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Selection and Bias in Prediction · Tutorial 777 of 989

Winner’s Curse in Polygenic Score Weights

Learn how selected effect estimates influence polygenic score weights, and why discovery-sample performance can overstate prediction in new data.

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What You'll Learn

  • Define winner’s curse in the effect estimates used as polygenic score weights
  • Derive how significance selection shifts the expected selected estimate
  • Calculate a score’s association and apparent accuracy in the discovery sample
  • Compare discovery-sample and independent-sample prediction under a simple model
  • Identify how linkage disequilibrium changes the simple orthogonal-variant calculations
  • Avoid treating a discovery-sample score association as independent validation

Why Selected GWAS Effects Matter for a Score

A polygenic score combines genetic variants using estimated effects as weights. When variants are selected because their GWAS results pass a significance threshold, the selected estimates tend to be unusually large in magnitude: some reflect genuine effects, and some reflect favorable sampling error. This winner’s curse matters beyond the individual effect estimates. It affects the score built from them and can make the score look more predictive when it is evaluated in the same data that selected and weighted its variants.

The distinction between discovery and evaluation is central. The discovery sample determines which variants enter the score and supplies their estimated weights. If the score is then associated with the phenotype in that same sample, the association reuses the data that produced those weights. Independent evaluation avoids that direct reuse, although the weights can still be noisy and selection can still affect which variants and effects the score carries forward.

Definition: For a variant with true effect \(\beta\), let \(\hat{\beta}\) be its GWAS effect estimate and \(SE\) its standard error. Winner’s curse is the tendency for \(\hat{\beta}\), conditional on passing a selection rule based on its observed association, to overstate the true effect in magnitude. A polygenic score uses such estimates as weights when it forms a weighted sum of genotypes.

Selection Changes the Distribution of the Weights

For a simple illustration, suppose the estimate has a normal sampling distribution, \(\hat{\beta}\sim N(\beta,SE^2)\), and a variant is selected if its signed test statistic exceeds a positive threshold \(c\): \(\hat{\beta}/SE>c\). Write \(\delta=\beta/SE\), and let \(\phi\) and \(\Phi\) be the standard normal density and cumulative distribution functions. Under this model,

$$ E(\hat{\beta}\mid \hat{\beta}/SE>c) = \beta+SE\frac{\phi(c-\delta)}{1-\Phi(c-\delta)}. $$

The added term is positive. It arises because the estimate must be high enough to cross the threshold; estimates with sufficiently negative sampling error cannot be selected by this positive-direction rule. With a two-sided rule, the signed conditional mean depends on the true effect and the direction of selection. In particular, across null variants the selected signs can be balanced even though the selected absolute estimates are large. Winner’s curse concerns the magnitude of estimates conditional on selection, not a claim that every selected signed estimate is biased upward.

A score for an individual with genotypes \(G_1,\ldots,G_m\) can be written as \(S=\sum_{j=1}^{m}\hat{\beta}_jG_j\). Inflated or noisy estimates affect both the relative weighting of variants and the variation of the resulting score. Thus a large score association in discovery data cannot, by itself, show that these weights will capture the same signal in new individuals.

Worked Example: A Selected Effect Estimate

Suppose a hypothetical variant has true effect \(\beta=0.08\) outcome units per allele and \(SE=0.05\) outcome units per allele. It is selected if its estimated effect is positive and its test statistic exceeds \(1.96\). The standardized true effect is \(\delta=0.08/0.05=1.60\).

Under the normal model, the conditional mean estimate among variants passing this positive-direction threshold is

$$ E(\hat{\beta}\mid \hat{\beta}/SE>1.96) = 0.08+0.05\frac{\phi(1.96-1.60)}{1-\Phi(1.96-1.60)} \approx 0.08+0.05\frac{0.3739}{0.3594} \approx 0.132. $$

The expected selected estimate is about \(0.132\) outcome units per allele, rather than the true \(0.08\). This is a conditional expectation over repeated studies under the stated model; it does not say that every selected estimate equals \(0.132\). It illustrates why using a selected estimate as a score weight can give the variant more influence than its true effect warrants.

