Why Tagging Can Transfer Poorly
A genome-wide association study may identify a variant because it is correlated with a nearby causal variant, rather than because the associated variant itself changes the trait. Such an associated variant is a tag: it carries information about the causal genotype through linkage disequilibrium (LD). If the correlation between tag and causal variants differs between a discovery population and a target population, the tag's predictive value can change even when the causal effect is identical in both populations.
This tutorial isolates that mechanism with one causal variant and one tag variant. It compares the population-level performance of the discovery tag weight when transferred to the target. The calculations assume standardized genotypes, so differences in genotype variance due to allele frequencies are not part of the comparison. They also assume a shared causal effect; the question here is how LD mismatch alone affects tagging.
A Single-Causal-Variant Model
Under this model, \(r_D\) and \(r_T\) can differ in both magnitude and sign. The sign depends on allele coding, so the tag and causal alleles must be oriented consistently when comparing correlations. A negative correlation is not automatically evidence of weaker tagging: its magnitude can be large, but it implies an association in the opposite direction for the same allele coding.
The tag's discovery association is obtained from its covariance with the outcome. Since the residual is uncorrelated with the tag, the population regression coefficient of \(Y_D\) on \(G_D\) is \(\operatorname{Cov}(G_D,Y_D)\), because the tag variance is one. The discovery weight is therefore
The transferred score is \(S_T=b_DG_T\): it uses the discovery weight unchanged, applied to the target tag genotype. Its performance depends on the target correlation between \(G_T\) and \(X_T\), not just on the association estimated in discovery.
Worked Example: A Smaller Positive LD Correlation in the Target
Suppose \(\beta=0.4\) outcome units per standard deviation of the causal genotype, \(\sigma^2=1\) outcome-units squared, \(r_D=0.8\), and \(r_T=0.5\). Assume the model conditions above hold. Calculate discovery and target squared-correlation performance for the transferred tag score.
The discovery weight is \(b_D=\beta r_D=(0.4)(0.8)=0.32\) outcome units per standard deviation of the tag. The outcome variance in either population is \(\operatorname{Var}(Y)=\beta^2+\sigma^2=0.16+1=1.16\) outcome-units squared.
In the target, \(\operatorname{Var}(S_T)=b_D^2=(0.32)^2=0.1024\), and \(\operatorname{Cov}(S_T,Y_T)=b_D\beta r_T=(0.32)(0.4)(0.5)=0.064\) outcome-units squared. Hence
The discovery squared correlation is \(\beta^2r_D^2/\operatorname{Var}(Y)=(0.16)(0.64)/1.16\approx0.08828\). The target value is lower; the ratio is \(0.03448/0.08828\approx0.3906\), also equal to \((0.5/0.8)^2\). These are proportions of outcome variance associated with the score in the one-predictor linear projection, not guarantees about an individual's prediction error.
Deriving the Transfer Result
For a nonconstant score, its squared correlation with the outcome is the squared covariance divided by the product of the two variances. In the target, the transferred score has variance \(b_D^2\), while its covariance with the outcome is \(b_D\beta r_T\). The outcome variance is \(\beta^2+\sigma^2\). Thus, whenever \(b_D\ne0\),
The same calculation in discovery, using \(S_D=b_DG_D\), gives \(R_D^2=\beta^2r_D^2/(\beta^2+\sigma^2)\), when the discovery score is nonconstant. Therefore the ratio is
These conditions are substantive. Since \(b_D=\beta r_D\), both are needed to ensure the discovery score has nonzero variance and \(R_D^2\) is nonzero. If \(\beta=0\), the discovery weight is zero and the score is constant; its correlation with the outcome is undefined, not zero, so the ratio cannot be formed. If \(r_D=0\), the tag has no discovery association in this model and the same problem occurs. If \(r_T=0\) while \(\beta r_D\ne0\), the target score is nonconstant but uncorrelated with the outcome, so \(R_T^2=0\) and the ratio is defined as zero.
The derivation cancels the squared discovery weight \(b_D^2\). That cancellation explains why the relative squared-correlation performance depends on the squared LD correlations, not on the numerical size of the nonzero transferred weight. The result applies to this one-tag model under the stated shared-effect and equal-variance assumptions; it is not a general formula for a multi-variant score or for settings where other model features differ.
Worked Example: LD Reversal and the Meaning of Squared Correlation
Suppose \(\beta=0.5\) outcome units per standard deviation, \(\sigma^2=0.75\) outcome-units squared, \(r_D=0.6\), and \(r_T=-0.3\). The shared-effect model gives \(\operatorname{Var}(Y)=0.25+0.75=1\) outcome-units squared.
