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Fine-Mapping Across Ancestries · Tutorial 646 of 989

Sample-Size Imbalance and Trans-Ancestry Power

Calculate ancestry-specific information and power, and distinguish strong association evidence from the additional information that LD differences provide for fine-mapping.

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What You'll Learn

  • Calculate genotype variance and association information from sample size and allele frequency
  • Combine independent ancestry-specific information under a shared-effect model
  • Estimate two-sided power from a noncentral normal statistic
  • Quantify how sample-size imbalance affects joint power and information shares
  • Explain how LD differences can help distinguish candidate causal variants
  • Identify when a large cohort may dominate evidence without resolving a locus

Why Sample-Size Balance Matters

Trans-ancestry fine-mapping combines evidence from cohorts whose sample sizes, allele frequencies, and linkage disequilibrium (LD) patterns may differ. A cohort with many more participants can contribute most of the association precision, even when smaller cohorts have useful LD differences. The distinction matters: strong evidence that a locus is associated is not automatically strong evidence that one variant has been distinguished from another.

A useful starting point is the information for an additive genetic effect. For a biallelic variant with allele frequency \(f\), assume genotype is coded as allele dosage 0, 1, or 2 and Hardy–Weinberg proportions give dosage variance \(V=2f(1-f)\). In a simple linear model with residual standard deviation \(\sigma\), sample size \(n\), and per-allele effect \(\beta\), the approximate standard error is \(\sigma/\sqrt{nV}\). Thus the noncentrality parameter of the association statistic is \(\beta\sqrt{nV}/\sigma\). This approximation assumes unrelated observations, a correctly specified additive model, and no unaccounted confounding or major loss of information from covariate adjustment.

Definition: In this tutorial, the approximate association information for an ancestry group \(k\) is \(I_k=n_k\,2f_k(1-f_k)/\sigma_k^2\). Under a shared per-allele effect \(\beta\), the squared noncentrality parameter is \(\lambda_k=\beta^2 I_k\). When residual variance is standardized to one, this reduces to \(I_k=n_k\,2f_k(1-f_k)\) and \(\lambda_k=\beta^2 n_k\,2f_k(1-f_k)\).

This expression shows why participant counts alone are not enough. A variant that is uncommon in a cohort has lower genotype variance and typically less information per participant. Under independent cohorts and the same effect on the modeled scale, information adds: \(I_{\text{joint}}=\sum_k I_k\), and the joint squared noncentrality is \(\lambda_{\text{joint}}=\sum_k\lambda_k\). These relations describe an idealized shared-effect analysis; they are not a guarantee that every fitted trans-ancestry model has that assumption.

Worked Example: Power Under a Large Sample-Size Imbalance

Worked Example: Power Under a Large Sample-Size Imbalance

Suppose two independent hypothetical cohorts test the same causal variant under a shared additive effect of \(\beta=0.03\) outcome standard deviations per allele. Residual variance is one, the allele frequency is \(f=0.05\) in both cohorts, and the sample sizes are \(n_A=39{,}800\) and \(n_B=1{,}000\). For a two-sided test at \(\alpha=0.05\), use the normal approximation with critical value \(z_{0.975}=1.96\). The assumptions are the standardized linear model above, independent cohorts, and a common effect.

The dosage variance in either cohort is \(V=2(0.05)(0.95)=0.095\). The noncentrality parameters are

$$ \delta_A=0.03\sqrt{39{,}800(0.095)}=\sqrt{3.4029}=1.8447, \qquad \delta_B=0.03\sqrt{1{,}000(0.095)}=\sqrt{0.0855}=0.2924. $$

For a two-sided test, approximate power at noncentrality \(\delta\) is \(P(Z>1.96)+P(Z<-1.96)\) for \(Z\sim N(\delta,1)\). Equivalently, it is \(1-\Phi(1.96-\delta)+\Phi(-1.96-\delta)\), where \(\Phi\) is the standard normal cumulative distribution function. Substitution gives

$$ \begin{aligned} \text{Power}_A &=1-\Phi(1.96-1.8447)+\Phi(-1.96-1.8447) \approx 0.4542,\\ \text{Power}_B &=1-\Phi(1.96-0.2924)+\Phi(-1.96-0.2924) \approx 0.05985. \end{aligned} $$

The approximate powers are 45.4% for cohort A and 5.985% for cohort B. Because the cohorts are independent and use the same effect and frequency, their information adds. The joint squared noncentrality is \(3.4029+0.0855=3.4884\), so

$$ \delta_{\text{joint}}=\sqrt{3.4884}=1.8677, \qquad \text{Power}_{\text{joint}} =1-\Phi(1.96-1.8677)+\Phi(-1.96-1.8677) \approx 0.4633. $$

Joint power is about 46.33%, only about 0.91 percentage points above cohort A alone. Cohort A supplies \(3.4029/3.4884\approx97.55\%\) of the total association information, while cohort B supplies about 2.45%. The small cohort adds information, but its contribution to power is limited by its sample size in this example.

