Deriving the LD Score Regression Slope

Why Should Association Statistics Depend on LD Score?

A SNP association statistic measures how strongly a tested SNP is associated with a phenotype. But the tested SNP need not itself affect the phenotype: it can also be associated because it is correlated, through linkage disequilibrium (LD), with other variants that do affect the phenotype. The more genetic variation a SNP tags, the larger its association statistic may be on average.

LD Score Regression makes this idea quantitative. Its central theoretical result is that, under a specified additive genetic model, the expected chi-square statistic at a SNP increases linearly with that SNP’s LD score. Deriving the relationship clarifies what its slope represents—and why that interpretation depends on assumptions about genetic effects, the LD reference, and association-test calibration.

Definition: For a tested SNP $j$ and a set of $M$ SNPs, the LD score is $l_j=\sum_{k=1}^{M} r_{jk}^{2}$, where $r_{jk}$ is the population correlation between the standardized genotypes at SNPs $j$ and $k$. The sum includes $k=j$, for which $r_{jj}^{2}=1$. Thus, $l_j$ measures the squared-correlation tagging of the SNP set by SNP $j$.

The Genetic Model and the Marginal Association

Consider a quantitative phenotype $Y$ standardized to have variance 1, and standardized genotypes $X_1,\ldots,X_M$, each with mean 0 and variance 1. Use the additive model

Y=\sum_{k=1}^{M} X_k\beta_k+\varepsilon,

where $\beta_k$ is the effect of SNP $k$ on the standardized phenotype and $\varepsilon$ is residual variation. Assume the residual is uncorrelated with the genotypes. Under the random-effects model used for this derivation, the effects have mean 0, are independent across SNPs, and have common variance $\sigma_\beta^2$. The expected additive genetic variance is then

h_g^2=\sum_{k=1}^{M}\operatorname{Var}(\beta_k)=M\sigma_\beta^2.

This definition is an expectation over the random effects. For a particular draw of effects, the realized genetic variance need not equal that expectation exactly. The equal-variance assumption means that, on average, each SNP contributes the same amount of effect variance; SNPs with zero effects can be represented by a random-effects distribution that includes zero-valued effects.

A marginal association at SNP $j$ reflects all SNP effects correlated with $X_j$. Because the genotypes are standardized, the population marginal regression coefficient is

\gamma_j=\operatorname{Cov}(X_j,Y)=\sum_{k=1}^{M} r_{jk}\beta_k.

Consequently, the SNP’s marginal effect is not generally just $\beta_j$. It combines the effects of variants tagged by that SNP, with the sign and magnitude of each contribution determined by the corresponding genotype correlation.

Worked Example: Calculating a Marginal Effect

Suppose a hypothetical tested SNP has correlations $r_{j1}=0.5$ and $r_{j2}=-0.2$ with two phenotype-associated SNPs. Their effects on a standardized phenotype are $\beta_1=0.04$ and $\beta_2=0.10$. Ignore other contributions for this calculation.

The marginal coefficient is the correlation-weighted sum of the two effects:

\gamma_j=r_{j1}\beta_1+r_{j2}\beta_2 =(0.5)(0.04)+(-0.2)(0.10) =0.020-0.020 =0.

In this hypothetical example the two contributions cancel, so the population marginal coefficient at the tested SNP is zero. This illustrates why an individual SNP’s statistic can depend on the signed effects and correlations, not simply on its LD score. The LD-score relationship concerns the expected statistic across random effect draws, not a guarantee about every particular locus.

From Marginal Effects to Expected Chi-Square

For a sufficiently large GWAS with $N$ independent, unrelated observations, a calibrated one-degree-of-freedom association test at SNP $j$ can be approximated by a standard-normal statistic $Z_j$. With standardized phenotype and genotype, its standard error for a small marginal effect is approximately $1/\sqrt{N}$, so its mean under that effect is approximately $\sqrt{N}\gamma_j$. In this approximation,

Z_j\approx N\!\left(\sqrt{N}\gamma_j,\,1\right), \qquad E[Z_j^2\mid \gamma_j]\approx 1+N\gamma_j^2.

