Extreme-Phenotype Sampling and Score Evaluation

Why extreme-phenotype sampling changes score evaluation

A study can recruit people with unusually high or low values of a quantitative phenotype to obtain a useful range of outcomes with fewer participants. For example, a hypothetical study might invite people from the upper and lower tails of a blood-pressure distribution to assess a polygenic score. The resulting sample can be informative, but it does not have the phenotype distribution of the population to which the score might later be applied.

That difference matters for evaluation. Selecting on the phenotype changes its variance, and because the score is associated with the phenotype, it also changes the score variance and their correlation. In symmetric two-tail sampling, the observed correlation can be larger in magnitude than the population correlation. It is not, by itself, a direct estimate of population predictive performance.

The calculations below concern selection based on a quantitative phenotype, not case-control sampling on a binary outcome. The core correction is derived under a joint normal model and a selection rule that depends only on the phenotype. Additional selection, such as participation that also depends on the score or on other characteristics, is outside that simple correction.

Model and selection rule

Assume \((Y,S)\) are jointly normally distributed, with \(\sigma_Y>0\), \(\sigma_S>0\), and \(-1<\rho<1\). Equivalently for the derivation, write the score as its linear conditional mean given \(Y\), plus an independent normal residual:

Selection is assumed to depend only on \(Y\), so the rule does not further select on \(\varepsilon\) once \(Y\) is known. Let \(A\) denote the selection event, and define the selected phenotype variance and its ratio to the source variance as

For symmetric selection of both tails, such as retaining individuals with standardized phenotype \(Z=(Y-\mu_Y)/\sigma_Y\) satisfying \(|Z|>a\), the selected mean remains \(\mu_Y\) and \(k>1\). A one-sided threshold or another selection rule can instead shift the mean, and \(k\) need not exceed one.

How tail selection changes correlation and variance

The phenotype variance in the selected sample is \(k\sigma_Y^2\) by definition. The score variance also changes: the part of \(S\) predictable from \(Y\) varies more when the selected phenotype varies more, while the residual variance remains unchanged under the stated assumptions. The score–phenotype covariance changes as well.

The correlation formula preserves the sign of \(\rho\). When \(k>1\), its magnitude exceeds the source correlation’s magnitude, except at the boundary cases \(\rho=0\) or \(|\rho|=1\). This is a consequence of changing the distribution on which the correlation is calculated; it does not mean that the score became more predictive in the source population.

Derivation of the selection formulas

The decomposition of \(S\) makes the effect of selection explicit. Conditional on \(A\), the residual \(\varepsilon\) remains independent of \(Y\) and has variance \(\sigma_S^2(1-\rho^2)\). Therefore, covariance with \(Y\) comes only from the linear part of \(S\):

For the score variance, the linear part has variance \(\rho^2\sigma_S^2 k\). Adding the independent residual variance gives

Dividing the covariance by the product of the selected standard deviations yields the selected correlation:

This derivation applies whether the selection event is symmetric or not, provided \(k\) is calculated using the variance within that event and the stated model assumptions hold. Symmetry is useful because it leaves the selected phenotype mean at the source mean and gives a simple threshold formula for \(k\).

Calculating the variance inflation for symmetric tails

Suppose \(Z=(Y-\mu_Y)/\sigma_Y\sim N(0,1)\), and the sample includes both tails, \(|Z|>a\), for \(a>0\). The selected fraction is \(2[1-\Phi(a)]\), where \(\Phi\) is the standard normal cumulative distribution function and \(\phi\) its density. By symmetry, the selected mean of \(Z\) is zero. Integrating the squared standardized phenotype over the two tails gives

The selected phenotype variance is therefore \(k\sigma_Y^2\). This threshold calculation requires a normal source phenotype and a selection rule that retains the tails at the stated cutoff. If selection probabilities are not known, or the phenotype distribution is not adequately represented by this model, this value of \(k\) should not be assumed.

Correcting selected-sample estimates

If \(k\) is known or estimated from a justified selection model, the correlation formula can be inverted. Let \(r_A\) be the observed selected-sample correlation. Substituting it for the model-implied selected correlation and solving for the source correlation gives

This is a model-based correction, not a guarantee that a finite-sample correlation is unbiased. It requires \(k-r_A^2(k-1)>0\); sample noise or model misspecification can produce values that violate the assumed relationship. Uncertainty in both the observed correlation and \(k\) should be reflected in inferential procedures.

Variance estimates can also be returned to the source scale. Under symmetric selection, the sample variance \(s_{Y,A}^2\) estimates the selected variance, so a plug-in estimate of the source phenotype variance is \(s_{Y,A}^2/k\). For the score, if a source correlation estimate is available, the model gives the plug-in correction \(s_{S,A}^2/[1+\widehat{\rho}^{\,2}(k-1)]\) for the source score variance. These corrections are approximate when based on sample estimates, and the score correction depends on the model-based correlation estimate.

Assumptions, interpretation, and a common pitfall

These corrections are appropriate only when the selection mechanism and target population are sufficiently well specified. In particular, the derivation assumes a joint normal score and phenotype, a linear conditional mean for \(S\) given \(Y\), and selection on \(Y\) alone. The threshold and source distribution must support the value of \(k\). If participation also depends on \(S\), or if the selected sample has additional filters, the formulas need not recover the source association. A selected-sample mean may also differ from the source mean under one-sided selection, even though the variance-based result uses the selected variance.

A common mistake is to report the correlation from a tail-enriched sample as if it were the correlation in the source population, reasoning that selecting extreme phenotypes merely makes the study more efficient. It can improve the spread of observed phenotypes, but it also changes the joint distribution used to compute the correlation. The correction above addresses that specific distributional change; it does not address other forms of bias or replace evaluation in a representative sample.