LD Score Regression Estimation from GWAS Summary Statistics

Why Regress Association Statistics on LD Scores?

A GWAS summary-statistics file contains association results for many tested SNPs, but it does not by itself reveal how much phenotypic variation is attributable to the analyzed SNP set. The association statistics are related to that quantity partly because a tested SNP can tag other variants through linkage disequilibrium (LD). SNPs that tag more variants tend, under a polygenic model, to have larger expected association statistics.

LD Score regression (LDSC) uses this relationship: it regresses association chi-square statistics on LD scores, then scales the fitted slope by the number of analyzed variants and the GWAS sample size. The resulting estimate is tied to the trait scale used to define SNP heritability. It is an estimate from summary statistics, not a direct calculation of the variance explained by a set of fitted SNP effects in the individuals.

Definition: For a tested SNP \(j\), its LD score is the sum of squared correlations between that SNP and the variants in a specified SNP set. Write it as \(\ell_j=\sum_{k=1}^{M}r_{jk}^{2}\), where \(r_{jk}\) is the allele-count correlation between SNPs \(j\) and \(k\), and \(M\) is the number of variants in the set. The LD score depends on the SNP set and the LD information used to calculate it.

The tested SNP is normally included in the sum, contributing \(r_{jj}^{2}=1\). Thus, a score is at least one when the SNP is in the set. A large score indicates that the tested SNP is correlated with many variants, or has stronger correlations with them; it is not itself a measure of heritability.

The Regression Model and Heritability Estimate

For each of \(m\) GWAS-tested SNPs, let \(Z_j\) be its association statistic and let \(\chi_j^2=Z_j^2\) be the one-degree-of-freedom chi-square statistic. Regress these chi-square statistics on the corresponding LD scores. A convenient model for stating the slope relationship is

E(\chi_j^2\mid \ell_j)\approx a+b\ell_j,\qquad b\approx \frac{n h_{\mathrm{SNP}}^2}{M}.

Here \(n\) is the GWAS sample size, \(M\) is the number of variants in the heritability SNP set, \(h_{\mathrm{SNP}}^2\) is SNP heritability on the specified trait scale, \(a\) is a fitted intercept, and \(b\) is the slope. The intercept is included in the regression; this tutorial uses it as a model component rather than interpreting it as a diagnostic. The estimate obtained from the slope \(\hat b\) is

\widehat{h_{\mathrm{SNP}}^2}=\frac{M}{n}\hat b.

The scale matters. For a quantitative trait, specify the phenotype scale on which its total variance and SNP heritability are defined. Rescaling a trait by a constant rescales both its genetic and total variance, so their ratio is unchanged; changing the definition of the trait or its variance scale can change the target. For a binary trait, an estimate on the observed case-control scale is not automatically an estimate on the liability scale.

Interpretation: The slope estimates the increase in expected association chi-square per unit increase in LD score under the model. Multiplying that slope by \(M/n\) converts it to an estimate of the proportion of phenotypic variance attributable to the specified SNP set on the specified trait scale.

Fitting the Regression

In ordinary least squares, the slope is the change in the fitted chi-square statistic per one-unit increase in LD score. In a weighted regression, observations receive weights \(w_j\), often chosen to improve the efficiency of the fit. The weighted slope, including a fitted intercept, is

\hat b= \frac{\sum_{j=1}^{m}w_j(\ell_j-\bar{\ell}_w)(\chi_j^2-\bar{\chi}_w^2)} {\sum_{j=1}^{m}w_j(\ell_j-\bar{\ell}_w)^2}, \qquad \bar{\ell}_w=\frac{\sum_{j=1}^{m}w_j\ell_j}{\sum_{j=1}^{m}w_j}, \quad \bar{\chi}_w^2=\frac{\sum_{j=1}^{m}w_j\chi_j^2}{\sum_{j=1}^{m}w_j}.

This formula makes the role of the data explicit: the slope is determined by how the association statistics vary with LD score after accounting for their weighted means. A practical analysis usually fits a weighted regression and uses LD scores calculated for a suitable SNP set. For the hand calculations below, the hypothetical data are deliberately small and the regression weights are given directly; they illustrate the conversion, not a recommended full GWAS analysis.