The first argument of $N(\mu,\sigma^2)$ is the mean and the second is the variance. The chi-square statistic is $Z_j^2$. Its expected value is the variance of $Z_j$ plus the square of its mean, giving $1+N\gamma_j^2$. The 1 is the sampling-noise contribution when the test is calibrated; the other term is the noncentrality contributed by the marginal genetic effect.

Now average over the random effects. Since effects at different SNPs are independent, have mean 0, and have variance $\sigma_\beta^2$, the cross-products between distinct effects have expectation 0. Therefore,

\begin{aligned} E[\gamma_j^2] &=E\left[\left(\sum_{k=1}^{M}r_{jk}\beta_k\right)^2\right]\\ &=\sum_{k=1}^{M}r_{jk}^{2}E[\beta_k^2] +\sum_{k\ne m}r_{jk}r_{jm}E[\beta_k\beta_m]\\ &=\sigma_\beta^2\sum_{k=1}^{M}r_{jk}^{2} =\frac{h_g^2}{M}l_j. \end{aligned}

Substituting this expected squared marginal effect into the association-statistic approximation yields the central relationship.

Theorem: Under the standardized additive random-effects model, with equal independent effect variances, calibrated association tests, and the large-sample approximation above, the expected chi-square statistic at SNP $j$, averaged over random effects and sampling, is approximately $$ E[\chi_j^2]\approx 1+\frac{N h_g^2}{M}l_j. $$ Thus the ideal regression slope of expected chi-square on LD score is $N h_g^2/M$. It is informative about SNP heritability when the assumptions supporting this relationship hold and the SNP set and LD scores correspond to the genetic model.

Worked Example: Expected Chi-Square at a Given LD Score

Suppose a hypothetical GWAS has $N=100{,}000$ individuals, the modeled SNP set contains $M=10{,}000$ SNPs, and its SNP heritability is $h_g^2=0.40$. Consider a SNP with $l_j=25$. Assume the model and calibration conditions used in the derivation apply.

First calculate the expected slope per unit of LD score:

\frac{N h_g^2}{M} =\frac{100{,}000(0.40)}{10{,}000} =4.

The expected chi-square statistic is therefore

E[\chi_j^2]\approx 1+(4)(25)=101.

This is an expectation across random effect draws and sampling, not a prediction that this particular SNP must have an observed chi-square of 101. A realized statistic can differ because the actual effects and sampling noise differ from their expectations.

Why the Slope Is Linear in LD Score

The derivation depends on two separate ingredients. First, LD transmits effects at other SNPs into a tested SNP’s marginal coefficient through $r_{jk}$. Second, squaring that coefficient and averaging over independent, mean-zero effects removes cross-products and leaves $r_{jk}^2$. Summing those squared correlations produces $l_j$. Equal effect variance lets the variance factor out of that sum.

The slope also has a direct scale interpretation. The expected effect variance per SNP is $h_g^2/M$. The contribution of a unit increase in LD score to expected chi-square is that variance multiplied by sample size $N$. Hence the slope is $N h_g^2/M$, and rearranging gives $h_g^2=M(\text{slope})/N$ in the idealized model. The relation is a theoretical justification for using the slope; it does not mean every observed point lies on a line.

Worked Example: Comparing Two LD Scores

Consider a hypothetical study with $N=50{,}000$, $M=5{,}000$, and $h_g^2=0.20$. Suppose two tested SNPs have LD scores 8 and 20. Assume both are tested with the same calibrated procedure and that the equal-variance model is appropriate.

The slope is

\frac{N h_g^2}{M} =\frac{50{,}000(0.20)}{5{,}000} =2.

For the SNP with $l_j=8$, the expected statistic is $1+(2)(8)=17$. For the SNP with $l_j=20$, it is $1+(2)(20)=41$. The difference is $41-17=24$, which also equals the slope times the LD-score difference: $2(20-8)=24$. This is a comparison of expected statistics under the model, not a claim that the higher-LD SNP must have the larger observed statistic in a particular dataset.