Information Is Not the Same as Participant Count

For a shared-effect association estimate with independent cohorts, the combined information is the sum of ancestry-specific information, not the simple sum of sample sizes unless genotype variance and residual variance are equal. The fraction of information supplied by cohort \(k\) is \(I_k/\sum_l I_l\). This fraction is often a clearer measure of dominance than the number of participants alone.

Worked Example: A Smaller Cohort With a More Common Allele

Consider hypothetical cohorts A and B with \(n_A=40{,}000\), \(f_A=0.05\), and \(n_B=1{,}000\), \(f_B=0.20\). Assume equal residual variance, so it cancels in comparing information. The dosage variances are \(V_A=2(0.05)(0.95)=0.095\) and \(V_B=2(0.20)(0.80)=0.32\). Their information quantities, omitting the common residual-variance divisor, are

$$ I_A=40{,}000(0.095)=3{,}800, \qquad I_B=1{,}000(0.32)=320. $$

The ratio is \(I_A/I_B=3{,}800/320=11.875\). Cohort A has 40 times as many participants, but its uncommon allele provides less information per participant than the allele in cohort B. That higher frequency narrows the gap: B contributes \(320/(3{,}800+320)\approx7.77\%\) of combined information, rather than the 2.44% it would contribute if the two cohorts had equal allele frequencies and the same residual variance. Still, A supplies about 92.23% of the information. Allele frequency changes the balance but does not erase the sample-size imbalance here.

This calculation concerns precision for a specified effect at a specified variant. If genotyping or imputation quality, covariate adjustment, or phenotype variance differs, the simple \(n\,2f(1-f)\) comparison may not describe effective information adequately. The model and data quality determine whether those factors need to be incorporated.

How LD Changes Fine-Mapping Information

Fine-mapping compares candidate causal variants, not merely the presence or absence of an association. The relevant question is how differently the candidates predict the observed association pattern. Ancestry-specific LD can make two candidates produce very similar patterns in one cohort and more distinct patterns in another. That second cohort may therefore contribute substantial discrimination even if it contributes relatively little total association information.

A simple normal-summary-statistic approximation makes the distinction explicit. Suppose, within cohort \(k\), the vector of standardized association statistics has covariance matrix \(R_k\), the ancestry-specific LD correlation matrix. Under a candidate causal variant \(j\) and a specified effect, its expected vector can be written in a simplified shared-effect model as \(\boldsymbol{\mu}_{kj}=\delta_k R_k[:,j]\), where \(R_k[:,j]\) is column \(j\) of the LD matrix and \(\delta_k\) is the expected standardized signal at the causal variant. This representation assumes correctly aligned alleles, a single causal signal for this comparison, and approximately normal summary statistics with unit marginal variance. It is an explanatory approximation, not a replacement for the full fine-mapping likelihood.

For two candidate variants \(j\) and \(l\), let \(\Delta_k=\boldsymbol{\mu}_{kj}-\boldsymbol{\mu}_{kl}\). Under normal distributions with equal covariance \(R_k\), the Kullback–Leibler divergence between the two mean models in cohort \(k\) is \(\tfrac12\Delta_k^T R_k^{-1}\Delta_k\). With independent cohorts, these divergences add across cohorts; if \(R_k\) is the identity matrix, this reduces to one half of the squared Euclidean distance between the expected summary-statistic patterns. Thus, a cohort can aid fine-mapping when its LD makes the candidate patterns different, even if its contribution to the overall association signal is modest.

Worked Example: A Smaller Cohort Can Separate LD Patterns

Worked Example: A Smaller Cohort Can Separate LD Patterns

Imagine two hypothetical candidate causal variants, \(j\) and \(l\), and one measured tag variant. Compare the expected tag association under the two candidates using the simplified model above. In cohort A, let \(n_A=20{,}000\), \(f_A=0.10\), and the correlation between the tag and either candidate be \(r_A=0.95\). In cohort B, let \(n_B=5{,}000\), \(f_B=0.20\), and let the correlations be \(r_{Bj}=0.20\) and \(r_{Bl}=0.80\). Assume a shared standardized per-allele effect \(\beta=0.03\), residual variance one, independent cohorts, and that the two candidates have equal expected direct signal strength for this illustration.