Worked Example: An Unweighted Slope Estimate

Suppose a hypothetical GWAS of a quantitative trait has \(n=10{,}000\) participants. The specified heritability SNP set contains \(M=100{,}000\) variants. Four tested SNPs have the following LD scores and chi-square statistics:

SNP	LD score \(\ell_j\)	\(\chi_j^2\)
1	100	1.40
2	200	1.80
3	300	2.20
4	400	2.60

The points lie exactly on the line \(\chi_j^2=1.00+0.00400\ell_j\). For example, when \(\ell_j=300\), the fitted value is \(1.00+(0.00400)(300)=2.20\). Thus, \(\hat b=0.00400\). The heritability estimate is

\widehat{h_{\mathrm{SNP}}^2} =\frac{100{,}000}{10{,}000}(0.00400) =10(0.00400) =0.0400.

Under the model and on the stated quantitative-trait scale, the estimated SNP heritability is \(0.0400\), or 4.00% of phenotypic variance. This example assumes a common sample size for the statistics, that the stated SNP set is the set represented by \(M\), and that the fitted slope is an appropriate summary of the association-statistic–LD-score relationship. With four points, it is only an arithmetic illustration, not evidence that those assumptions hold in an actual study.

Worked Example: A Weighted Regression

Consider another hypothetical quantitative-trait GWAS, with \(n=20{,}000\) and \(M=50{,}000\). Three SNPs have LD scores \(50,100,150\) and chi-square statistics \(1.10,1.20,1.30\), respectively. Suppose their regression weights are \(1,2,1\). The weighted means are

\bar{\ell}_w=\frac{1(50)+2(100)+1(150)}{1+2+1}=100, \qquad \bar{\chi}_w^2=\frac{1(1.10)+2(1.20)+1(1.30)}{4}=1.20.

Substitution into the weighted-slope formula gives the numerator \((1)(50-100)(1.10-1.20)+(2)(100-100)(1.20-1.20)+(1)(150-100)(1.30-1.20)=10\). The denominator is \((1)(50-100)^2+(2)(100-100)^2+(1)(150-100)^2=5{,}000\). Therefore, \(\hat b=10/5{,}000=0.00200\). The heritability estimate is

\widehat{h_{\mathrm{SNP}}^2} =\frac{50{,}000}{20{,}000}(0.00200) =2.5(0.00200) =0.00500.

The estimate is \(0.00500\), or 0.500% of phenotypic variance on the specified scale. In this example the weighted points happen to lie on a straight line, so the fitted slope is also easy to check directly: each increase of 50 LD-score units corresponds to a 0.10 increase in chi-square, giving \(0.10/50=0.00200\). As in any application, the interpretation depends on the sample size, SNP set, trait scale, and validity of the regression relationship.

Why the Slope Scales With Sample Size

The slope-to-heritability conversion can be checked with a short scaling argument. Under the polygenic approximation, if SNP heritability \(h_{\mathrm{SNP}}^2\) is spread across \(M\) variants, the average contribution per variant is proportional to \(h_{\mathrm{SNP}}^2/M\). A tested SNP tags variants in proportion to their squared correlations, summarized by its LD score. The association signal therefore grows approximately with \(n(h_{\mathrm{SNP}}^2/M)\ell_j\). In the regression, that coefficient of \(\ell_j\) is the slope, so \(b\approx n h_{\mathrm{SNP}}^2/M\). Solving this relation for heritability yields the estimator used above:

b\approx\frac{n h_{\mathrm{SNP}}^2}{M} \quad\Longrightarrow\quad \frac{M}{n}b\approx h_{\mathrm{SNP}}^2 \quad\Longrightarrow\quad \widehat{h_{\mathrm{SNP}}^2}=\frac{M}{n}\hat b.

This scaling also explains why the inputs must agree. The sample size belongs to the association statistics being regressed, while \(M\) and the LD scores must refer to a consistent SNP set and heritability target. Substituting a variant count from a different set changes the conversion factor, not merely the way the result is reported.

