a Score–Phenotype Association

Consider a hypothetical source population with two independent binary variables: score indicator \(S\) and an unmeasured factor \(U\), each equally likely to be 0 or 1. Let the phenotype be \(Y=U\), so the score has no causal effect on the phenotype. Suppose a person is included in the study if \(S=1\) or \(U=1\).

In the source population, all four \((S,U)\) pairs have probability \(1/4\). Independence gives \(\operatorname{Cov}(S,Y)=\operatorname{Cov}(S,U)=0\). Among included people, the three possible pairs \((1,0)\), \((0,1)\), and \((1,1)\) are equally likely, each with conditional probability \(1/3\).

In the selected sample, \(E[S]=2/3\), \(E[Y]=E[U]=2/3\), and \(E[SY]=1/3\). Hence \(\operatorname{Cov}(S,Y)=1/3-(2/3)(2/3)=-1/9\). Because \(S\) is binary, \(E[S^2]=E[S]=2/3\), so \(\operatorname{Var}(S)=2/3-4/9=2/9\). The selected-sample regression slope is therefore \((-1/9)/(2/9)=-0.5\) phenotype units per score-indicator unit.

The selected sample has a negative predictive association, although the source-population association is zero and the score does not cause the phenotype. Selection has made \(S\) and \(U\) statistically dependent: among people with \(S=0\), inclusion requires \(U=1\), whereas those with \(S=1\) can have either value of \(U\). Since \(Y=U\), this dependence induces the observed score–phenotype association.

What Prediction Can and Cannot Establish

A properly evaluated score can establish that it helps predict a specified phenotype in a specified target population, using a stated performance measure and evaluation design. It can also motivate further work on transportability or causal mechanisms. But an association, an \(R^2\), or a reduction in prediction error does not by itself show that the score is a causal exposure, that its variants act through a particular biological mechanism, or that a score effect would persist under a change in population or environment.

A causal claim needs a defined causal contrast and a design or model whose assumptions support identifying it. For genetic questions, relevant concerns include population structure, relatedness and family context, selection, and the interpretation of the genetic variation represented by a score. Evidence that addresses one concern does not automatically resolve the others. In particular, an accurate prediction in an unrelated-person sample is not equivalent to a within-family causal comparison.

This tutorial is not defining gene–environment effect modification. The narrower point is that predictive performance and causal interpretation answer distinct questions, even when prediction varies across populations or settings. Such variation can reflect changes in predictor–outcome relationships without, by itself, specifying a causal interaction.

Common Mistake

Wrong reasoning: “The score predicts the phenotype well in an independent evaluation sample, so the score’s genetic effects must be causal.” Out-of-sample evaluation can show that a predictive relationship generalizes to the evaluation population; it does not rule out ancestry-related signal, correlated family or environmental factors, or selection.

Correction: State the prediction target and the evidence the evaluation provides. Then separately ask what causal contrast is being claimed and whether the study supports it. Treat predictive performance as evidence of forecast utility in its target setting—not as a stand-alone causal estimate.

Key takeaway: Predictive validity describes how well a score forecasts outcomes under a particular population distribution and sampling process. Causal evidence requires additional assumptions and a defined causal contrast. A score can predict well because it tracks ancestry, family environments, or selection, without identifying causal genetic effects.

Check Your Understanding

Answer in terms of the population and causal interpretation involved.

  1. What does a positive least-squares slope for phenotype regressed on score establish, and what does it not establish?
  2. In the ancestry example, why is the pooled slope nonzero even though there is no within-group score–phenotype association?
  3. In the family-environment example, identify the shared factor that creates predictive association and state why the slope is not a causal effect of the score.
  4. In the selection example, calculate in words why the included participants with \(S=0\) must have \(U=1\), while those with \(S=1\) need not.
  5. Why should predictive performance always be described with reference to a target population and sampling process?
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