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Statistical Fine-Mapping and Credible Sets · Tutorial 712 of 795

Summary-Statistic Fine-Mapping With LD

Learn to connect marginal association results to joint genetic effects through the LD matrix and to set up a fine-mapping likelihood from summary statistics.

Advanced 10 min read

What You'll Learn

  • Distinguish marginal regression effects from joint effects in a multi-variant model.
  • Derive how the LD matrix maps joint effects to expected marginal associations.
  • Express the Gaussian summary-statistic likelihood using LD and marginal statistics.
  • Work through numerical examples of tagging, joint-effect recovery, and relative likelihood.
  • Recognize the assumptions and numerical limitations of summary-statistic fine-mapping.

From Marginal Associations to a Joint Model

Fine-mapping often begins with a vector of association statistics, one for each variant, and an LD matrix describing correlations among those variants. The statistics are marginal: each variant is tested separately, without adjusting for the others. The causal model, however, describes effects jointly. LD links the two: a variant can have a strong marginal association because it is correlated with a variant whose joint effect is nonzero.

The goal here is to make that link explicit and show what information the summary statistics and LD matrix provide. The examples use a quantitative phenotype and a small number of variants so the calculations can be checked directly. In practice, the same relationships apply to much larger sets of variants, provided the summary statistics and LD matrix use compatible variants and conventions.

Definition: Let \(y\) be a centered quantitative phenotype vector for \(n\) individuals and let \(X\) be the \(n \times p\) matrix of centered genotype dosages. Write the joint linear model as \(y=X\beta+\varepsilon\), with \(\varepsilon \sim N(0,\sigma^2 H)\), where \(H=I_n-\mathbf{1}\mathbf{1}^{\mathsf T}/n\). The vector \(\beta\) contains joint effects: each coefficient describes the association of one variant with the phenotype while the other variants are included in the model. Assume \(X\) is centered, so \(HX=X\). For standardized genotype columns, \(X^{\mathsf T}X/n=R\), where \(R\) is the LD correlation matrix.

The centered-error specification matters. Because \(y\) and the columns of \(X\) are centered, the model requires centered errors as well. The covariance \(\sigma^2H\) gives errors no component in the constant-vector direction. Although this covariance is singular, the quantities used below are well-defined: \(X^{\mathsf T}HX=X^{\mathsf T}X\).

For variant \(j\), the marginal regression uses \(y\) and only column \(x_j\). When \(x_j^{\mathsf T}x_j=n\), its estimated slope is \(t_j=x_j^{\mathsf T}y/n\). Collect these slopes in \(t=X^{\mathsf T}y/n\). The joint least-squares estimate, when \(R\) is invertible, is \(\hat{\beta}=R^{-1}t\). Thus \(t\) and \(\hat{\beta}\) are generally not the same: \(t\) is the vector of separate, one-variant-at-a-time effects, while \(\hat{\beta}\) adjusts for correlation among variants.

The Summary-Statistic Relationship

Under the stated model, \(t\) is a linear transformation of \(y\). Its expectation and covariance follow from the mean and covariance of \(y\):

$$ E(t)=\frac{X^{\mathsf T}X}{n}\beta=R\beta, \qquad \operatorname{Cov}(t) =\frac{X^{\mathsf T}(\sigma^2H)X}{n^2} =\frac{\sigma^2}{n}R. $$

Therefore, with fixed \(X\),

$$ t \sim N\left(R\beta,\frac{\sigma^2}{n}R\right). $$

This is the central summary-statistic relationship. Joint effects are transformed by LD into expected marginal effects. The off-diagonal entries matter: if two variants are correlated, a joint effect at one variant can induce a marginal association at the other. Under the null \(\beta=0\), the marginal estimates are also correlated, with covariance \(\sigma^2R/n\).

If \(\sigma\) is treated as known, define \(z=\sqrt{n}\,t/\sigma\). Then

$$ z \sim N\left(\frac{\sqrt{n}}{\sigma}R\beta,R\right). $$

In applications, reported \(z\)-statistics usually use an estimated standard error and may rely on approximations, such as a large-sample normal approximation. The displayed distribution is the Gaussian model with known \(\sigma\); it explains why LD appears both in the mean under nonzero effects and in the covariance of the statistics. If the phenotype has also been scaled so that \(\sigma\) is approximately one, the mean is often written approximately as \(\sqrt{n}R\beta\).

