Worked Analysis of Two Correlated Signals

Why Analyze Two Correlated Variants Together?

At a locus, two variants can each show strong marginal association with a phenotype because their genotypes are correlated. That pattern alone does not show that each carries distinct association evidence: one variant may partly tag the other. A joint analysis asks a more specific question: does either variant contribute association evidence after accounting for the other?

This tutorial works through that question using marginal tests, joint regression, a nested-model test, and linkage disequilibrium (LD) diagnostics. The goal is to evaluate evidence for two statistical signals at a locus, not to establish which variant is causal.

Definition: In a two-variant joint regression, a variant’s conditional association is its regression association with the phenotype while the other included variant is held fixed. A second signal is supported statistically when adding the second variant provides evidence of improved fit beyond the model containing the first, subject to the model and data assumptions.

Models and Assumptions

Let \(Y_i\) be a quantitative phenotype for individual \(i\), and let \(X_{i1}\) and \(X_{i2}\) be that individual’s genotype dosages for two variants. Dosages may be allele counts or imputed expected allele counts. The dosage scale must be consistent when interpreting coefficients. For clarity, the main example uses centered, standardized genotypes, so each has mean zero and variance one.

\begin{aligned} \text{Reduced model:}\quad &Y_i=\beta_0+\beta_1X_{i1}+\varepsilon_i,\\ \text{Full model:}\quad &Y_i=\beta_0+\beta_1X_{i1}+\beta_2X_{i2}+\varepsilon_i. \end{aligned}

The joint coefficient \(\beta_2\) describes the expected phenotype difference associated with a one-unit increase in variant 2 dosage among individuals with the same variant 1 dosage, under the linear model. It is a conditional association, not a causal effect. The corresponding interpretation applies to \(\beta_1\) after adjusting for variant 2.

For the usual regression tests, assume the model is correctly specified, observations are independent, residuals have mean zero and constant variance conditional on the genotypes, and the genotypes are not perfectly collinear. Normally distributed residuals give exact small-sample \(t\) and \(F\) reference distributions; with large samples, these tests are often used as approximations. In genetic association analyses, population structure, relatedness, and other covariates may require adjustment. Omitting necessary adjustment can make the regression assumptions and resulting tests unreliable.

How Correlation Changes Marginal Slopes

A marginal regression includes one variant at a time. If both standardized genotypes are included in the phenotype-generating linear model, their marginal slopes combine the two joint effects according to their correlation. Write \(r=\operatorname{Corr}(X_1,X_2)\). With centered genotypes of variance one and an error uncorrelated with both genotypes,

\begin{aligned} \operatorname{Cov}(X_1,Y)&=\beta_1+r\beta_2,\\ \operatorname{Cov}(X_2,Y)&=\beta_2+r\beta_1. \end{aligned}

Because each genotype has variance one, these covariances are also the respective marginal regression slopes. Thus, a variant with no independent contribution can still have a nonzero marginal slope if it is correlated with a variant that does contribute.

Worked Example: Marginal and Joint Results at One Locus

Consider a hypothetical analysis of \(n=1000\) unrelated individuals. Both genotypes are centered and standardized, their correlation is \(r=0.80\), and the joint model has coefficients \(\beta_1=0.30\) and \(\beta_2=0.25\). Suppose the residual variance is \(0.96\), with residuals independent of the genotypes. For this constructed example, use \(999\) as the genotype sum-of-squares factor in the model-based standard-error calculations.

Check the phenotype variance. The genetic component has variance \(\beta_1^2+\beta_2^2+2r\beta_1\beta_2\). Adding the residual variance gives

\operatorname{Var}(Y)=0.30^2+0.25^2+2(0.80)(0.30)(0.25)+0.96 =0.09+0.0625+0.12+0.96=1.2325.

Calculate the marginal slopes. The first variant’s marginal slope is \(0.30+(0.80)(0.25)=0.50\). The second’s is \(0.25+(0.80)(0.30)=0.49\). Both can therefore appear strongly associated in separate regressions even though their joint coefficients differ.

