Why LD Differences Can Help Fine-Mapping
When two variants are strongly correlated, either one may produce a similar pattern of marginal association across a locus. Association evidence can then be difficult to attribute to one candidate rather than the other. Across ancestries, correlations between variants can differ. If a candidate pair is less correlated in one ancestry, the association patterns expected under the two causal hypotheses may be more distinct there.
The important question is not simply whether LD differs between ancestries. It is whether the differences separate the particular candidate variants being compared, and whether the relevant ancestry contributes enough association information for that separation to matter. A numerical measure can make this reasoning explicit, provided its assumptions are kept in view.
Association Patterns Under Competing Candidates
Consider one ancestry with \(m\) variants and an ancestry-specific LD correlation matrix \(R\). Its entry \(r_{jk}\) is the correlation between standardized genotypes at variants \(j\) and \(k\). Let \(e_c\) denote the \(m\)-vector with a 1 in position \(c\) and 0 elsewhere. Under a single-causal-variant approximation, suppose the vector of marginal association \(z\)-statistics has distribution
where \(\kappa\) is the signal-strength scale for that ancestry. For example, in a simple standardized phenotype model, \(\kappa\) is approximately \(\sqrt{n}\beta\), where \(n\) is sample size and \(\beta\) is the standardized causal effect. The approximation uses the LD matrix as the covariance of the marginal \(z\)-statistics and assumes that the candidate causal variants are tested and harmonized consistently.
The mean vector under candidate \(c\) is proportional to column \(c\) of \(R\). Thus, under \(c\), the association at another variant \(j\) is expected to track its correlation with \(c\). Under candidate \(d\), the expected vector instead follows column \(d\). If those columns are similar, the candidates predict similar association patterns.
For the comparison below, assume the same \(\kappa\) under the two candidate hypotheses within an ancestry. This isolates the role of LD in distinguishing them. It does not assert that real effect sizes are known or identical across ancestries. In a trans-ancestry analysis, each ancestry can have its own signal-strength scale and LD matrix.
A Pairwise Information Measure
For two multivariate normal distributions with a common positive-definite covariance matrix \(R\) and means \(\mu_c\) and \(\mu_d\), their Kullback–Leibler divergence from the distribution under \(c\) to the distribution under \(d\) is
Here the divergence is the expected log density ratio under the first distribution. With the model above, the difference in means is \(\kappa R(e_c-e_d)\). Substitution gives
The last equality follows because \(e_c-e_d\) has a 1 at \(c\), a −1 at \(d\), and zeros elsewhere. Its quadratic form is \(r_{cc}+r_{dd}-2r_{cd}=2(1-r_{cd})\), since the diagonal entries of a correlation matrix are 1. This result says that, under the stated model, the information separating two candidates is larger when their correlation is lower and when the signal-strength scale is larger.
Justification of the Normal-Model Formula
The divergence formula can be obtained directly from the normal log densities. Write \(Z=\mu_c+\varepsilon\) under candidate \(c\), where \(E(\varepsilon)=0\) and \(\operatorname{Cov}(\varepsilon)=R\). Let \(A=R^{-1}\). The log density ratio is
Under \(c\), the second quadratic term has expectation \(\operatorname{tr}(AR)\). The first has expectation \(\operatorname{tr}(AR)+(\mu_c-\mu_d)^TA(\mu_c-\mu_d)\): its random centered component contributes the same trace, and its mean offset contributes the displayed quadratic form. Taking expectations therefore cancels the trace terms and leaves one half of the quadratic form. Positive definiteness is needed here to use the ordinary inverse and the usual nonsingular normal density.
Worked Example: A Lower-Correlation Ancestry Adds Separation
Suppose two hypothetical ancestries have pairwise correlations \(r_{cd}=0.95\) and \(r_{cd}=0.20\) for the same two candidate variants. Let their signal-strength weights \(\kappa_a^2\) be 10 and 4, respectively. Assume the two ancestry-specific summary-statistic vectors are independent conditional on the candidate, and each follows the model above with a positive-definite LD matrix.
For independent ancestries, the joint log-likelihood ratio is the sum of the ancestry-specific log-likelihood ratios, so their expected values add. The total pairwise information is
The first ancestry contributes 0.50 nats, while the second contributes 3.20 nats. In this hypothetical comparison, the lower correlation in the second ancestry produces a much more distinct pair of expected association patterns, even though its signal-strength weight is smaller. The value 3.70 is an expected log-likelihood separation under the model; it is not a posterior probability or a guarantee that the data will identify the causal variant.
Combining Ancestries and Reading the Weights
If ancestry-specific summary-statistic vectors are conditionally independent and each has a common covariance \(R_a\) under the competing candidates, the joint log-likelihood ratio is the sum of the ancestry-specific ratios. Consequently, the pairwise information is
where \(r_{a,cd}\) is the correlation between candidates \(c\) and \(d\) in ancestry \(a\). In the simple standardized setup, \(\kappa_a^2\) is approximately \(n_a\beta_a^2\). The weights therefore express how strongly that ancestry’s data can contribute to distinguishing the candidates; LD contrast alone is not the whole measure.
This sum is symmetric in \(c\) and \(d\) under the equal-covariance model, even though KL divergence is generally directional. The symmetry here follows because the two distributions share a covariance matrix, making the quadratic separation the same in either direction. A value of zero occurs when every weighted ancestry has \(r_{a,cd}=1\), so the model predicts no separation for the pair. Larger values mean greater expected separation in log-likelihood terms.
