Why Decompose Portability?
A tag variant can show a strong association in one ancestry group and a weaker association in another for several distinct reasons. The causal effect may differ, the tag may capture the causal variant differently because linkage disequilibrium (LD) differs, the causal allele may have a different frequency, or one group may simply provide less precise evidence. These explanations have different implications: a smaller test statistic does not by itself establish that the underlying effect is smaller.
This tutorial decomposes the expected strength of a tag-variant association across ancestry strata. The decomposition is a way to organize evidence, not a causal attribution procedure: the components can be estimated with uncertainty, and their interpretation depends on the genetic and outcome model.
Model and Decomposition
Consider ancestry stratum \(k\), with a quantitative outcome \(Y\), a causal genotype dosage \(G_C\), and a tag genotype dosage \(G_T\). Assume the genotypes count copies of consistently harmonized alleles. For clarity, suppose there is one causal variant and the outcome model, after adjustment for appropriate covariates, is
where the residual has mean zero, variance \(\sigma_k^2\), and is uncorrelated with the genotypes. The model assumes no remaining confounding of the genotype-outcome association within each stratum. Let \(v_{C,k}=\operatorname{Var}(G_C)\), \(v_{T,k}=\operatorname{Var}(G_T)\), and let \(r_k\) be the correlation between the two genotype dosages. With random mating and Hardy–Weinberg equilibrium for a biallelic variant of allele frequency \(p\), its dosage variance is \(2p(1-p)\). In an analysis with covariates, the variances and correlation should correspond to the appropriately residualized genotypes.
The marginal regression coefficient at the tag is \(\beta_{T,k}=\operatorname{Cov}(G_T,Y)/\operatorname{Var}(G_T)\). Under the stated single-causal-variant model, it is related to the causal effect by \(\beta_{T,k}=\beta_{C,k}\operatorname{Cov}(G_T,G_C)/v_{T,k}\). Define the signed noncentrality parameter for the tag association as the mean of its large-sample \(Z\)-statistic when the residual variance is treated as known:
The absolute value \(|\lambda_k|\) describes expected signal magnitude; the sign records the direction of the harmonized association. In an actual study, the observed \(Z\)-statistic fluctuates around \(\lambda_k\). Thus, this decomposition concerns expected strength, not a claim that every observed test statistic equals its expectation.
Why the Formula Holds
The derivation makes clear what is assumed and what is being separated. First, the linear model gives \(\operatorname{Cov}(G_T,Y)=\beta_{C,k}\operatorname{Cov}(G_T,G_C)\), because the residual is uncorrelated with \(G_T\). Therefore the tag’s marginal regression coefficient is \(\beta_{C,k}\operatorname{Cov}(G_T,G_C)/v_{T,k}\). For a sample of \(n_k\) independent observations, the standard error of that coefficient is approximately \(\sigma_k/\sqrt{n_kv_{T,k}}\). Dividing the coefficient by this standard error gives the first expression for \(\lambda_k\). Finally, \(\operatorname{Cov}(G_T,G_C)=r_k\sqrt{v_{T,k}v_{C,k}}\); substituting cancels \(\sqrt{v_{T,k}}\), giving the second expression.
The cancellation does not mean tag-allele frequency is irrelevant. It affects the possible LD with the causal genotype and therefore \(r_k\); it also affects the covariance representation of the tag association. The factorization says that once LD is expressed as a correlation, tag variance is not a separate multiplicative factor. If only allele frequencies and marginal tag estimates are available, it may not be possible to assign a difference uniquely to tag frequency versus LD.
Taking a ratio between two strata, A and B, gives a useful accounting identity when neither expected signal is zero:
Each ratio answers a different question: did sample size change; did the causal effect change; did tagging change; did causal-genotype variance change; or did residual variability change? Their product, not any one factor alone, describes the model-based change in expected strength. If an effect or LD estimate is zero, a ratio involving it is undefined; compare the two strata on the original scale or use the product expression for each stratum instead.
Worked Example: Accounting for a Weaker Expected Signal
Suppose hypothetical estimates for two ancestry strata give the following model-based quantities. In stratum A, \(n_A=1{,}000\), \(\beta_{C,A}=0.20\) outcome units per allele, \(r_A=0.80\), \(v_{C,A}=0.42\), and \(\sigma_A=1.00\) outcome units. In stratum B, \(n_B=800\), \(\beta_{C,B}=0.16\), \(r_B=0.50\), \(v_{C,B}=0.18\), and \(\sigma_B=1.20\). Assume the same outcome scale and compatible allele coding in both strata.
For A,
For B,
The expected-strength ratio is \(0.800/3.279\approx0.244\). Its components are \(\sqrt{800/1{,}000}=0.8944\) for sample size, \(0.16/0.20=0.800\) for causal effect, \(0.50/0.80=0.625\) for LD, \(\sqrt{0.18/0.42}=0.6547\) for causal-genotype variation, and \(1.00/1.20=0.8333\) for residual variation. Multiplying them gives approximately \(0.244\), consistent with the ratio of the two noncentrality parameters. In this hypothetical comparison, all five factors reduce expected signal in B; the calculation alone does not establish why the effects or LD differ.
Frequency and LD in a Covariance Representation
A complementary expression can make frequency changes more explicit. Let \(p_T\) and \(p_C\) denote allele frequencies and \(D\) the haplotype disequilibrium, defined as the frequency of the haplotype carrying both specified alleles minus \(p_Tp_C\). Under random pairing of haplotypes, dosage covariance is \(2D\), and dosage variances are \(2p_T(1-p_T)\) and \(2p_C(1-p_C)\). Thus the marginal tag coefficient and expected strength can be written
This form displays tag frequency and haplotype association together. It is algebraically equivalent to the correlation form because \(r=D/\sqrt{p_T(1-p_T)p_C(1-p_C)}\). Neither representation creates a unique causal allocation from summary statistics alone; the data and model determine which quantities can be estimated.
