Allele-Frequency Differences and Score Variance

Why Allele Frequencies Change a Score Distribution

A polygenic score applies a set of fixed weights to the allele dosages carried by each person. Even if those weights are held constant, the score need not have the same distribution in two populations. The reason is that the dosage distribution at a locus depends on its allele frequency: a common effect allele is counted more often on average, while the variability in its dosage is greatest at an intermediate frequency.

This distinction matters when interpreting score comparisons. A difference in score variance can arise from allele-frequency differences alone; it does not, by itself, show that the weights are less accurate in one population. To understand the connection with prediction, we first derive the score’s mean and variance and then state the conditions under which variance contributes to explained outcome variation.

Definition: For \(m\) genetic variants, let \(G_j\) be the number of copies (0, 1, or 2) of the designated effect allele at variant \(j\), and let \(\beta_j\) be its fixed score weight. The uncentered polygenic score is \(S=\sum_{j=1}^{m}\beta_jG_j\). In a population where the effect-allele frequency at variant \(j\) is \(p_j\), \(E(G_j)\) and \(\operatorname{Var}(G_j)\) describe the dosage mean and variance in that population.

One Locus: The Role of Hardy–Weinberg Equilibrium

At a single biallelic locus, assume Hardy–Weinberg equilibrium (HWE): the genotype probabilities for 0, 1, and 2 copies of the effect allele are \((1-p)^2\), \(2p(1-p)\), and \(p^2\), respectively. Equivalently, the two allele copies can be treated as independent draws with effect-allele probability \(p\), so the dosage \(G\) has a binomial distribution with two trials.

G\sim\operatorname{Binomial}(2,p),\qquad E(G)=2p,\qquad \operatorname{Var}(G)=2p(1-p).

For a one-variant score \(S=\beta G\), multiplying a random variable by \(\beta\) multiplies its mean by \(\beta\) and its variance by \(\beta^2\). Thus:

E(S)=2\beta p,\qquad \operatorname{Var}(S)=2\beta^2p(1-p).

The mean changes monotonically with \(p\) when the allele coding and weight are fixed. The variance does not: \(p(1-p)\) is largest at \(p=0.5\), and becomes smaller as \(p\) approaches 0 or 1. A frequency difference can therefore change the score’s location and spread in distinct ways.

Worked Example: One Variant in Two Hypothetical Populations

Suppose a score uses one effect allele with weight \(\beta=0.4\) score units per allele. In hypothetical Population A, the effect-allele frequency is \(p_A=0.2\); in hypothetical Population B, it is \(p_B=0.5\). Assume HWE in each population.

In Population A, the dosage probabilities for 0, 1, and 2 copies are \(0.8^2=0.64\), \(2(0.2)(0.8)=0.32\), and \(0.2^2=0.04\). The score values are 0, 0.4, and 0.8. Using the dosage formulas:

E(S_A)=2(0.4)(0.2)=0.16\text{ score units},\qquad \operatorname{Var}(S_A)=2(0.4)^2(0.2)(0.8)=0.0512\text{ score units}^2.

In Population B, the dosage probabilities are \(0.25\), \(0.50\), and \(0.25\), with the same score values. The calculations are:

E(S_B)=2(0.4)(0.5)=0.40\text{ score units},\qquad \operatorname{Var}(S_B)=2(0.4)^2(0.5)(0.5)=0.0800\text{ score units}^2.

The fixed weight produces a higher mean and a higher variance in Population B. Both conclusions follow from the stated frequencies and HWE assumption; neither implies that the weight has changed.

From One Variant to a Polygenic Score

For a score made from several variants, the mean is the sum of the weighted dosage means. The variance includes both each variant’s contribution and any covariance between dosage counts at different variants. This follows from the general variance identity for a weighted sum.

E(S)=\sum_{j=1}^{m}\beta_jE(G_j),\qquad \operatorname{Var}(S)=\sum_{j=1}^{m}\beta_j^2\operatorname{Var}(G_j) +2\sum_{j<k}\beta_j\beta_k\operatorname{Cov}(G_j,G_k).

If every variant is in HWE, then \(E(G_j)=2p_j\) and \(\operatorname{Var}(G_j)=2p_j(1-p_j)\). If, in addition, the dosage counts are independent across variants, all covariance terms are zero. Under those conditions:

E(S)=2\sum_{j=1}^{m}\beta_jp_j,\qquad \operatorname{Var}(S)=2\sum_{j=1}^{m}\beta_j^2p_j(1-p_j).

Independence is a simplifying assumption, not a consequence of HWE at each variant separately. When dosage counts are correlated, the covariance terms must be included; depending on the signs of the weights and covariances, they can increase or decrease the total variance. The formula makes clear why a comparison based only on the individual allele frequencies can be incomplete if those terms are appreciable.

Derivation: Write \(S-E(S)=\sum_j\beta_j\{G_j-E(G_j)\}\). Squaring and taking expectations gives the sum of the diagonal terms \(\beta_j^2\operatorname{Var}(G_j)\) plus two copies of each cross-product expectation \(\beta_j\beta_k\operatorname{Cov}(G_j,G_k)\), for \(j<k\). This yields the variance identity above. If dosage counts are independent, each cross-product expectation is zero, leaving only the diagonal terms.

Worked Example: Same Weights, Different Score Variances

Consider two variants with weights \(\beta_1=1\) and \(\beta_2=0.5\) score units per allele. In hypothetical Population A, let \(p_1=0.5\) and \(p_2=0.4\). In hypothetical Population B, let \(p_1=0.2\) and \(p_2=0.2\). Assume HWE at both variants and independent dosage counts within each population.