Calculate marginal residual variances and standard errors. In a one-variant regression, the residual variance is the phenotype variance minus the squared marginal slope. Thus, for variant 1 it is \(1.2325-0.50^2=0.9825\), and for variant 2 it is \(1.2325-0.49^2=0.9924\). Using \(SE=\sqrt{\text{residual variance}/999}\),

SE_1=\sqrt{0.9825/999}=0.03136,\qquad SE_2=\sqrt{0.9924/999}=0.03152.

The corresponding marginal statistics are \(0.50/0.03136=15.94\) and \(0.49/0.03152=15.55\). These values are strong marginal evidence against a zero marginal slope; they do not, by themselves, establish two distinct signals.

Fit the joint model. With both genotypes included, the residual variance is \(0.96\). After accounting for variant 1, the variance in variant 2 remaining is \(1-r^2=1-0.80^2=0.36\). The standard error for either joint coefficient in this symmetric standardized design is

SE(\hat{\beta}_j)=\sqrt{\frac{0.96}{999(0.36)}}=0.05167.

The joint statistics are \(0.30/0.05167=5.806\) for variant 1 and \(0.25/0.05167=4.839\) for variant 2. The second coefficient remains clearly different from zero after adjustment in this hypothetical analysis. The marginal slopes were larger because each included some association aligned with the other correlated genotype.

Testing Whether the Second Variant Adds Evidence

A direct way to assess the second signal is to compare the reduced model, containing variant 1, with the full model, containing variants 1 and 2. The null hypothesis is \(H_0:\beta_2=0\) in the full model; the alternative is \(H_a:\beta_2\ne 0\). This is a nested-model test because the reduced model is obtained by setting the added coefficient to zero.

Nested-model test: If \(RSS_R\) and \(RSS_F\) are the residual sums of squares for the reduced and full models, \(q\) is the number of added parameters, and \(df_F\) is the full model’s residual degrees of freedom, then

F=\frac{(RSS_R-RSS_F)/q}{RSS_F/df_F}.

Here \(q=1\), since one variant is added, and under \(H_0\) the statistic has an \(F\) distribution with \(1\) and \(df_F\) degrees of freedom under the usual linear-model assumptions. For a single added coefficient, this partial \(F\) statistic equals the square of the coefficient’s \(t\) statistic. A small p-value is evidence that the added variant improves fit beyond the reduced model; it is not the probability that \(H_0\) is true.

The role of LD in this test follows from residualization. Regress \(X_2\) on \(X_1\), and denote the residual by \(X_{2\cdot1}\). In the population, the residual variance is \(\operatorname{Var}(X_2)(1-r^2)\) for standardized variants. Consequently, as \(|r|\) approaches one, less variation in variant 2 remains to identify its conditional coefficient. Its standard error grows, even if its underlying coefficient does not change.

\operatorname{Var}(X_{2\cdot1})=1-r^2,\qquad SE(\hat{\beta}_2)\ \text{is proportional to}\ \frac{1}{\sqrt{1-r^2}}.

This identity explains why high LD can make a genuine conditional effect difficult to detect. It also motivates checking the LD and precision diagnostics before interpreting a weak conditional result as evidence against a second signal.

Worked Example: A Conditional Test Under High LD

Suppose a hypothetical sample has \(n=500\), two standardized variants with \(r=0.95\), and a full-model residual variance estimate of \(1\). After fitting variant 1, the estimated coefficient for variant 2 is \(0.20\). Use \(499\) as the genotype sum-of-squares factor. The residual genotype variance is \(1-0.95^2=0.0975\), so

SE(\hat{\beta}_2)=\sqrt{\frac{1}{499(0.0975)}}= \sqrt{\frac{1}{48.6525}}=0.1434.

The conditional statistic is \(t=0.20/0.1434=1.395\). With \(df=497\) in the full model, the equivalent partial test is \(F=t^2=1.395^2=1.946\), with \(1\) and \(497\) degrees of freedom. Its two-sided p-value is approximately \(0.164\). At a conventional significance level of \(0.05\), this analysis does not provide statistically significant evidence that variant 2 improves fit after variant 1 is included.