Worked Example: Frequency and LD Determine the Tag Coefficient
Consider hypothetical haplotype frequencies \(p_T=0.20\), \(p_C=0.30\), and joint-allele frequency \(0.10\). Then \(D=0.10-(0.20)(0.30)=0.0400\). The tag dosage variance is \(2(0.20)(0.80)=0.32\), and the dosage covariance is \(2D=0.0800\). If the causal per-allele effect is \(\beta_C=0.40\) outcome units, then
The same result follows from the correlation expression. Here \(r=0.0400/\sqrt{(0.20)(0.80)(0.30)(0.70)}\approx0.2182\), and \(v_C=2(0.30)(0.70)=0.42\). Therefore \(0.40(0.2182)\sqrt{0.42/0.32}\approx0.100\). The tag association is smaller than the causal effect because the tag is an imperfect proxy. A different frequency or haplotype distribution could change the tag coefficient even if the causal effect remained fixed.
Separating Signal From Sampling Precision
The factor \(\sqrt{n_k}/\sigma_k\) describes precision in the simple model. A smaller sample or greater residual variability increases the standard error and reduces expected \(Z\)-scale signal, even if the underlying marginal coefficient is unchanged. More generally, precision also depends on the genotype variance, covariate adjustment, relatedness, missingness, and the effective number of independent observations. In those settings, use the standard error from the fitted model rather than assuming the simple formula is exact.
Worked Example: Same Association, Different Sample Size
Suppose a hypothetical tag has the same marginal coefficient, \(\beta_T=0.10\) outcome units per allele, genotype variance \(v_T=0.32\), and residual standard deviation \(\sigma=1.00\) in two strata. Stratum A has \(n_A=400\), while stratum B has \(n_B=100\). The approximate standard error in A is
In B it is \(SE_B=1/\sqrt{100(0.32)}=1/\sqrt{32}=0.1768\). The corresponding expected \(Z\)-scale strengths are \(0.10/0.08839\approx1.131\) and \(0.10/0.1768\approx0.5657\). The second is half the first, matching \(\sqrt{100/400}=0.5\). The weaker expected statistic in B reflects lower sampling precision, not a changed marginal coefficient. In real data, the observed estimates also vary randomly, so this calculation does not predict their exact observed \(Z\)-statistics.
Assumptions, Diagnostics, and Alternatives
Before interpreting a decomposition, check that the strata use harmonized effect alleles, comparable outcome units, compatible covariate adjustment, and estimates of LD and allele frequency from populations relevant to the association samples. Inspect standard errors, allele counts, missingness, and imputation quality in each stratum. Sparse genotypes or poorly estimated LD can make frequency and tagging contributions especially unstable. Local ancestry, residual population structure, relatedness, or different phenotype measurement can also undermine the simple within-stratum model.
A diagnostic mismatch is informative: if the observed marginal effects are inconsistent with the single-causal-variant prediction, possible explanations include multiple causal variants, inaccurate LD estimates, ancestry-dependent causal effects, or model misspecification. The decomposition should then be treated as an approximation, not forced to account for the discrepancy. Effect heterogeneity tests address whether effects differ beyond sampling error; they do not by themselves identify whether LD or causal-effect differences produced changing tag estimates.
| Approach | What it summarizes | Main caution |
|---|---|---|
| Compare ancestry-stratified \(Z\)-statistics | Observed evidence relative to its standard error | Combines underlying association and precision |
| Compare marginal effect estimates and confidence intervals | Association size on the stated effect scale, with uncertainty | Tag effects can change when LD changes, even if causal effects do not |
| Decompose model-based noncentrality | Expected signal changes attributed to specified model factors | Requires credible estimates and the assumed causal/LD model |
| Test effect heterogeneity | Evidence against a common effect under the test’s model | Does not isolate frequency, LD, and precision as explanations |
The decomposition is most useful when its inputs are estimable and the model is plausible. It is not a substitute for comparing effect scales, checking calibration, or accounting for uncertainty. A point estimate of each component can look precise while uncertainty in LD, causal-variant identity, and effect estimation remains substantial. Also, for binary outcomes the linear-model formula is not generally exact; a suitable generalized-model approximation and its scale must be specified before making the same interpretation.
Common Pitfall
Check Your Understanding
Answer each question, then compare your reasoning with the solutions.
- If \(n\), \(\beta_C\), \(r\), and \(\sigma\) are unchanged but causal-genotype variance is multiplied by four, by what factor does \(|\lambda|\) change?
- Under the single-causal model, what happens to \(\beta_T\) if \(r=0\), while \(\beta_C\) is nonzero?
- Two groups have equal marginal coefficients and genotype variances, but one group has four times the sample size. What is the ratio of their expected \(Z\)-scale strengths, assuming equal residual standard deviations?
- Why does a smaller observed \(Z\)-statistic not, by itself, establish a smaller causal effect?
Solutions. (1) It doubles, because the variance enters as its square root: \(\sqrt{4}=2\). (2) The tag coefficient is zero under this model, since \(\beta_T=\beta_C r\sqrt{v_C/v_T}\); the tag is uncorrelated with the causal genotype. (3) The group with four times the sample size has twice the expected strength, because the factor is \(\sqrt{4}=2\). (4) A \(Z\)-statistic depends on the estimate divided by its standard error. Sample size, residual variability, allele frequency, LD, and random sampling variation can change it without a change in causal effect.