The expected direct signal squared is \(\delta_k^2=n_k\,2f_k(1-f_k)\beta^2\). For A, it is \(20{,}000(0.18)(0.0009)=3.24\); for B, it is \(5{,}000(0.32)(0.0009)=1.44\). Cohort A therefore has 2.25 times the association information for the specified effect.

In A, the expected tag signal differs between the candidates by \(\delta_A(0.95-0.95)=0\). The tag cannot distinguish them in this illustration. In B, the expected difference is \(\delta_B(0.20-0.80)=-0.60\delta_B\). Its squared magnitude is \(0.36(1.44)=0.5184\), and its contribution to the two-model normal divergence is \(0.5184/2=0.2592\). A contributes no such difference through this tag, despite its stronger association signal. The example isolates one summary statistic to make the comparison transparent; real fine-mapping uses the full pattern of statistics and LD across variants. It shows why “the largest cohort dominates association information” does not always mean “the largest cohort alone determines which candidate is favored.”

Deriving the Additive Information Rule

The information-addition rule follows directly from independent normal observations in the simple regression model. Suppose the genotype dosage \(G_i\) has mean zero after centering, variance \(V\), and the outcome satisfies \(Y_i=\beta G_i+\varepsilon_i\), with independent errors of variance \(\sigma^2\). The log likelihood's quadratic term for \(\beta\) has curvature \(\sum_i G_i^2/\sigma^2\). For a large sample, \(\sum_iG_i^2\) is approximately \(nV\), giving information \(I\approx nV/\sigma^2\). For independent ancestry groups, the log likelihoods add, so their curvatures add:

$$ I_{\text{joint}} \approx \sum_k \frac{n_k V_k}{\sigma_k^2} =\sum_k \frac{n_k\,2f_k(1-f_k)}{\sigma_k^2}. $$

When all groups estimate the same \(\beta\), the approximate variance of the joint estimate is \(1/I_{\text{joint}}\). This derivation justifies why larger sample size, greater genotype variance, and lower residual variance increase precision. It also states the limits: if effects differ by ancestry, or if observations are not independent, this shared-effect calculation does not by itself represent the appropriate model.

Common Mistakes and Practical Interpretation

  • Equating sample size with information. The reasoning “the cohort with twice as many people always has twice the information” ignores allele frequency and residual variance. Compare \(n_k2f_k(1-f_k)/\sigma_k^2\), subject to the model assumptions.
  • Assuming joint power is the sum of cohort powers. Power is a nonlinear function of the noncentrality parameter. Under the shared-effect approximation, add information or squared noncentralities first, then compute joint power from the resulting statistic.
  • Treating association strength as fine-mapping resolution. A large signal can be consistent with several highly correlated candidates. Resolution depends on how candidate variants predict different association patterns, including through ancestry-specific LD.
  • Assuming a small cohort must be irrelevant. A small cohort may contribute little to total association precision but still help distinguish candidates if its LD pattern differs. Its value depends on the contrast between candidate predictions, not just on its sample count.
  • Interpreting power as a posterior probability. Power is the probability, under a specified effect and model, of rejecting the null at the chosen threshold. It is not the probability that a candidate variant is causal or that a particular result is true.

In practice, report both the ancestry-specific association information or precision and the way LD changes candidate predictions. A large cohort may dominate the joint likelihood's overall signal, while smaller cohorts contribute a useful but limited refinement. Conversely, if a smaller cohort has little effective information or its LD does not separate candidates, its presence may barely change fine-mapping. These are questions about the structure of the data and the assumed model, not a fixed advantage of any ancestry group.

Key takeaway: Sample size, allele frequency, and residual variance determine much of the association precision through \(n\,2f(1-f)/\sigma^2\). Fine-mapping adds another consideration: how differently each ancestry's LD makes competing causal variants predict the observed statistics.

Check Your Understanding

Use the stated assumptions in each question and distinguish association precision from variant discrimination.

  1. With equal residual variance and sample size, which has greater approximate additive-effect information: a variant with \(f=0.05\) or one with \(f=0.20\)? Show the two dosage variances.
  2. Two independent cohorts have noncentrality parameters 1.2 and 0.8 under a shared-effect model. What quantity is added to obtain the joint noncentrality, and why is joint power not the sum of the two powers?
  3. A very large cohort has strong association statistics, but two candidate variants have nearly identical LD patterns in it. What does this imply about its association evidence and its ability to distinguish the candidates?
  4. Explain how a smaller cohort could improve fine-mapping while contributing only a small fraction of total association information.
  5. Name two assumptions that support the information-addition calculation and describe why violating either could change the interpretation.
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