Summary-Statistic Pathway Analysis with LD Reference Panels

Why LD Enters a Summary-Statistic Pathway Test

A pathway analysis based on GWAS summary statistics does not have individual genotypes available for recomputing gene scores or their joint distribution. Yet variant statistics are not generally independent: nearby variants can be correlated through linkage disequilibrium (LD), and gene-level scores can inherit that correlation. A reference panel supplies an estimate of the genotype correlations needed to combine variant statistics and represent dependence among genes.

The reference panel is not a substitute for the association results. The summary statistics describe the observed evidence; the LD panel supplies covariance information. If the panel represents a different ancestry or otherwise has different LD from the GWAS, calculations that depend on this covariance may change. The changes can affect individual gene scores and the pathway statistic, so the resulting p-values may also be miscalibrated.

Definition: Let \(z\) be a vector of signed, standardized variant association statistics for \(m\) aligned variants. Let \(R\) be their \(m \times m\) LD correlation matrix, with entries \(r_{ij}\). For a specified gene, a weighted gene score is \(U_g=a_g^\mathsf{T}z_g\), where \(a_g\) is a prespecified vector of variant weights and \(z_g\) contains the gene's variant statistics. When the null covariance of \(z_g\) is \(R_{gg}\), the standardized gene score is \(T_g=U_g/\sqrt{a_g^\mathsf{T}R_{gg}a_g}\).

This formulation assumes that the variant statistics are signed and coded to the same effect alleles as the genotypes used to estimate LD. The weights and the variants assigned to each gene must also be defined before calculating the score. A two-sided p-value alone does not retain the sign of a variant statistic and is not enough to reconstruct this weighted score. In applications using a different gene-level statistic, its corresponding covariance calculation must be used rather than treating it as if it were this weighted score.

From Variant Statistics to a Pathway Statistic

For several genes, write their standardized scores as a vector \(t\). Under the null, their covariance is not necessarily the identity matrix. Genes can share correlated variants, or their assigned variants can be in LD across gene boundaries. A covariance matrix derived from the reference panel accounts for these relationships.

For clarity, suppose each gene score is a linear combination of the same \(m\) variant statistics. Let row \(g\) of a matrix \(L\) contain the gene's weights after standardizing its score. Then \(t=Lz\), and if the null covariance of \(z\) is \(R\), the covariance of the gene scores is

C = LRL^\mathsf{T}.

For a pathway containing a specified subset of genes, use the corresponding subvector \(t_S\) and submatrix \(C_{SS}\). A covariance-aware quadratic statistic is

Q_S = t_S^\mathsf{T} C_{SS}^{+} t_S,

where \(C_{SS}^{+}\) is the Moore–Penrose pseudoinverse. If \(C_{SS}\) is full rank, the pseudoinverse is the ordinary inverse. This is a self-contained test: its null hypothesis is that the mean vector of the pathway's gene scores is zero. It does not test whether the pathway is more associated than genes outside the pathway.

Theorem: If \(t_S\) is multivariate normal with mean zero and covariance \(C_{SS}\), and that covariance is treated as known, then under the self-contained null \(Q_S\) has a chi-square distribution with \(df=\operatorname{rank}(C_{SS})\). This reference distribution is conditional on the specified covariance model; estimating LD from a finite or mismatched panel can make the nominal p-value inaccurate.

The result follows by diagonalizing the covariance matrix. Let \(C_{SS}=V\Lambda V^\mathsf{T}\), where the columns of \(V\) are orthonormal eigenvectors and the positive eigenvalues are \(\lambda_1,\ldots,\lambda_r\), with \(r=\operatorname{rank}(C_{SS})\). Under the null, the nonzero components of \(V^\mathsf{T}t_S\), divided by their standard deviations \(\sqrt{\lambda_j}\), are \(r\) independent standard normal variables \(Z_j\). Therefore,

Q_S = \sum_{j=1}^{r}\frac{(V_j^\mathsf{T}t_S)^2}{\lambda_j} = \sum_{j=1}^{r} Z_j^2 \sim \chi^2_r.