Worked Example: Checking the Implied Association Signal

Suppose a hypothetical GWAS has \(n=50{,}000\), a heritability SNP set of \(M=1{,}000{,}000\) variants, and a fitted slope \(\hat b=0.00150\). The implied estimate on the specified trait scale is

\widehat{h_{\mathrm{SNP}}^2} =\frac{1{,}000{,}000}{50{,}000}(0.00150) =20(0.00150) =0.0300.

As an arithmetic check, the fitted association-statistic increase attributable to the slope over an LD-score difference of 200 is \((0.00150)(200)=0.300\). For instance, if the fitted intercept were 1.05, the fitted chi-square value at LD score 200 would be \(1.05+0.300=1.35\). This calculation checks the units of the slope and fitted line; it does not establish that the assumptions hold or provide a separate heritability estimate. The \(0.0300\) estimate represents 3.00% of phenotypic variance on the stated scale.

Conditions and Limits of Interpretation

The relationship between slope and heritability is an approximation that depends on the genetic and statistical setting. Before treating the converted slope as an estimate of SNP heritability, check the following conditions:

Consistent target and SNP set. The \(M\) variants used in the conversion should correspond to the SNP set defining the heritability target and the LD-score calculation.
Comparable association statistics. The regressed statistics should be on a common basis, including a compatible sample size \(n\). If sample size varies substantially by SNP, a single-\(n\) conversion may not represent the data correctly.
Appropriate LD information. The LD scores should represent the relevant correlations among variants. Poor correspondence between the LD used and the GWAS population can distort the relationship.
Suitable regression relationship. The method relies on association statistics increasing with LD score as predicted by the model. A source of inflation that varies with LD score can affect the slope, rather than acting only through a constant baseline.
Trait scale stated clearly. The result estimates heritability for the phenotype scale used in the analysis. It should not be silently relabeled as a different scale.

A fitted slope has sampling uncertainty, which carries through the conversion: if the standard error of the slope is \(SE(\hat b)\), then for fixed \(M\) and \(n\), the standard error of the converted estimate is \((M/n)SE(\hat b)\). A noisy slope can be negative, producing a negative unconstrained estimate. Since a variance proportion is not negative, such a result is not evidence of negative genetic variance; it indicates that sampling variation or model limitations prevent the fitted slope from giving a meaningful positive estimate. It should not be silently changed to zero without stating that a constraint was imposed.

Common Mistake

A frequent error is to report the fitted slope itself as SNP heritability. The slope is on a different scale: it describes change in expected chi-square per LD-score unit and is approximately \(n h_{\mathrm{SNP}}^2/M\). The heritability estimate is obtained only after multiplying by \(M/n\). For example, the slope \(0.00150\) in the preceding example is not heritability of \(0.00150\); the specified \(M/n=20\) gives an estimate of \(0.0300\). Another mistake is to change \(M\) while keeping the slope fixed and assume the estimate is unchanged. Because \(M\) defines the target and appears in the conversion, changing it changes what is being estimated as well as the numerical conversion.

Key takeaway: LDSC estimates SNP heritability by fitting the association chi-square–LD-score slope and converting it as \(\widehat{h_{\mathrm{SNP}}^2}=(M/n)\hat b\). Report the trait scale, sample size, and SNP set that define this estimate, and interpret it only when the regression assumptions are credible.

Check Your Understanding

Use the definitions and calculations above to answer these questions.

A GWAS has \(n=25{,}000\), \(M=100{,}000\), and fitted slope \(0.00200\). What is the slope-based SNP heritability estimate, and how would you state it as a percentage of phenotypic variance on the specified scale?
What does an LD score summarize, and why is it not itself a heritability measure?
Explain why a fitted slope must be multiplied by \(M/n\) before it is reported as SNP heritability.
If sample size varies across SNPs, why might using one common \(n\) in the conversion be inappropriate?
A noisy regression produces a negative slope. Why is it incorrect to describe the result as negative genetic variance?