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SNP Heritability Estimation · Tutorial 522 of 989

Deriving Haseman–Elston Regression from Pairwise Covariance

Follow the genetic covariance model from its assumptions to the pairwise regression equation, and learn what its slope and intercept represent.

Advanced 10 min read

What You'll Learn

  • Derive the expected phenotype cross-product for a pair of distinct individuals
  • Identify the assumptions that make genomic relationship the predictor in the regression
  • Interpret the regression slope as a genetic variance component
  • Relate the slope to SNP heritability when the GRM has unit diagonal
  • Understand how GRM scaling and shared residual covariance affect the equation

From Genetic Covariance to a Pairwise Regression

Haseman–Elston regression connects two levels of a genetic model. At the individual level, a genomic relationship matrix describes how genetic effects covary across people. At the pair level, the product of two people’s centered phenotypes carries information about that covariance. The regression equation follows by taking the expected value of that product under the model.

The key is to distinguish a model-implied expectation from an observed pairwise product. A particular pair’s product can differ substantially from its expectation because phenotypes contain residual variation. The regression uses many pairs to relate those noisy products to genomic similarity.

Definition: Let \(y_i\) be the phenotype of individual \(i\), and let \(\mathbf{x}_i^\mathsf{T}\boldsymbol{\beta}\) be its mean under specified fixed effects. Define the centered phenotype \(r_i=y_i-\mathbf{x}_i^\mathsf{T}\boldsymbol{\beta}\). Let \(K\) be the genomic relationship matrix for the SNP set, with entry \(K_{ij}\) describing the genomic relationship between individuals \(i\) and \(j\). In the additive random-effect model, \(r_i=g_i+e_i\), where \(g_i\) is the additive genetic effect and \(e_i\) is the residual.

Assumptions and the Pairwise Expectation

For the basic derivation, assume that the genetic effects have mean zero and covariance matrix \(\sigma_g^2K\), and that residuals have mean zero, variance \(\sigma_e^2\), and are uncorrelated across individuals. Assume also that genetic effects and residuals are uncorrelated. In matrix notation, these assumptions are

$$ \operatorname{Cov}(\mathbf{g})=\sigma_g^2K,\qquad \operatorname{Cov}(\mathbf{e})=\sigma_e^2I,\qquad \operatorname{Cov}(\mathbf{g},\mathbf{e})=0. $$

Here \(\sigma_g^2\) is the genetic variance component for the SNP set under the chosen GRM scaling, and \(\sigma_e^2\) is the residual variance. The derivation needs these covariance assumptions and zero means; normality is not required. Since \(r_i=g_i+e_i\), expand the cross-product and take its expectation:

$$ \begin{aligned} E[r_i r_j] &=E[(g_i+e_i)(g_j+e_j)]\\ &=E[g_i g_j]+E[g_i e_j]+E[e_i g_j]+E[e_i e_j]\\ &=\sigma_g^2K_{ij}+\sigma_e^2 I_{ij}. \end{aligned} $$

The two mixed terms vanish because the genetic and residual effects are uncorrelated and have mean zero. The final equality uses the specified covariance matrices. For a pair of distinct individuals, \(I_{ij}=0\), so the expression reduces to

$$ E[r_i r_j]=\sigma_g^2K_{ij},\qquad i\ne j. $$

This is the central link: under the model, the expected cross-product is proportional to genomic relationship, and the proportionality constant is the genetic variance component. For a single pair, \(r_i r_j\) is not itself an estimate of \(\sigma_g^2K_{ij}\); the equality is about its expectation over the random effects.

Worked Example: Expected Covariance for One Pair

Suppose a hypothetical cohort is modeled with \(\sigma_g^2=0.60\) phenotype-units squared, \(\sigma_e^2=1.40\) phenotype-units squared, and \(K_{12}=0.20\) for two distinct individuals. Under the stated model, their expected residual-phenotype cross-product is

Formula: \(E[r_1r_2]=\sigma_g^2K_{12}\) for \(1\ne2\).

Substitution: \(E[r_1r_2]=(0.60)(0.20)=0.12\) phenotype-units squared.

The residual variance does not appear in this pair’s expectation because the residuals are assumed uncorrelated across distinct individuals. This does not mean the residual variance is zero: it contributes to each individual’s phenotype and to the variability of the observed product.