The rank, rather than automatically the number of genes, determines the degrees of freedom. For example, perfectly redundant gene scores make the covariance matrix singular and contribute fewer independent directions of evidence. Near-singularity can also make estimates unstable, so the chosen numerical treatment of small eigenvalues should be reported.

Worked Example: An LD-Adjusted Score for One Gene

Suppose a hypothetical gene contains two variants with signed summary-statistic values \(z_g=(2.4,1.6)^\mathsf{T}\). Use equal weights \(a_g=(1,1)^\mathsf{T}\). Assume the appropriate LD correlation between the variants is \(r=0.5\), so

R_{gg}= \begin{pmatrix} 1 & 0.5\\ 0.5 & 1 \end{pmatrix}.

The unstandardized score is \(U_g=2.4+1.6=4.0\). Its null variance under this LD model is

a_g^\mathsf{T}R_{gg}a_g = (1,1) \begin{pmatrix} 1 & 0.5\\ 0.5 & 1 \end{pmatrix} \begin{pmatrix} 1\\ 1 \end{pmatrix} = 3.0.

Thus \(T_g=4.0/\sqrt{3.0}=2.309\). If a standard normal approximation is used for this signed score, its two-sided nominal p-value is approximately \(2\Phi(-2.309)=0.0209\). This calculation uses the stated weights and assumes the variant statistics have the specified null covariance.

Now suppose a mismatched panel gives \(r=0.8\), while the summary statistics and weights remain unchanged. The variance estimate becomes \(2+2(0.8)=3.6\), so the calculated score is \(4.0/\sqrt{3.6}=2.108\), with a two-sided nominal p-value of about \(0.0350\). The difference is not a change in the observed association statistics; it arises from changing the LD-based variance used to standardize them.

How Reference-Panel Mismatch Changes Pathway Results

Mismatch can enter at two levels. LD among variants assigned to one gene affects the variance used to standardize that gene's score. LD between variants assigned to different genes affects the covariance between their scores, which enters the pathway statistic. These are related but distinct effects: a panel may alter the marginal gene scores, the joint covariance, or both.

The direction of a change is not universal. Greater estimated correlation can reduce the effective amount of independent evidence when scores point in the same direction, but the quadratic statistic also depends on the full pattern of scores and correlations. A changed p-value should therefore not be summarized as “more LD always weakens enrichment” or “less LD always strengthens it.”

Worked Example: Between-Gene LD Changes the Pathway Test

Consider a hypothetical pathway with two genes, each represented by one signed gene-level statistic. Suppose the observed vector is \(t_S=(2,2)^\mathsf{T}\). Since each gene has one statistic, its variance is one; the reference panel determines their correlation \(r\), giving \(C_{SS}=\begin{pmatrix}1&r\\r&1\end{pmatrix}\). For \(|r|<1\), direct inversion gives

Q_S = \frac{t_1^2-2rt_1t_2+t_2^2}{1-r^2}.

With a panel estimate \(r=0.2\), the calculation is

Q_S = \frac{2^2-2(0.2)(2)(2)+2^2}{1-0.2^2} = \frac{6.4}{0.96} = 6.667.

The covariance matrix has rank 2, so the null reference distribution is \(\chi^2_2\). Its upper-tail p-value is \(\exp(-6.667/2)=0.0357\). With a mismatched panel estimating \(r=0.8\), the same observed statistics give

Q_S = \frac{2^2-2(0.8)(2)(2)+2^2}{1-0.8^2} = \frac{1.6}{0.36} = 4.444,

and the nominal p-value is \(\exp(-4.444/2)=0.1084\), again using \(df=2\). Thus the same pair of observed gene statistics leads to a different pathway result when the assumed cross-gene correlation changes. Neither p-value is guaranteed to be calibrated if its reference-panel correlation is a poor estimate of the GWAS correlation.