Summary-statistic likelihood: With \(R=X^{\mathsf T}X/n\), \(t=X^{\mathsf T}y/n\), and known \(\sigma^2\), the part of the Gaussian log likelihood that depends on \(\beta\) is proportional to
$$ \ell(\beta) =\frac{n}{\sigma^2}\left(\beta^{\mathsf T}t-\frac{1}{2}\beta^{\mathsf T}R\beta\right)+C, $$

where \(C\) does not depend on \(\beta\). This expression uses the same information about \(\beta\) as the individual-level Gaussian likelihood when \(X^{\mathsf T}X\), \(X^{\mathsf T}y\), and the variance information are available. Fine-mapping methods can combine this likelihood with a model or prior for which variants have nonzero effects. The LD matrix is not an optional correction applied after the analysis: it is part of the likelihood.

Why the Relationship Holds

The joint model gives \(E(y)=X\beta\). A marginal slope for variant \(j\) is the projection \(x_j^{\mathsf T}y/n\). Taking its expectation gives

$$ E(t_j) =\frac{x_j^{\mathsf T}X\beta}{n} =\sum_{k=1}^{p}\frac{x_j^{\mathsf T}x_k}{n}\beta_k =\sum_{k=1}^{p}R_{jk}\beta_k. $$

So a marginal slope is a weighted sum of joint effects, with weights given by correlations between the tested variant and all variants in the model. This is why a noncausal variant may show a marginal association when it tags a causal variant. It is also why interpreting each marginal slope as that variant’s independent effect can be misleading.

The likelihood expression follows by expanding the squared residual norm. Since \(y\) and \(X\) are centered, \(X^{\mathsf T}HX=X^{\mathsf T}X\). Ignoring terms independent of \(\beta\), the Gaussian log likelihood is

$$ -\frac{1}{2\sigma^2}(y-X\beta)^{\mathsf T}H(y-X\beta) = \frac{1}{\sigma^2}\left(\beta^{\mathsf T}X^{\mathsf T}y -\frac{1}{2}\beta^{\mathsf T}X^{\mathsf T}X\beta\right)+C. $$

Substituting \(X^{\mathsf T}y=nt\) and \(X^{\mathsf T}X=nR\) yields the stated summary-statistic likelihood. The omitted term includes the phenotype’s quadratic form, which is constant with respect to \(\beta\). Thus the reduction is justified for likelihood comparisons over \(\beta\) when the required scale and design summaries are compatible.

Worked Example: One Joint Effect Produces Two Marginal Associations

Suppose there are two standardized genotype columns with \(n=1{,}000\) and LD matrix \(R=\begin{pmatrix}1&0.8\\0.8&1\end{pmatrix}\). Let the joint effects be \(\beta=(0.20,0)^{\mathsf T}\) phenotype standard deviations per genotype standard deviation, and let \(\sigma=1\). The matrix is a valid two-variant correlation matrix because its eigenvalues are \(1.8\) and \(0.2\), both positive.

The expected marginal slopes are \(R\beta\):

$$ \begin{pmatrix}1&0.8\\0.8&1\end{pmatrix} \begin{pmatrix}0.20\\0\end{pmatrix} = \begin{pmatrix}0.20\\0.16\end{pmatrix}. $$

The second variant has no joint effect in this model, yet its expected marginal slope is \(0.16\), because its correlation with the first variant is \(0.8\). The expected \(z\)-statistics under the known-\(\sigma\) model are

$$ \frac{\sqrt{1{,}000}}{1} \begin{pmatrix}0.20\\0.16\end{pmatrix} = \begin{pmatrix}6.325\\5.060\end{pmatrix}. $$

These are expected values, not guaranteed observed statistics. The example illustrates why marginal evidence alone does not identify which correlated variant carries the joint effect.

Worked Example: Recovering Joint Effects From Marginal Slopes

Suppose two variants have \(R=\begin{pmatrix}1&0.8\\0.8&1\end{pmatrix}\), and the observed marginal slopes are \(t=(0.12,0.10)^{\mathsf T}\). Assume both slopes and effects are expressed in the same standardized units. Since the eigenvalues of \(R\) are positive, \(R\) is invertible. The joint least-squares estimate is

$$ \hat{\beta}=R^{-1}t =\frac{1}{1-0.8^2} \begin{pmatrix}1&-0.8\\-0.8&1\end{pmatrix} \begin{pmatrix}0.12\\0.10\end{pmatrix}. $$

The determinant is \(1-0.64=0.36\). The numerator vector is \((0.12-0.08,\;0.10-0.096)^{\mathsf T}=(0.04,0.004)^{\mathsf T}\), so

$$ \hat{\beta} = \begin{pmatrix}0.04/0.36\\0.004/0.36\end{pmatrix} = \begin{pmatrix}0.1111\\0.01111\end{pmatrix}. $$

The marginal slopes were \(0.12\) and \(0.10\), but after adjustment the joint estimates are about \(0.1111\) and \(0.01111\). This inversion is an algebraic illustration, not a recommendation to use unregularized joint estimates as final fine-mapping conclusions. The eigenvalues \(1.8\) and \(0.2\) show that the matrix is less stable than an identity matrix: small changes in the input can have a larger effect in the low-eigenvalue direction.