Why the Equation Is a Regression

For each distinct pair \(i,j\), the model gives an expected response \(E[r_ir_j]\) and a predictor \(K_{ij}\). Across pairs, these expectations lie on a line through the origin with slope \(\sigma_g^2\). The corresponding moment-regression equation is

$$ r_i r_j \approx \sigma_g^2 K_{ij},\qquad i<j. $$

The approximation sign matters: the left side is observed and noisy, while the right side is its model-implied expectation. In an analysis, the regression slope is obtained by matching observed pairwise products to genomic relationships across the chosen pairs. This tutorial’s equation is a consequence of the covariance model, not a claim that every observed point lies exactly on its line.

An intercept can be included when the pairwise products are expected to have a shared covariance component even at \(K_{ij}=0\). For example, if the residuals have a common off-diagonal covariance \(\tau\), then for distinct individuals

$$ E[r_i r_j]=\tau+\sigma_g^2K_{ij}. $$

In that model, the expected intercept is \(\tau\), and the expected slope remains \(\sigma_g^2\). Under the basic independent-residual model, \(\tau=0\), so the model-implied intercept is zero. An estimated intercept can nevertheless differ from zero in a finite sample. It should not automatically be interpreted as residual variance: the residual variance appears on the diagonal of the covariance matrix, not in the off-diagonal expectation for distinct individuals.

Worked Example: Reading the Expected Regression Line

Consider a hypothetical model with genetic variance \(\sigma_g^2=0.50\) phenotype-units squared. For distinct pairs, suppose the genomic relationships are \(0.00\), \(0.10\), and \(0.30\). Under independent residuals, the respective expected cross-products are

Formula: \(E[r_ir_j]=\sigma_g^2K_{ij}\).

Substitution and arithmetic: \(0.50(0.00)=0.00\), \(0.50(0.10)=0.05\), and \(0.50(0.30)=0.15\), each in phenotype-units squared.

These expectations follow the line \(E[r_ir_j]=0.50K_{ij}\). If instead the model included a shared off-diagonal residual covariance \(\tau=0.02\) phenotype-units squared, the three expectations would be \(0.02\), \(0.07\), and \(0.17\). That model changes the intercept, not the genetic slope.

From the Slope to SNP Heritability

A variance component’s numerical value depends on the scale of the GRM. To connect the slope to SNP heritability, use a GRM normalized so that \(K_{ii}=1\). The marginal variance of a centered phenotype for an individual is then

$$ \operatorname{Var}(r_i) =\operatorname{Var}(g_i)+\operatorname{Var}(e_i) =\sigma_g^2K_{ii}+\sigma_e^2 =\sigma_g^2+\sigma_e^2. $$

For this model and SNP set, SNP heritability is the fraction of that marginal variance attributable to the additive genetic component. Thus, once the regression slope identifies \(\sigma_g^2\), the corresponding variance ratio is

$$ h^2_{\mathrm{SNP}}= \frac{\sigma_g^2}{\sigma_g^2+\sigma_e^2}. $$

The pairwise regression supplies the genetic variance component through the slope; it does not, by itself, turn that slope into a heritability. The residual variance is also needed for the denominator. The unit-diagonal condition matters: if \(K_{ii}\) is not one, the genetic contribution to individual \(i\)’s marginal variance is \(\sigma_g^2K_{ii}\), so the displayed ratio cannot be used without accounting for the GRM scale.

Worked Example: Converting a Slope to a Variance Ratio

Suppose a hypothetical pairwise model has a GRM with unit diagonal, a genetic variance component of \(0.40\) phenotype-units squared, and a residual variance of \(1.60\) phenotype-units squared. The expected regression slope for distinct pairs is \(0.40\) phenotype-units squared per unit of genomic relationship. The SNP heritability under this model is

Formula: \(h^2_{\mathrm{SNP}}=\sigma_g^2/(\sigma_g^2+\sigma_e^2)\).

Substitution: \(h^2_{\mathrm{SNP}}=0.40/(0.40+1.60)=0.40/2.00=0.20\).

The model attributes 20% of the marginal phenotype variance, measured on this phenotype scale and for this SNP set, to the additive genetic component. This interpretation is conditional on the model and its variance decomposition; it is not a statement that 20% of an individual’s phenotype is genetically determined.

