Why Cross-Trait LD Score Regression Works
Two traits can share genetic influences even when their association signals appear at different nearby variants. If a causal variant affects both traits, a correlated marker can show association with both because of linkage disequilibrium (LD). A marker correlated with many variants can therefore accumulate more cross-trait association signal than a marker correlated with few variants.
Cross-trait LD Score regression (LDSC) uses this pattern: it regresses the product of the two traits’ association \(z\)-statistics at each marker on that marker’s LD score. Under a specified genetic-effects model, the expected product is linear in LD score. The slope is proportional to genetic covariance; it is not itself the genetic correlation.
Model and Assumptions
Let \(G_k\) be the genotype at variant \(k\), standardized to have mean zero and variance one. For traits \(t=1,2\), use the additive model \(Y_t=\sum_{k=1}^{M}G_k\beta_{tk}+\epsilon_t\), where \(\beta_{tk}\) is the effect of variant \(k\) on trait \(t\), and \(\epsilon_t\) is the residual. The traits are standardized to variance one, so their genetic variances are their SNP heritabilities \(h_1^2\) and \(h_2^2\), and their genetic covariance is on the standardized trait scale.
The derivation uses an idealized random-effects model. Effects have mean zero; each variant has the same effect variances \(\operatorname{Var}(\beta_{1k})=\sigma_1^2\) and \(\operatorname{Var}(\beta_{2k})=\sigma_2^2\), and same-variant cross-trait covariance \(\operatorname{Cov}(\beta_{1k},\beta_{2k})=\sigma_{12}\). Crucially, effects at distinct variants are uncorrelated within each trait and across traits: for \(k\ne q\), \(\operatorname{Cov}(\beta_{tk},\beta_{tq})=0\), and \(\operatorname{Cov}(\beta_{1k},\beta_{2q})=0\). These conditions ensure that genetic variances and covariances do not include unmodeled cross-variant contributions.
Assume the genotype correlations are represented by the LD reference used to calculate \(l_j\), and that marginal association estimates are well approximated by their large-sample normal form. For the basic slope interpretation, variants are analyzed on a compatible scale and the sample sizes \(N_1,N_2\) are treated as fixed. Any approximately LD-independent contribution to the expected cross-trait \(z\)-score product can be represented by an intercept \(a\). Its causes and interpretation are a separate issue; here the focus is the LD-dependent slope.
Deriving the LD-Dependent Cross-Product
Write \(r_{jk}=\operatorname{Corr}(G_j,G_k)\). The population marginal effect of variant \(j\) on trait \(t\) is the association of \(G_j\) with the genetic component of that trait. Since the genotypes are standardized, this marginal effect is \(m_{tj}=\sum_{k=1}^{M}r_{jk}\beta_{tk}\). The product of the two marginal effects contains terms for every pair of variants:
Taking expectation over the effect distribution, all terms with \(k\ne q\) have expectation zero by the cross-variant cross-trait assumption. The terms with \(k=q\) each have expected product \(\sigma_{12}\). Thus the remaining sum is \(\sigma_{12}\sum_k r_{jk}^2=\sigma_{12}l_j\).
With standardized phenotypes and genotypes, a marginal association \(z\)-statistic is approximately \(\sqrt{N_t}\) times its marginal effect, apart from sampling noise. Consequently, the expected product of \(z\)-statistics has an LD-dependent component \(\sqrt{N_1N_2}\sigma_{12}l_j\). Including an LD-independent contribution gives the regression relationship:
From Effect Covariance to Genetic Covariance
The conversion from the effect covariance to the trait-level genetic covariance requires care. For a given set of effects, the genetic value is \(G\beta_t\). Its variance over genotypes involves the genotype correlation matrix, so one cannot simply add effect variances unless the cross-variant contributions are addressed.
In this model, the expected genetic variance is \(E[\beta_t^\mathsf{T}R\beta_t]\), where \(R\) is the genotype correlation matrix. Its diagonal entries are one. The off-diagonal terms have expectation zero because effects at distinct variants are uncorrelated. The result is \(M\sigma_t^2\). Applying the same argument to the two traits gives expected genetic covariance \(M\sigma_{12}\).
These equalities depend on the zero cross-variant effect covariances, not merely on standardized genotypes. Combining the covariance expression with the slope formula gives the usual slope-to-covariance conversion. Genetic correlation is genetic covariance divided by the square root of the two genetic variances:
Here \(\hat b\) is the fitted cross-trait LDSC slope, and \(\hat h_1^2,\hat h_2^2\) are estimates on the same standardized trait and variant-set scale. Equivalently, under this homogeneous model, \(r_g=\sigma_{12}/(\sigma_1\sigma_2)\). The covariance conversion and the correlation calculation are distinct steps: multiplying the slope by \(M/\sqrt{N_1N_2}\) gives covariance, while dividing by the genetic standard deviations produces correlation.