Worked Example: Evaluating a Joint-Effect Candidate

Let \(n=400\), \(\sigma^2=1\), \(R=\begin{pmatrix}1&0.5\\0.5&1\end{pmatrix}\), and observed marginal slopes \(t=(0.08,0.06)^{\mathsf T}\). Compare the candidate \(\beta=(0.08,0)^{\mathsf T}\) with the null \(\beta=(0,0)^{\mathsf T}\). The matrix is a valid correlation matrix, with eigenvalues \(1.5\) and \(0.5\).

The log-likelihood difference from the null is

$$ \ell(\beta)-\ell(0) =n\left(\beta^{\mathsf T}t-\frac{1}{2}\beta^{\mathsf T}R\beta\right). $$

For the candidate, \(\beta^{\mathsf T}t=0.08(0.08)+0(0.06)=0.0064\), and \(\beta^{\mathsf T}R\beta=0.08^2=0.0064\). Therefore

$$ \ell(\beta)-\ell(0) =400(0.0064-0.0032)=1.28. $$

The candidate’s likelihood is \(e^{1.28}\), approximately \(3.597\), times the null likelihood, conditional on the stated Gaussian model and fixed variance. This is a likelihood comparison, not a posterior probability: posterior probabilities also require prior probabilities or prior distributions over effects and configurations.

Assumptions, Diagnostics, and Alternatives

The derivation relies on a linear quantitative-trait model, correctly centered phenotype and genotype columns, and a common set and ordering of variants in \(t\) and \(R\). The reported effects and genotypes must also use compatible allele coding and scaling. The Gaussian likelihood uses a specified residual variance; replacing it with an estimate or using association \(z\)-statistics introduces estimation or approximation considerations. The LD matrix should describe the genotype correlations relevant to the association statistics. When these assumptions fail, the numerical likelihood may not represent the data-generating model.

Before fitting a fine-mapping model, check that \(R\) is symmetric, has diagonal entries near one if it is a correlation matrix, and is positive semidefinite up to small numerical rounding error. Check that its variant labels and order match the summary statistics, and verify that effect alleles and sample sizes are defined consistently. A discrepancy in those inputs can create an apparent pattern of joint effects that is an artifact of incompatible summaries. Extreme or unstable joint estimates can also signal near-collinearity: small eigenvalues of \(R\) mean that some linear combinations of effects are weakly identified.

Summary-statistic methods are attractive because they avoid moving individual-level genotypes and can reduce computation. When individual-level \(X\) and \(y\) are available, direct modeling can retain details such as sample-specific covariates and missingness. A likelihood based on \(R\) and \(t\) is a compact alternative when the required summaries are available and the model assumptions are acceptable. Using marginal \(p\)-values alone is more limited: \(p\)-values typically discard effect direction and scale, and without LD they do not encode the joint dependence needed to distinguish correlated variants.

A common mistake is to treat each marginal slope as the corresponding joint effect. That reasoning ignores the relation \(E(t)=R\beta\). The correction is to account for LD in the model: under a joint-effect model, the expected marginal slope at each variant is a combination of effects across correlated variants. A second mistake is to read a likelihood ratio as the probability that a variant is causal. It is not; a posterior probability additionally depends on the prior model.

Key takeaway: For standardized genotypes, the marginal-effect vector has expectation \(R\beta\), not generally \(\beta\). The LD matrix both maps joint effects to marginal associations and describes their sampling dependence, so it must be used together with compatible summary statistics in fine-mapping.

Check Your Understanding

Answer each question, then compare with the solutions below.

  1. If \(R=I\), what is the relationship between the expected marginal slopes and joint effects?
  2. For two variants with correlation \(0.6\) and joint effects \((0.10,0)^{\mathsf T}\), what are the expected marginal slopes?
  3. Why can two marginal \(z\)-statistics be correlated even under the null hypothesis?
  4. What extra ingredient is needed to turn a likelihood comparison into a posterior probability for a causal configuration?

Solutions. (1) Since \(R=I\), \(E(t)=R\beta=\beta\). (2) Multiply by the LD matrix: \(\begin{pmatrix}1&0.6\\0.6&1\end{pmatrix}\begin{pmatrix}0.10\\0\end{pmatrix}=\begin{pmatrix}0.10\\0.06\end{pmatrix}\). (3) Under the null, \(\operatorname{Cov}(t)=\sigma^2R/n\), so nonzero LD gives correlated sampling errors and hence correlated statistics. (4) A prior probability or prior distribution over causal configurations and effects is needed; likelihood alone is not posterior probability.

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