Linkage Disequilibrium and Correlated Gene Statistics

Why Gene Statistics Are Not Automatically Independent

A pathway test may combine scores for genes that lie near one another, share mapped variants, or are influenced by correlated variants. The resulting gene statistics can therefore move together even under the null hypothesis of no association. If a test treats those statistics as independent, it can assign the wrong variability to the pathway score and produce a miscalibrated p-value.

Linkage disequilibrium (LD) is the non-random correlation of alleles at different variant sites in a population. In association analysis, LD can induce correlation between the corresponding standardized variant statistics. When those variant statistics contribute to more than one gene score—or to different genes whose variants are in LD—the dependence can carry through to gene-set analysis. The central task here is to quantify that dependence and account for it in the null calibration.

Definition: Let \(Z_j\) be a standardized association statistic for variant \(j\), with \(Z=(Z_1,\ldots,Z_m)^T\). Under a null model, suppose \(Z\) is approximately multivariate normal with mean zero and covariance matrix \(R\). Thus \(R_{jk}=\operatorname{Cov}(Z_j,Z_k)\), and \(R_{jj}=1\). For standardized score statistics from the same analysis, \(R_{jk}\) is often approximately the signed correlation between the corresponding genotypes, under the usual score-test assumptions. It is not generally the squared correlation \(r^2\).

This approximation assumes that the variant statistics use consistent allele coding, that the null model and its covariates have been handled appropriately, and that the normal approximation for the statistics is reasonable. The sign of a correlation matters: changing the coded allele for a variant changes the sign of its standardized statistic and its correlations, though a consistently defined test remains unchanged. The formulae below describe linear gene scores. For a nonlinear statistic, such as a maximum or a sum of squared statistics, its null dependence must be calibrated for that statistic rather than inferred from the linear formula.

From Variant LD to Gene-Score Covariance

A useful general representation of a linear gene score is a weighted sum of variant statistics. Let \(a_g\) be the vector of weights for gene \(g\), including zero weights for variants not used in that gene’s score. Define \(U_g=a_g^T Z\). These weights might encode a chosen mapping or aggregation rule; they do not, by themselves, imply that every contributing variant has a causal effect on that gene.

\operatorname{Var}(U_g)=a_g^T R a_g, \qquad \operatorname{Cov}(U_g,U_h)=a_g^T R a_h.

The cross-gene covariance depends both on the LD structure and on which variants contribute to each score. Even if two genes use different variants, their scores may covary when those variants are in LD. If a variant contributes to both genes, that shared contribution can create covariance even without LD between distinct variants.

When \(a_g^T R a_g>0\), one may standardize the score as \(G_g=U_g/\sqrt{a_g^T R a_g}\), giving it null variance one. Its covariance with another standardized gene score is \(\operatorname{Cov}(G_g,G_h)=a_g^T R a_h/\sqrt{(a_g^T R a_g)(a_h^T R a_h)}\). Standardizing each gene does not make the gene scores independent.

Key result: If a gene-set statistic is a weighted sum \(T=\sum_g b_g G_g\), then its null variance is \(\operatorname{Var}(T)=\sum_g b_g^2\operatorname{Var}(G_g)+2\sum_{g<h}b_gb_h\operatorname{Cov}(G_g,G_h)\). The independence calculation drops the cross-gene terms. It is correct only when those terms are zero (or when their sum is zero for the particular weights).

For equal weights and unit-variance standardized gene scores, the variance is the number of genes plus twice the sum of their pairwise covariances. Positive covariance increases the null variance relative to the independence calculation; negative covariance decreases it. Consequently, assuming independence is not guaranteed to be anti-conservative in every setting. Its direction of error depends on the signed covariance and on the gene-set weights.

Worked Examples

Worked Example: LD Inflates the Variance of a Two-Variant Sum

Suppose two hypothetical standardized variant statistics have null correlation \(0.80\). Under the null, their covariance matrix is

R= \begin{pmatrix} 1 & 0.80\\ 0.80 & 1 \end{pmatrix}.

Consider the sum \(S=Z_1+Z_2\). Its null variance is

\operatorname{Var}(S) = \begin{pmatrix}1&1\end{pmatrix} R \begin{pmatrix}1\\1\end{pmatrix} =1+1+2(0.80)=3.60.

If the statistics were incorrectly treated as independent, the assigned variance would be \(1+1=2\). Now suppose the hypothetical observed statistics are \(Z_1=2.5\) and \(Z_2=2.5\), so \(S=5.0\). Under the LD-aware normal null, the standardized sum is

\frac{5.0}{\sqrt{3.60}}=2.635.

Its one-sided upper-tail normal p-value is approximately \(0.0042\). Under the independence calculation, it would instead be \(5.0/\sqrt{2}=3.536\), with a one-sided p-value of approximately \(0.0002\). In this hypothetical example, ignoring positive LD makes the evidence look stronger than the covariance-aware calculation supports. Neither p-value is the probability that the null hypothesis is true.

The same calculation applies at the gene level, but the gene covariance must reflect all variant contributions. In particular, the statement “these are different genes” is not a justification for assuming their scores are independent.

Worked Example: Two Gene Scores Share Dependence Through LD

Consider three hypothetical variant statistics with null correlation matrix

R= \begin{pmatrix} 1&0.60&0.30\\ 0.60&1&0.10\\ 0.30&0.10&1 \end{pmatrix}.

This is a valid correlation matrix (its determinant is \(0.576>0\)). Define gene A’s score as \(U_A=Z_1+Z_2\), and gene B’s score as \(U_B=Z_3\). Their null variances are \(\operatorname{Var}(U_A)=1+1+2(0.60)=3.20\) and \(\operatorname{Var}(U_B)=1\). Their covariance is

\operatorname{Cov}(U_A,U_B) =\operatorname{Cov}(Z_1+Z_2,Z_3) =0.30+0.10=0.40.