The discovery weight is \(b_D=(0.5)(0.6)=0.3\). In the target, \(\operatorname{Cov}(G_T,Y_T)=\beta r_T=(0.5)(-0.3)=-0.15\), so \(\operatorname{Cov}(S_T,Y_T)=(0.3)(-0.15)=-0.045\), while \(\operatorname{Var}(S_T)=0.3^2=0.09\). The squared correlation is
In discovery, \(R_D^2=\beta^2r_D^2/\operatorname{Var}(Y)=(0.25)(0.36)/1=0.09\), so the ratio is \(0.0225/0.09=0.25=(-0.3/0.6)^2\). The target score's correlation has reversed sign, but its squared correlation is positive. The sign reversal matters if the discovery weight is used as a fixed prediction rule: in the target, that score points in the direction opposite to the outcome association. Squared correlation alone does not report this directional mismatch.
What “Explained Variance” Means Here
In this model, \(R^2\) means the squared correlation between the score and the outcome. Equivalently, it is the fraction of outcome variance accounted for by the best linear projection of the outcome on that score, with an intercept. It allows the projection slope to be optimal in the population being evaluated. It is not necessarily the performance of every prediction rule that applies the discovery coefficient without adjustment.
For example, with centered outcome and score, a frozen prediction rule that predicts \(S_T\) has mean squared error \(\operatorname{Var}(Y_T)-2\operatorname{Cov}(S_T,Y_T)+\operatorname{Var}(S_T)\). A negative target covariance can make that error exceed the error from predicting the mean, even though the score's squared correlation is nonnegative. Consequently, state which metric is being compared: squared correlation, an optimally recalibrated linear prediction, or error from a fixed score rule. The LD-ratio result above is specifically for squared-correlation performance.
Worked Example: No Target Tagging and a Defined Zero Ratio
Let \(\beta=0.3\) outcome units per standard deviation, \(\sigma^2=0.91\) outcome-units squared, \(r_D=0.5\), and \(r_T=0\). Then \(\operatorname{Var}(Y)=0.09+0.91=1\) outcome-units squared, and the discovery weight is \(b_D=(0.3)(0.5)=0.15\). This is nonzero, so the transferred score has variance \(0.15^2=0.0225\).
Because the target tag is uncorrelated with the causal genotype, \(\operatorname{Cov}(S_T,Y_T)=b_D\beta r_T=(0.15)(0.3)(0)=0\), giving \(R_T^2=0^2/[(0.0225)(1)]=0\). Discovery performance is \(R_D^2=(0.3^2)(0.5^2)/1=0.0225\). Thus \(R_T^2/R_D^2=0\), matching \(r_T^2/r_D^2=0/0.25=0\). Unlike the case of a zero discovery weight, the target score here is variable, and its squared correlation is well-defined.
Common Mistakes and Practical Interpretation
- Using only the LD magnitude informally. A smaller \(|r_T|\) than \(|r_D|\) implies lower squared-correlation performance under this model, but the sign also determines whether the target association direction agrees with the discovery weight.
- Claiming the ratio when the discovery performance is zero. A zero \(\beta\) or zero \(r_D\) makes \(b_D=0\); the score is constant and its correlation is undefined. The ratio formula requires both \(\beta\ne0\) and \(r_D\ne0\).
- Calling the result a universal prediction-loss formula. The ratio compares squared correlations under a single causal variant, standardized genotypes, and shared effects and residual variance. It does not automatically describe fixed-rule prediction error or multi-variant scores.
- Confusing LD mismatch with a causal effect change. Here \(\beta\) is fixed across populations by assumption. The transfer difference arises because the tag captures the causal genotype differently; effect-size heterogeneity is a distinct source of portability change.
The practical lesson is that a GWAS tag is predictive only to the extent that it tracks the causal variant in the population where the score is used. Under the stated single-tag model, squared-correlation transfer is proportional to \(r_T^2\); comparing it with discovery performance requires a nonzero discovery association. For real scores, this calculation is a useful mechanism-level result, not a substitute for evaluating the score in the intended target population.
Check Your Understanding
Use the model assumptions stated above when answering.
- If \(\beta=0.4\), \(r_D=0.5\), and \(r_T=0.25\), what is the target-to-discovery ratio of squared-correlation performance, assuming the ratio is defined?
- Why does \(r_D=0\) prevent use of the ratio, even if \(r_T\ne0\)?
- Can a tag have positive squared-correlation performance in the target while its association direction is opposite to discovery? Explain using the LD correlation signs.
- What does \(R^2\) measure in this tutorial, and why can it differ in implication from the error of a frozen prediction rule?
- Which assumptions let the derivation attribute the performance change to LD mismatch rather than to other sources?