Assumptions That Make the Slope Informative

The slope has the SNP-heritability interpretation above only to the extent that the derivation’s assumptions are reasonable. In particular:

Effects are adequately represented by the model. The effects have mean zero, are independent across SNPs, and share a common variance. These conditions make the expected squared marginal effect proportional to the ordinary LD score.
The modeled SNP set is coherent. The number $M$, the heritability $h_g^2$, and the SNPs used to compute $l_j$ must refer to a compatible set. Changing the set can change the heritability target and the LD scores.
LD scores represent the study population. Correlations from a mismatched reference population can misrepresent which genetic effects each tested SNP tags.
Association statistics have the assumed sampling behavior. The approximation uses a calibrated null variance of 1 and a large-sample standard error of approximately $1/\sqrt{N}$. Dependence, population structure, or other sources of statistic inflation can violate this simple baseline.
The approximation is appropriate. The derivation uses a large-sample association approximation and a linear additive phenotype model. It is not an exact finite-sample identity for every trait, test, or study design.

These are conditions for interpreting the slope, not merely technical details. If they fail, a fitted line may still summarize the observed association statistics, but its slope need not equal $N h_g^2/M$. The expected relationship also does not establish that any individual association is causal.

Worked Example: Unequal Effect Variances Change the Relationship

Consider a hypothetical simplified genetic model with two effect-bearing SNPs. Suppose their independent effects have variances $0.003$ and $0.001$, respectively, so the expected total additive variance is $h_g^2=0.004$. For two tested markers, suppose the squared correlations with these two SNPs are $(0.9,0.1)$ for marker A and $(0.2,0.8)$ for marker B. Focus on these two contributions to each marker’s LD score; both sums of squared correlations equal 1.

Because the two effect variances are unequal, the expected squared marginal effect is a variance-weighted sum. For marker A it is

E[\gamma_A^2]=(0.9)(0.003)+(0.1)(0.001)=0.0028.

For marker B it is

E[\gamma_B^2]=(0.2)(0.003)+(0.8)(0.001)=0.0006+0.0008=0.0014.

If $N=10{,}000$, the corresponding expected statistics from these contributions are approximately $1+10{,}000(0.0028)=29$ and $1+10{,}000(0.0014)=15$. Yet the two markers have the same summed squared correlations to the two effect-bearing SNPs. Their different expected statistics arise because marker A tags more of the SNP with larger effect variance.

This toy calculation shows why total heritability alone is not enough when effects have nonuniform variances: the relevant quantity is $\sum_k r_{jk}^2\operatorname{Var}(\beta_k)$, not a common variance multiplied by $\sum_k r_{jk}^2$. The simple slope follows when the common-variance assumption makes that factorization valid.

Common Mistake and Key Takeaway

Common mistake: “A SNP with a higher LD score must have a larger association statistic, so the statistic at that SNP reveals its causal effect.” This reasoning confuses a relationship between expected statistics across random effects with a deterministic SNP-by-SNP rule. A specific SNP’s marginal effect depends on the signed effects and correlations of the variants it tags; contributions can reinforce or cancel. Sampling variation also changes the observed statistic.

The correction is that LD Score Regression’s ideal slope is derived from an average under an additive, equal-variance random-effects model. Under the stated conditions, higher LD score corresponds to higher expected chi-square, with slope $N h_g^2/M$. The slope is informative because averaging removes the random cross-products and leaves the sum of squared correlations. Departures from the assumptions can alter that relationship.

Key takeaway: Under calibrated large-sample association testing and independent, mean-zero SNP effects with a common variance, $E[\chi_j^2]\approx 1+(N h_g^2/M)l_j$. The LD-score slope reflects SNP heritability through this derivation only when the genetic-effect, SNP-set, LD-reference, and sampling assumptions are sufficiently appropriate.

Check Your Understanding

Use the model and assumptions stated in this tutorial.

Why does the marginal effect at a tested SNP involve effects from other SNPs in LD with it?
Which assumption makes the expected squared marginal effect proportional to the ordinary LD score?
For $N=80{,}000$, $M=20{,}000$, and $h_g^2=0.10$, calculate the ideal slope per unit of LD score.
Why is the relationship between LD score and chi-square an expected relationship rather than a guarantee for each SNP?
How can unequal effect variances make two SNPs with similar LD scores have different expected association statistics?