Worked Example: Small LD Differences Can Add Little
Consider a hypothetical pair of candidates with correlations 0.94 and 0.90 in two ancestries. Let each have signal-strength weight \(\kappa_a^2=8\). Both correlations are high, though the second ancestry has a different value. Assume positive-definite LD matrices and the same single-causal-variant comparison model in each ancestry.
The total information is
If the second ancestry had instead had correlation 0.94, the total would be \(8(0.06)+8(0.06)=0.96\) nats. Changing that correlation from 0.94 to 0.90 adds 0.32 nats in this example. It does add information, but the difference between the ancestry-specific correlations is not, by itself, a measure of strong fine-mapping resolution: both candidates remain highly correlated in both ancestries.
Using the Full LD Matrices for Several Candidates
For a pair of candidates, the relevant matrix entry is their correlation. With several candidates, it is useful to inspect the pattern of correlations rather than focus only on whether matrices differ overall. The same pairwise formula can be applied to each candidate pair, using its correlation in each ancestry. This produces a map of which pairs are predicted to be hard or relatively easy to distinguish under the model.
Worked Example: Comparing Three Candidate Pairs
Suppose two hypothetical ancestries have the following LD matrices for three standardized variants. Assume each matrix is positive definite, the ancestry-specific signal-strength weights are both 5, and the same single-causal effect scale applies to all three candidate hypotheses within each ancestry.
For example, the determinant of \(R_1\) is \(1+2(0.90)(0.20)(0.25)-0.90^2-0.20^2-0.25^2=0.1775\), and the determinant of \(R_2\) is \(1+2(0.25)(0.75)(0.30)-0.25^2-0.75^2-0.30^2=0.3975\). Their leading principal minors are positive, so these symmetric matrices are positive definite and the inverse-based result applies.
For candidates 1 and 2, the correlations are 0.90 and 0.25. Thus
For candidates 1 and 3, the correlations are 0.20 and 0.75, giving
For candidates 2 and 3, the correlations are 0.25 and 0.30, giving
Under these assumptions, candidates 2 and 3 have the greatest expected pairwise separation, and candidates 1 and 2 the least. The matrices show why it is not enough to say that the ancestries have different LD: the information comparison depends on the correlations for the specific candidate pair, along with the ancestry-specific weights.
Singular LD Matrices and Scope of the Measure
An LD matrix need not be positive definite. Perfectly correlated variants, for instance, can make it singular. In that case the ordinary inverse \(R^{-1}\) and a full-dimensional normal density do not exist, so the inverse-based formula must not be applied as written. For the model here, however, the difference in means is \(\kappa R(e_c-e_d)\), which lies in the range of \(R\). The two normal distributions therefore have the same affine support. Their divergence on that support can be written using the Moore–Penrose pseudoinverse \(R^+\):
The second equality follows from \(RR^+R=R\). In this particular mean model, the resulting pairwise expression is still \(\kappa^2(1-r_{cd})\). Perfectly correlated candidates have \(r_{cd}=1\) and zero separation. In practice, singularity or near-singularity also calls for care in estimating LD and interpreting fine-mapping results; the formula is a model-based comparison, not a remedy for indistinguishable variants.
The derivation is most direct for a single causal variant and a common covariance across the candidate hypotheses. With multiple causal variants, ancestry-specific effects, model misspecification, or estimated rather than known LD, a complete likelihood may have additional structure. The pairwise calculation is then a useful way to understand one source of resolution, not a substitute for the specified fine-mapping model.
Common Mistakes and Practical Interpretation
- Equating different LD matrices with high information. The wrong reasoning is that any ancestry-to-ancestry change in LD must improve resolution. The correction is to examine the correlation between each candidate pair and weight its contrast by the strength of the relevant association data.
- Ignoring signal strength. A low candidate-pair correlation in an ancestry may produce little practical separation if its \(\kappa_a^2\) is small. Conversely, an ancestry with a more modest LD contrast can still contribute meaningfully when its signal is stronger.
- Calling the divergence a probability. A value in nats is an expected log-likelihood separation under a specified model. It is not the probability that a candidate is causal, nor a posterior inclusion probability.
- Using \(R^{-1}\) for a singular matrix. The formula with an ordinary inverse requires positive definiteness. For a singular matrix, use the support-aware pseudoinverse result when its conditions hold, or use an appropriate model that handles the singularity.
- Interpreting fine-mapping as causal proof. Even substantial separation between two modeled candidates does not establish that either candidate is causal. The result is conditional on the candidate set, model, harmonization, LD inputs, and signal assumptions.
Check Your Understanding
Use the stated model and show the steps in your reasoning.
- For one ancestry, \(r_{cd}=0.60\) and \(\kappa^2=3\). What is the pairwise information in nats?
- Two ancestries have \(r_{cd}=0.80\) and \(0.30\), with \(\kappa_a^2=2\) in each. Calculate the total information and identify which ancestry contributes more.
- Why does a difference between ancestry-specific LD matrices not, by itself, guarantee useful resolution for a particular pair of candidates?
- State the matrix condition needed for the derivation that uses \(R^{-1}\). What change is required if \(R\) is singular?
- Explain why an information value in nats is not a posterior probability that either candidate is causal.