How the Discovery Sample Can Make a Score Look Better

The following derivation uses a deliberately simple setting so the mechanism is visible. Assume the selected genotype columns are centered and scaled to have sample variance one, and are mutually orthogonal in the discovery sample. Suppose the weights are the joint ordinary least-squares estimates for those columns. Let \(Y\) be the centered outcome, with sample variance one, and let \(X\) be the matrix of selected genotype columns. Then \(X'X/n=I\), where \(I\) is the identity matrix, and the estimated weight vector is \(\hat{\boldsymbol{\beta}}=X'Y/n\). The discovery score vector is \(S=X\hat{\boldsymbol{\beta}}\).

$$ \operatorname{Cov}_{D}(S,Y) = \hat{\boldsymbol{\beta}}'X'Y/n = \hat{\boldsymbol{\beta}}'\hat{\boldsymbol{\beta}}, \qquad \operatorname{Var}_{D}(S) = \hat{\boldsymbol{\beta}}'X'X\hat{\boldsymbol{\beta}}/n = \hat{\boldsymbol{\beta}}'\hat{\boldsymbol{\beta}}. $$

Here the subscript \(D\) denotes the discovery sample. Since the outcome variance is one, the squared sample correlation between the score and outcome is

$$ R_D^2 = \frac{\operatorname{Cov}_{D}(S,Y)^2} {\operatorname{Var}_{D}(S)\operatorname{Var}_{D}(Y)} = \hat{\boldsymbol{\beta}}'\hat{\boldsymbol{\beta}}. $$

The same-sample fit is therefore directly tied to the squared size of the estimated weights. Selection favors variants with large observed estimates, and those estimates are also what construct the score. This reuse produces an optimistic discovery-sample association. The identity is exact under the stated orthogonal, joint-fit setup; it is not a universal formula for every GWAS score. In particular, marginal GWAS estimates and correlated genotypes do not generally satisfy these equalities.

Worked Example: One Weight in Discovery and Evaluation

Consider hypothetical standardized genotypes \(G\) and an outcome \(Y\), both with variance one. Suppose the true outcome model has effect \(\beta=0.10\) outcome standard deviations per genotype standard deviation, but the discovery estimate is \(\hat{\beta}=0.30\), with \(SE=0.08\). The discovery statistic is \(0.30/0.08=3.75\), so the variant passes a two-sided \(1.96\) threshold. For this calculation, assume the estimate is the one-variant least-squares estimate in discovery, so the sample covariance of \(G\) and \(Y\) is \(0.30\).

The score is \(S=0.30G\). In discovery, its variance is \(0.30^2=0.09\), and its covariance with the outcome is \(0.30(0.30)=0.09\). Thus the discovery correlation is \(0.09/\sqrt{0.09(1)}=0.30\), giving \(R_D^2=0.30^2=0.09\), or 9%.

Now consider an independent evaluation sample in which the true effect and the genotype and outcome variances are as specified, and assume the same genotype scale. The score variance remains \(0.09\), but its covariance with the outcome is \(0.30(0.10)=0.03\). Its population correlation is \(0.03/\sqrt{0.09(1)}=0.10\), and its \(R^2\) is \(0.10^2=0.01\), or 1%. These are population calculations for the hypothetical model, not a guarantee about any particular evaluation sample. The contrast shows how reusing the discovery estimate can make association look stronger than the association implied by the true effect.

What Changes in an Independent Sample

For a fixed score evaluated in a new population, let \(\boldsymbol{\beta}\) be the vector of true effects, \(\mathbf{w}\) the weights estimated in discovery, and \(\Sigma_G\) the genotype covariance matrix in the evaluation population. Assume a linear outcome model \(Y=\boldsymbol{\beta}'\mathbf{G}+\varepsilon\), with residual \(\varepsilon\) uncorrelated with the genotypes. Then

$$ \operatorname{Cov}(S,Y)=\mathbf{w}'\Sigma_G\boldsymbol{\beta}, \qquad \operatorname{Var}(S)=\mathbf{w}'\Sigma_G\mathbf{w}, \qquad R^2= \frac{(\mathbf{w}'\Sigma_G\boldsymbol{\beta})^2} {\operatorname{Var}(S)\operatorname{Var}(Y)}. $$

These expressions describe an association for the chosen weights in the evaluation population. If genotypes are uncorrelated and standardized there, \(\Sigma_G=I\), so the covariance is \(\sum_j w_j\beta_j\) and the score variance is \(\sum_j w_j^2\). Under that simplifying assumption, a selected weight can be large because of its sampling error without contributing a correspondingly large true effect to the covariance. Its squared size still contributes to score variance.

In an independent sample, the discovery noise that selected a weight is not reused to calculate the evaluation phenotype. Consequently, selection alone does not create a positive expected association with a wholly null outcome when the evaluation data and genotypes are independent of discovery and there is no genuine genetic association. For nonzero effects, however, noisy selected weights can reduce how well the score aligns with the true effects. Which way a particular realized evaluation estimate moves also depends on its sampling variability.