That result is not proof that the conditional coefficient is zero. The LD diagnostic is substantial: the variance inflation factor for either standardized genotype is \(1/(1-r^2)=1/0.0975=10.26\). The high correlation leaves little genotype variation that can distinguish the variants’ conditional contributions, resulting in a large standard error. The data are therefore compatible with considerable uncertainty about the second coefficient.

Reading LD Diagnostics Alongside the Test

The squared correlation \(r^2\) summarizes how much of the variation in one standardized genotype is predictable from the other in a two-variant setting. The corresponding variance inflation factor is \(1/(1-r^2)\). These quantities describe the design’s capacity to separate conditional coefficients; they do not indicate whether either variant is causal.

A useful interpretation combines three pieces of information: the marginal results, the conditional coefficient and its uncertainty, and the LD between variants. Marginal results describe each variant on its own. The joint regression asks whether each association remains after adjustment. LD diagnostics explain how much precision is available to answer that conditional question.

Worked Example: A Conditional Signal With Moderate LD

Consider a hypothetical, prespecified two-variant analysis with \(n=800\), standardized genotype correlation \(r=0.60\), full-model residual variance \(1\), and conditional estimate \(\hat{\beta}_2=0.12\). Use \(799\) as the genotype sum-of-squares factor. The residual genotype variance is \(1-0.60^2=0.64\), giving

SE(\hat{\beta}_2)=\sqrt{\frac{1}{799(0.64)}}= \sqrt{\frac{1}{511.36}}=0.04422.

The conditional statistic is \(t=0.12/0.04422=2.714\). The partial \(F\) statistic is \(2.714^2=7.366\), with \(1\) and \(797\) degrees of freedom, giving a two-sided p-value of approximately \(0.0068\). The variance inflation factor is \(1/0.64=1.563\), much lower than in the preceding example. This analysis provides nominal evidence that variant 2 adds association beyond variant 1 under the stated model. It does not establish a causal variant, and the p-value should be interpreted in light of any broader testing or selection that led to this analysis.

A Practical Interpretation Sequence

Check the analysis inputs.
Confirm the phenotype, genotype coding, sample, covariates, and missing-data handling. Verify that the two variants have enough observed genotype variation for a joint model.

Review marginal associations.
Report each variant’s marginal estimate and uncertainty as single-variant results. Do not use two small marginal p-values alone as proof of two signals.

Fit and test the joint model.
Report conditional coefficients and uncertainty, then compare the reduced and full models to test whether adding the second variant improves fit.

Interpret the test with LD.
Use \(r\), \(r^2\), or the variance inflation factor to describe how well the data can distinguish the variants’ conditional effects. Treat weak conditional evidence cautiously when LD is high.

Common Mistakes

Counting marginal associations as separate signals. Two small marginal p-values can arise because correlated variants tag the same underlying association. Compare conditional results before concluding that the locus has two statistical signals.
Calling a nonsignificant conditional test proof of no second signal. A large standard error, especially under high LD, can make the test inconclusive. Report the estimate and uncertainty rather than converting a nonsignificant result into a zero-effect claim.
Ignoring the model being tested. The nested test asks whether the added variant improves fit conditional on the included variant and covariates. Changing the model changes the question.
Interpreting a conditional coefficient causally. Regression adjustment identifies a conditional statistical association under its assumptions. It does not, by itself, establish that either variant causes the phenotype.

Key takeaway: Evidence for two signals comes from the joint analysis, especially the test of whether the second variant adds information beyond the first. Marginal associations show where to look; conditional estimates, nested-model tests, and LD diagnostics show how strongly the data distinguish the two variants statistically.

Check Your Understanding

Try these questions before reviewing the worked calculations.

For standardized genotypes correlated at \(r=0.70\), with joint coefficients \(\beta_1=0.20\) and \(\beta_2=0.10\), what is the marginal slope for variant 1 under the stated linear model?
What null hypothesis is tested when variant 2 is added to a model already containing variant 1?
Why does the conditional standard error tend to increase as the absolute genotype correlation approaches one?
If a conditional test has a p-value above \(0.05\), what can and cannot be concluded about the second coefficient?
Why do two strong marginal associations not necessarily demonstrate two distinct statistical signals?