GRM Scaling and the Meaning of the Slope

The relationship predictor’s scale and the variance component’s scale are linked. Suppose a GRM is multiplied by a positive constant \(c\), producing \(K^*=cK\). The same genetic covariance matrix can be written either as \(\sigma_g^2K\) or as \(\sigma_g^{2*}K^*\), provided

$$ \sigma_g^{2*}=\frac{\sigma_g^2}{c}, \qquad \sigma_g^{2*}K^*=\frac{\sigma_g^2}{c}(cK)=\sigma_g^2K. $$

Consequently, a slope expressed per unit of \(K^*\) is \(\sigma_g^2/c\), not \(\sigma_g^2\) on the original scale. The covariance model has not changed; its component has been re-expressed using a differently scaled predictor. A heritability calculation must use a variance component and a marginal variance that correspond to the same GRM convention.

Worked Example: Rescaling the Relationship Matrix

Suppose a hypothetical covariance model uses \(K\) with unit diagonal and genetic variance \(\sigma_g^2=0.30\) phenotype-units squared. Its off-diagonal covariance is \(0.30K_{ij}\). Now define \(K^*=2K\). To preserve the covariance, the new component must be

Formula: \(\sigma_g^{2*}=\sigma_g^2/2\).

Substitution: \(\sigma_g^{2*}=0.30/2=0.15\) phenotype-units squared.

For a pair with \(K_{ij}=0.10\), the original covariance is \(0.30(0.10)=0.03\) phenotype-units squared. The rescaled relationship is \(K^*_{ij}=0.20\), giving \(0.15(0.20)=0.03\) phenotype-units squared as well. The covariance is invariant, but the slope’s numerical value changes with the predictor’s scale.

Derivation as a Sequence of Steps

1
Remove the modeled mean.
Write \(r_i=y_i-\mathbf{x}_i^\mathsf{T}\boldsymbol{\beta}\), so that the random part of the phenotype is \(g_i+e_i\).
2
Specify covariance components.
Assume \(\operatorname{Cov}(\mathbf{g})=\sigma_g^2K\), \(\operatorname{Cov}(\mathbf{e})=\sigma_e^2I\), and zero covariance between genetic effects and residuals.
3
Take a distinct-pair expectation.
Expand \(E[r_ir_j]\); the mixed terms vanish, and the residual term is zero when \(i\ne j\). Therefore \(E[r_ir_j]=\sigma_g^2K_{ij}\).
4
Match expectations across pairs.
Regress or otherwise match the observed products \(r_ir_j\) to \(K_{ij}\). The model identifies the genetic variance component as the slope on the GRM’s chosen scale.

When the Derivation Can Mislead

The regression equation depends on the covariance assumptions, not merely on the fact that a GRM has been computed. If residuals from related individuals share environmental influences, their off-diagonal covariance need not be zero. If that covariance is constant across pairs, it contributes an intercept; if it varies with genomic relationship, it can be difficult to distinguish from the genetic contribution using this equation alone. Population structure, misspecified fixed effects, or a mismatched GRM can also make the assumed relationship between phenotype covariance and \(K_{ij}\) inadequate.

There is also an important distinction between pairwise expectations and independent observations. Pairs that share an individual are statistically dependent because they contain that individual’s phenotype. The covariance derivation explains the regression target; it does not establish that all pairwise products are independent or justify treating them as such for uncertainty calculations.

Common pitfall: “The regression slope is always SNP heritability.” This is incorrect. Under the model, the slope is the genetic variance component per unit of GRM. Heritability is a variance ratio requiring the residual variance and a compatible GRM scale. With a unit-diagonal GRM, the ratio is \(\sigma_g^2/(\sigma_g^2+\sigma_e^2)\); with a differently scaled GRM, the component must first be interpreted on that scale.

Check Your Understanding

Use the covariance model and the distinct-pair equation to answer these questions.

  1. For distinct individuals, why does \(\sigma_e^2\) not appear in \(E[r_ir_j]\) under the basic independent-residual model?
  2. If \(\sigma_g^2=0.25\) phenotype-units squared and \(K_{ij}=0.12\), what is the expected cross-product for that pair?
  3. What does the regression slope represent when the predictor is the GRM entry \(K_{ij}\)?
  4. How does multiplying every GRM entry by three change the component needed to preserve the same genetic covariance matrix?
  5. Under what GRM condition does \(\sigma_g^2/(\sigma_g^2+\sigma_e^2)\) give the marginal SNP heritability in the model?
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