Standardize the gene scores: \(G_A=U_A/\sqrt{3.20}\) and \(G_B=U_B\). Each has null variance one, but their covariance is \(0.40/\sqrt{3.20}=0.2236\). For an equal-weight set statistic \(T=G_A+G_B\), the LD-aware variance is \(1+1+2(0.2236)=2.447\), with standard deviation \(1.564\).

Suppose the hypothetical observed variant statistics are \(Z_1=1.5\), \(Z_2=1.0\), and \(Z_3=2.5\). Then \(G_A=(1.5+1.0)/\sqrt{3.20}=1.398\), \(G_B=2.5\), and \(T=3.898\). The LD-aware standardized statistic is \(3.898/\sqrt{2.447}=2.491\), giving a one-sided normal p-value of about \(0.0064\). Treating the two standardized gene scores as independent instead gives \(3.898/\sqrt{2}=2.756\), with a p-value of about \(0.0029\). The difference arises even though the gene scores are defined separately: their contributing variant statistics are correlated.

Calibrating a Gene-Set Statistic

For a linear set statistic, one direct approach is to calculate its null variance from the covariance matrix of the gene scores. If the standardized gene-score vector is \(G\), the set statistic is \(T=b^T G\), and its null covariance matrix is \(\Sigma_G\), then \(\operatorname{Var}(T)=b^T\Sigma_G b\). Under a suitable multivariate normal approximation, \(T/\sqrt{b^T\Sigma_G b}\) is approximately standard normal under the null. A tail probability can then be calculated from that calibrated statistic.

A second approach is to generate or obtain null realizations that preserve the relevant dependence among variants and recalculate the full gene and set statistics in each realization. The resulting reference distribution can be used directly, which is especially useful when gene scores or set statistics are nonlinear and a simple covariance formula is insufficient. The null procedure must correspond to the analysis being tested; preserving LD alone does not correct a misspecified association model or other sources of bias.

The following hypothetical calculation shows the size of the variance correction when gene-level covariances are available. Suppose three standardized gene statistics have null pairwise covariances \(0.50\), \(0.40\), and \(0.20\), and the set statistic is their unweighted sum. The variance is

\operatorname{Var}(T) =3+2(0.50+0.40+0.20) =5.20.

Worked Example: Correcting a Three-Gene Set Test

Suppose the hypothetical observed standardized gene statistics are \(2.0\), \(1.8\), and \(1.6\), so their sum is \(T=5.4\). Using the covariance-aware variance above gives \(SE(T)=\sqrt{5.20}=2.280\), and the standardized statistic is \(5.4/2.280=2.368\). Its one-sided normal p-value is approximately \(0.0089\).

An independence calculation would assign variance \(3\), standard error \(\sqrt{3}=1.732\), and standardized statistic \(5.4/1.732=3.118\), with a one-sided p-value of approximately \(0.00091\). In this hypothetical case, independence substantially understates null variability. The p-values are conditional on the stated normal approximation and covariance values; they are not interchangeable with a null distribution generated under a different model.

Why the Covariance Terms Matter

A pathway result is interpreted against a null distribution for the chosen set statistic. If that distribution is too narrow because it omits positive dependence, a statistic that is ordinary under the true null can appear unusually large. If relevant covariances are negative, omitting them can instead make the assumed distribution too wide. Thus the independence assumption is not merely a computational shortcut: it changes the reference distribution and potentially the conclusion.

The covariance calculation also clarifies why LD is not the only possible source of dependence. A variant mapped to two genes contributes to both scores, and overlapping variant weights can induce dependence directly. Conversely, LD between variants matters only insofar as the variants contribute to the gene scores under consideration. The relevant object is the covariance of the actual statistics being combined, not a general claim that all nearby genes are correlated by the same amount.

Common Mistakes and Checks

Assuming distinct genes are independent. Different gene labels do not remove correlation between their contributing variant statistics. Calculate or calibrate cross-gene covariance.
Using \(r^2\) as the covariance of signed statistics. \(r^2\) is nonnegative and discards allele-orientation information. For signed standardized statistics, the corresponding covariance is generally related to the signed genotype correlation under the stated null approximation.
Adding only gene variances. For a sum, the variance includes twice each pairwise covariance, with the appropriate weights. Leaving out these terms is justified only when their total contribution is zero.
Assuming LD must make a test anti-conservative. Positive covariance often increases the variance of a positively weighted sum, but negative covariance or different weights can reduce it. Determine the sign and size of the terms.
Applying the linear-score formula to every gene statistic. The covariance expression \(a_g^T R a_h\) is for linear scores. Nonlinear statistics need their own null covariance or an appropriate joint null calibration.

Key takeaway: LD can propagate from variant statistics to gene statistics and then to gene-set statistics. For a linear set score, its null variance includes cross-gene covariance terms; an independence-based calibration is reliable only when those terms are negligible or cancel for the chosen weights.

Check Your Understanding

Answer each question using the covariance structure of the statistic being tested.

Two standardized variant statistics have null correlation \(0.5\). What is the null variance of their unweighted sum, and what variance would an independence assumption assign?
Why can two separately defined gene scores be correlated even if each gene score has null variance one?
For an equal-weight sum of three unit-variance gene scores, explain how the pairwise covariances enter the set statistic’s variance.
Why is squared genotype correlation not generally a substitute for the covariance of signed standardized association statistics?
When is the linear formula for the covariance between gene scores insufficient for calibrating a pathway statistic?