For Population A, the score mean is:

E(S_A)=2(1)(0.5)+2(0.5)(0.4)=1.00+0.40=1.40\text{ score units}.

Its variance is the sum of the two weighted dosage variances:

\operatorname{Var}(S_A)=2(1)^2(0.5)(0.5)+2(0.5)^2(0.4)(0.6) =0.500+0.120=0.620\text{ score units}^2.

For Population B, the mean is:

E(S_B)=2(1)(0.2)+2(0.5)(0.2)=0.400+0.200=0.600\text{ score units}.

The variance is:

\operatorname{Var}(S_B)=2(1)^2(0.2)(0.8)+2(0.5)^2(0.2)(0.8) =0.320+0.080=0.400\text{ score units}^2.

Thus, with these same weights and assumptions, Population A has both the higher mean score (1.40 versus 0.600 score units) and the higher variance (0.620 versus 0.400 score units squared). The variance comparison is not inferred from the mean comparison: each follows from its own formula and calculation.

When Score Variance Affects Predictive Performance

Score variance is one part of predictive performance, but the relationship depends on how the score relates to the outcome. A useful illustrative model in population \(k\) is \(Y_k=a_k+S_k+\varepsilon_k\), where \(a_k\) is a population-specific intercept, the score is the true additive genetic component in outcome units, and the residual \(\varepsilon_k\) has mean zero, is uncorrelated with \(S_k\), and has variance \(\tau^2\). Assume the same \(\tau^2\) in the populations being compared. Under this model, the variance of the outcome is \(\operatorname{Var}(Y_k)=\operatorname{Var}(S_k)+\tau^2\), and the fraction of outcome variance explained by the score is:

R_k^2=\frac{\operatorname{Var}(S_k)} {\operatorname{Var}(S_k)+\tau^2}.

With the residual variance fixed, a larger score variance gives a larger \(R^2\) in this particular model. But this is a conditional statement, not a general rule that a population with greater score variance must have better predictions. The model assumes the score is correctly weighted, measured in outcome units, and uncorrelated with the residual. If the weights do not represent the outcome relationship equally well, if residual variances differ, or if the outcome’s variance changes, score variance alone does not determine predictive performance. Correlation-based performance measures also depend on the outcome distribution, not just on the score’s spread.

Worked Example: A Conditional Link to Explained Variance

Use the two-population score variances from the preceding example. For illustration, suppose each score is the true genetic component of a quantitative outcome, measured in outcome units, and \(Y_k=a_k+S_k+\varepsilon_k\). Assume the residual is uncorrelated with the score and has variance \(\tau^2=1.00\) outcome units squared in each population. These assumptions are deliberately stronger than merely having a polygenic score.

For Population A, \(\operatorname{Var}(S_A)=0.620\), so:

R_A^2=\frac{0.620}{0.620+1.00} =\frac{0.620}{1.620}=0.3827.

For Population B, \(\operatorname{Var}(S_B)=0.400\), so:

R_B^2=\frac{0.400}{0.400+1.00} =\frac{0.400}{1.400}=0.2857.

Under this stipulated model, the score explains 38.27% of outcome variance in Population A and 28.57% in Population B. This difference is caused by the score-variance difference while residual variance and the score-to-outcome relationship are held fixed. It is not evidence that allele-frequency differences alone generally cause a particular portability gap.

Interpreting Score Comparisons Carefully

An uncentered score’s mean depends on allele frequencies and the chosen allele coding. Subtracting a population-specific mean changes the location but not the variance. Dividing by a population-specific standard deviation also changes the scale: the resulting standardized scores have variance one by construction. Such standardization can help describe relative standing within each population, but standardized values no longer retain the same raw score scale across populations.

A score distribution is not fully described by its mean and variance. Two populations can have the same mean and variance but different genotype combinations and therefore different score distributions. Nor does a different score variance by itself establish a difference in rank ordering, calibration, or accuracy. Those are separate predictive questions requiring outcome data and suitable performance summaries.

Common Mistakes

Assuming more frequent alleles always mean greater score variance. Under HWE, dosage variance is \(2p(1-p)\), which is largest at \(p=0.5\), not at the largest possible \(p\).
Comparing variances while omitting covariance terms. The sum of single-variant contributions is sufficient only when dosage counts are independent (or when the covariance terms otherwise sum to zero).
Interpreting a score-mean shift as a change in predictive quality. The uncentered mean can shift because allele frequencies differ; this alone says nothing about whether the score distinguishes higher from lower outcomes well.
Claiming that greater score variance must mean greater predictive performance. That conclusion follows only under additional assumptions, such as the illustrative model with a correctly specified score and equal residual variance.

Key takeaway: With fixed weights and HWE, allele frequencies determine each variant’s expected dosage and its contribution to score variance. The score’s total variance also depends on cross-variant covariance. Differences in that variance can affect explained outcome variation under a specified model, but they are not, on their own, a verdict on predictive performance.

Check Your Understanding

Answer each question using the assumptions stated in the question.

Under HWE, what are the mean and variance of an effect-allele dosage with frequency \(p\)? At what frequency is the variance largest?
A single-variant score has weight \(\beta=0.3\) and effect-allele frequency \(p=0.4\). Write the formulas for its mean and variance before substituting values.
Why is HWE at each of two variants not, by itself, enough to set the covariance between their dosages to zero?
Two populations have identical weights but different score variances. Name one additional assumption needed before concluding that their explained outcome variances differ in the same direction.
What happens to score variance if a constant is subtracted from every person’s score? What happens if each score is then divided by its population-specific standard deviation?