Worked Example: Two Selected Weights

Suppose a hypothetical quantitative outcome has variance one, and two standardized genotype variables are uncorrelated in both samples. Their true effects are \(0.12\) and \(0.04\) outcome standard deviations per genotype standard deviation. In discovery, a joint fit gives estimates \(w_1=0.28\) and \(w_2=0.10\); suppose both have \(SE=0.04\), so their absolute statistics are \(7.00\) and \(2.50\), above a two-sided threshold of \(1.96\). Treat the given estimates and variances as the hypothetical data for this illustration.

The discovery score variance is \(0.28^2+0.10^2=0.0884\). Under the orthogonal joint-fit assumptions used above, the discovery \(R^2\) is therefore \(0.0884\), or 8.84%.

In an independent population, the score-outcome covariance is

$$ (0.28)(0.12)+(0.10)(0.04)=0.0336+0.0040=0.0376. $$

The score variance remains \(0.0884\), so its independent-sample \(R^2\) is

$$ R^2=\frac{0.0376^2}{(0.0884)(1)} =\frac{0.00141376}{0.0884} \approx 0.0160. $$

Thus the model’s independent-sample \(R^2\) is about 1.60%, compared with 8.84% in discovery under the idealized joint-fit calculation. The difference reflects the fact that the estimated weights are larger than the true effects, while the independent covariance depends on the true effects. It is not a universal ratio between discovery and evaluation performance.

Linkage Disequilibrium and the Scope of the Simple Calculation

The orthogonal-variant model isolates the selection-and-reuse mechanism, but genetic variants can be correlated through linkage disequilibrium. With correlated genotypes, \(\Sigma_G\) need not be the identity matrix: a weight at one variant can contribute to the score’s covariance through its correlation with another variant. The independent-sample formulas using \(\Sigma_G\) accommodate that covariance when the stated linear model and residual assumption hold.

In discovery data, the exact score-outcome covariance is determined by the observed genotype-outcome cross-products and the chosen weights. If the weights are a joint least-squares fit to the included variants, the fitted-score identities follow from the least-squares equations, with the genotype covariance matrix retained rather than replaced by \(I\). If weights are marginal GWAS estimates, or selection and fitting use different specifications, the orthogonal joint-fit identity need not hold. The basic warning remains: performance measured in data that helped select variants or estimate weights is not an independent assessment of generalization.

Why the Distinction Matters

A discovery-sample score association combines biological signal with the consequences of selecting and weighting variants using the same outcome data. The score can still be useful, and selection does not imply that every selected variant is false. Rather, the observed discovery estimates are not a neutral description of the true effects once selection has occurred. Reporting an in-sample association as though it were expected performance in new individuals confuses fit to the discovery data with evidence of transportable prediction.

Independent evaluation should apply the already specified score weights to genotypes in data not used to select variants or estimate those weights. The distinction also applies when interpreting a score’s apparent explained variance: it is a sample-specific summary, and its interpretation depends on how the score was constructed and evaluated. For binary outcomes, the relevant performance measure and its interpretation require additional care; the linear \(R^2\) calculations here are for a quantitative outcome with the stated variance model.

Key takeaway: Significance selection can make GWAS estimates too large in magnitude, and using those estimates in the same sample can make a polygenic score’s association look especially strong. Separate discovery from evaluation, and interpret discovery performance in light of how the variants and weights were chosen.

Common Mistakes

  • “Every selected effect is biased upward.” Selection tends to inflate magnitude; signed estimates can be positive or negative, and null variants selected by a two-sided rule can have balanced signs.
  • “A strong discovery association proves the score predicts well elsewhere.” The discovery phenotype helped choose or weight the score. Independent evaluation is needed to assess performance without that direct reuse.
  • “The discovery \(R^2\) identity applies to every PRS.” The simple identity here assumes orthogonal standardized variants and joint least-squares weights. Correlated variants and marginal GWAS weights require attention to genotype covariance and the actual estimation procedure.
  • “Winner’s curse means there is no real signal.” Some selected effects may be genuine. The issue is that selected estimates can overstate their magnitudes, which can distort score weights and apparent performance.

Check Your Understanding

Use the assumptions in each question, and distinguish discovery-sample quantities from independent-sample quantities.

  1. In the positive-threshold normal model, why does conditioning on \(\hat{\beta}/SE>c\) increase the conditional mean above \(\beta\)?
  2. A score uses weights estimated in the same sample in which its association is reported. What source of optimism does this reuse create?
  3. With uncorrelated standardized genotypes, weights \(w_1,w_2\), and true effects \(\beta_1,\beta_2\), write the independent-sample score-outcome covariance.
  4. Why can a selected weight increase score variance without making a correspondingly large contribution to independent-sample covariance?
  5. Which assumptions are needed for the discovery-sample identity \(R_D^2=\hat{\boldsymbol{\beta}}'\hat{\boldsymbol{\beta}}\) derived here?
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