Worked Example: Combining Within-Gene and Between-Gene LD

Consider a hypothetical pathway with gene A represented by two variants with \(z_A=(2,1)^\mathsf{T}\), and gene B represented by one variant with \(z_B=1.5\). Use equal weights for gene A. In one panel, the within-A correlation is \(0.5\), and each A variant has correlation \(0.2\) with the B variant. These values give

R= \begin{pmatrix} 1 & 0.5 & 0.2\\ 0.5 & 1 & 0.2\\ 0.2 & 0.2 & 1 \end{pmatrix}.

The standardized gene A score is \(t_A=(2+1)/\sqrt{2+2(0.5)}=3/\sqrt{3}=1.732\), and gene B's score is \(t_B=1.500\). The covariance between the unstandardized gene A score and gene B's score is \(0.2+0.2=0.4\). After standardization, the gene-score correlation is \(c=0.4/\sqrt{3}=0.2309\). Therefore \(t=(1.732,1.500)^\mathsf{T}\) and \(C_{SS}=\begin{pmatrix}1&c\\c&1\end{pmatrix}\).

Using the two-gene quadratic formula,

Q_S = \frac{(3/\sqrt{3})^2+(1.500)^2-2(0.2309)(3/\sqrt{3})(1.500)} {1-(0.2309)^2} \approx \frac{4.050}{0.9467} = 4.278.

The rank is 2, giving a nominal \(\chi^2_2\) p-value of \(\exp(-4.278/2)=0.1178\). If another panel instead estimates within-A correlation \(0.8\) and correlation \(0.6\) between each A variant and B, then \(t_A=3/\sqrt{3.6}=1.581\), and the standardized cross-gene correlation is \(c=1.2/\sqrt{3.6}=0.6325\). The resulting calculation is

Q_S = \frac{(3/\sqrt{3.6})^2+(1.500)^2-2(0.6325)(3/\sqrt{3.6})(1.500)} {1-(0.6325)^2} \approx \frac{1.750}{0.6000} = 2.917,

with nominal p-value \(\exp(-2.917/2)=0.2326\) on 2 degrees of freedom. Both the gene A standardization and the cross-gene covariance changed. The comparison illustrates why a pathway result cannot be attributed to one altered gene score alone.

Assumptions and Practical Interpretation

The chi-square calculation is a model-based approximation, not an automatic property of any collection of summary statistics. Its usual justification relies on signed, standardized statistics whose null distribution is approximately multivariate normal, with covariance adequately represented by \(R\). The covariance estimate should be based on variants with compatible allele coding, variant identifiers, and genome build. Differences in ancestry, sample composition, imputation, or variant availability can make the reference LD a poor representation of the GWAS LD. The analysis also treats the estimated covariance as fixed; finite panel size introduces uncertainty not reflected in the simple chi-square reference distribution.

The interpretation is correspondingly specific. A small p-value indicates that the pathway's gene-score vector is unusual under the stated zero-mean, LD-based null model. It does not establish that the pathway is causal, that every gene in it is associated, or that the pathway is more associated than genes outside it. A competitive claim requires a different estimand and test.

Common Mistake

Wrong reasoning: “The summary-statistic z-scores did not change, so the pathway result must not change when I use a different reference panel.” This ignores that both the gene-score standardization and the covariance used in the pathway statistic depend on LD. Correction: Keep the observed statistics distinct from the reference-panel inputs. Recalculate the gene variances and gene-score covariance under the chosen panel, and interpret the resulting statistic and p-value as conditional on that LD model.

Key takeaway: In this workflow, summary statistics provide the observed gene-level evidence and the LD reference panel supplies the covariance model. Because the pathway statistic uses both, ancestry- or otherwise mismatched LD can change gene scores, pathway statistics, and nominal calibration.

Check Your Understanding

Answer each question by referring to the stated model and assumptions.

Why does a signed variant statistic contain information needed for a weighted gene score that a two-sided p-value alone does not provide?
For two variants with equal weights and LD correlation \(r\), write the null variance of their summed score.
What null hypothesis does the covariance-aware quadratic pathway statistic test, and what does it not test?
If two genes have perfectly redundant scores, how does that affect the degrees of freedom of the quadratic statistic?
Name two sources of reference-panel mismatch that could make the nominal pathway p-value unreliable.