Worked Example: Predicting a Cross-Trait Product at One Marker
Suppose a hypothetical analysis has \(M=4\) variants, \(N_1=10{,}000\), and \(N_2=16{,}000\). Let \(\sigma_{12}=0.000020\). For a tested marker \(j\), suppose its correlations with the four variants are \(1, 0.5, 0,\) and \(-0.2\). Use an illustrative LD-independent intercept \(a=0.10\) to calculate the expected product.
First compute the LD score. The sign of a correlation disappears on squaring:
The slope implied by the model is \(\sqrt{N_1N_2}\sigma_{12}\). Here \(\sqrt{10{,}000\times16{,}000}=\sqrt{160{,}000{,}000}\approx12{,}649.1\), so \(b\approx12{,}649.1(0.000020)=0.2530\). Therefore:
This is a model-based expected product, not a claim that a measured product at this marker must equal 0.4264. Sampling variation can make an observed product higher or lower.
Worked Example: Converting a Fitted Slope to Genetic Correlation
Consider a hypothetical weighted regression whose fitted line passes through the four points \((l_j,z_{1j}z_{2j})=(1,0.35),(2,0.60),(3,0.85),(4,1.10)\). These points lie on \(0.10+0.25l_j\), so the fitted intercept is \(0.10\) and the slope is \(\hat b=0.25\). Suppose \(M=4\), \(N_1=10{,}000\), \(N_2=16{,}000\), and separate univariate analyses give \(\hat h_1^2=0.00020\) and \(\hat h_2^2=0.00032\).
Convert the slope to genetic covariance using \(\sqrt{N_1N_2}\approx12{,}649.1\):
The genetic standard deviation product is \(\sqrt{0.00020\times0.00032}=\sqrt{0.000000064}\approx0.0002530\). Hence:
On these hypothetical standardized scales, the estimated genetic correlation is positive and about 0.313. The slope of 0.25 is not that correlation; it first had to be converted to covariance, then standardized by the two genetic standard deviations.
Worked Example: Interpreting a Negative Slope
Suppose a hypothetical analysis uses \(M=100{,}000\) variants, \(N_1=50{,}000\), and \(N_2=80{,}000\). Its fitted slope is \(\hat b=-0.0200\), and the heritability estimates are \(\hat h_1^2=0.12\) and \(\hat h_2^2=0.08\). Since \(\sqrt{50{,}000\times80{,}000}\approx63{,}245.6\), the estimated genetic covariance is:
The denominator for the genetic correlation is \(\sqrt{0.12\times0.08}=\sqrt{0.0096}\approx0.09798\). Thus:
The negative estimate indicates that, in this model and on the specified scales, genetic effects that increase one trait tend on average to decrease the other. The estimated magnitude is within the allowable correlation range: \(|-0.3227|<1\). In practice, uncertainty in the slope and heritability estimates also matters; this calculation alone does not establish that the genetic correlation differs from zero.
Fitting the Regression and Using Its Result
In practice, the regression uses association statistics at many tested variants, with \(z_{1j}z_{2j}\) as the response and \(l_j\) as the predictor. LDSC implementations typically use weighted regression to account for differences in the variability of the response across markers. The fitted slope is then converted to genetic covariance using the variant count and sample-size scaling, and to genetic correlation using trait-specific heritability estimates.
This interpretation rests on the model, the LD scores, the sample-size scaling, and a compatible definition of the analyzed variant set. If the genetic effects do not follow the stated covariance pattern, or if the LD scores do not represent the relevant LD, the fitted slope need not correspond to the intended covariance through this simple conversion. It is also important that the heritabilities used in the denominator refer to the same traits and compatible genetic variant set. These requirements are part of the estimand, not merely bookkeeping.
Common Mistake: Treating the Slope as the Correlation
A common but incorrect argument is: “The regression slope is 0.25, so the genetic correlation is 0.25.” The slope depends on sample sizes and on the per-variant effect covariance. It is not standardized by the two genetic standard deviations. The correction is to convert slope to genetic covariance, then divide by \(\sqrt{h_1^2h_2^2}\), as shown in the worked example.
Another important pitfall is to claim \(h_t^2=M\sigma_t^2\) from standardized genotypes alone. In general, the genetic variance includes terms involving \(\operatorname{Cov}(\beta_{tk},\beta_{tq})\) for distinct variants, multiplied by genotype correlations. The equality follows here only because the model explicitly assumes those distinct-variant effect covariances are zero. Without that assumption or another model that eliminates the cross-variant contributions, both the heritability conversion and the resulting genetic-correlation conversion require further justification.
Check Your Understanding
Try these questions before moving on.
- Why does the LD score use squared genotype correlations, and what is \(l_j\) if the correlations are \(1, 0.3,\) and \(-0.4\)?
- Which assumption makes the \(k\ne q\) terms disappear when deriving \(E[m_{1j}m_{2j}]\)?
- If \(\hat b=0.10\), \(M=20{,}000\), and \(\sqrt{N_1N_2}=40{,}000\), what is the corresponding genetic covariance estimate?
- Why is a fitted slope not, by itself, an estimate of genetic correlation?
- What extra effect-covariance condition is needed to obtain \(h_t^2=M\sigma_t^2\) in the model described here?