Why Work Through a Pedigree Covariance Matrix?
A pedigree covariance matrix translates family relationships and shared environments into the expected pattern of similarity among quantitative phenotypes. Its diagonal entries describe the variance of each person’s phenotype; its off-diagonal entries describe the covariance between pairs. Calculating the entries carefully makes clear which part of the covariance comes from genetic relatedness, which comes from a shared household, and which applies only to an individual.
The goal here is to construct and interpret that matrix for small pedigrees with specified kinship coefficients and household memberships. The matrix is a model-based expectation, not the covariance calculated directly from a small set of observed phenotypes. No Gaussian distributional assumption is needed for the calculations below.
Model and Assumptions
For individual \(i\), write the quantitative phenotype as a mean plus three random components: an additive genetic effect, a household effect, and an individual residual. A mean or other fixed-effect contribution does not affect the covariance once it is treated as fixed.
Here \(Y_i\) is the phenotype of individual \(i\); \(\mu_i\) is its fixed mean; \(g_i\) is the additive genetic effect; \(h_{c(i)}\) is the effect for the household \(c(i)\) to which the individual belongs; and \(e_i\) is the individual residual. The variance components are \(\sigma_A^2\) for additive genetic effects, \(\sigma_H^2\) for household effects, and \(\sigma_e^2\) for residuals.
The model assumes the genetic, household, and residual components are mutually independent as component vectors. In particular, genetic effects across individuals are not assumed independent: their covariance is specified by pedigree kinship. Household effects for different households have zero covariance, and residuals are independent across individuals. Each component has mean zero. The examples use noninbred individuals and unrelated pedigree founders; kinship values are supplied or follow from those relationships.
Let \(I\{c(i)=c(j)\}\) equal 1 when individuals \(i\) and \(j\) belong to the same household and 0 otherwise. A person shares a household effect with themself, just as they have their own genetic effect and residual. Thus, for any pair \(i,j\), the expected phenotype covariance is
For \(i=j\), the genetic term is \(\sigma_A^2\) for a noninbred individual, the household term is present, and the residual variance is present. For different people, the residual term is zero. Their covariance can nevertheless be nonzero because of genetic relatedness, a shared household, or both.
Justifying the Entry Formula
The formula follows by expanding the covariance of the phenotype sums. Fixed means contribute no covariance. Mutual independence of the component vectors makes covariances between different types of components zero; for example, \(\operatorname{Cov}(g_i,h_{c(j)})=0\). Within the genetic component, however, \(\operatorname{Cov}(g_i,g_j)=2\phi_{ij}\sigma_A^2\), which need not be zero for relatives. The household term is \(\sigma_H^2\) if both people have the same household effect and zero otherwise. Residual independence gives \(\operatorname{Cov}(e_i,e_j)=0\) for distinct people and \(\sigma_e^2\) when \(i=j\). Adding the nonzero terms gives the stated expression.
If \(K\) is the matrix with entries \(K_{ij}=2\phi_{ij}\), \(C\) is the household-membership matrix with \(C_{ij}=I\{c(i)=c(j)\}\), and \(I\) is the identity matrix, the entire expected covariance matrix is
The ordering of people determines the row and column ordering, and the matrix is symmetric because each component covariance is symmetric. Its diagonal entries are variances in squared phenotype units; its off-diagonal entries are covariances in the same squared units. A positive covariance means that deviations above the respective means tend to occur together under the model. It does not establish that one person’s phenotype causes another’s.
Worked Example: Two Parents and Their Child in One Household
Consider a hypothetical nuclear family ordered as father, mother, and child. Assume the parents are unrelated and noninbred, and the child is their biological offspring. All three live in the same household. Let \(\sigma_A^2=4\), \(\sigma_H^2=2\), and \(\sigma_e^2=3\), in squared phenotype units.
The kinship coefficients on the diagonal are \(1/2\). The father and mother have kinship 0; each parent and the child have kinship \(1/4\). Therefore the additive genetic covariance for a parent and child is \(2(1/4)(4)=2\). Every pair shares the household effect, contributing 2, while only a diagonal entry includes residual variance 3.
For the father’s variance, the three terms are \(2(1/2)(4)=4\), \(2\), and \(3\), giving \(4+2+3=9\). The mother and child have the same diagonal variance. The father–mother covariance is \(2(0)(4)+2=2\). Each parent–child covariance is \(2(1/4)(4)+2=4\). Thus the expected covariance matrix is
The entry 9 on the father’s diagonal is his phenotype variance. The entry 2 for the parents is entirely household covariance in this model: they are genetically unrelated, but share a household. The entry 4 for a parent and child combines genetic covariance 2 and household covariance 2. No off-diagonal entry includes residual variance because these are distinct individuals.
Worked Example: Full Siblings and Half Siblings in Two Households
Suppose four hypothetical offspring are ordered \(A,B,C,D\). Individuals \(A\) and \(B\) are full siblings; \(C\) and \(D\) are full siblings; and each member of the first sibling pair is a half sibling of each member of the second pair. This can occur when two unrelated, noninbred parents have children in one household, and one parent later has children with a different, unrelated, noninbred partner in another household. The first pair shares household 1, and the second pair shares household 2. Let \(\sigma_A^2=6\), \(\sigma_H^2=1.5\), and \(\sigma_e^2=2\).
Every individual has diagonal kinship \(1/2\), so each diagonal genetic variance is \(2(1/2)(6)=6\). Adding household variance 1.5 and residual variance 2 gives \(6+1.5+2=9.5\). Full siblings have kinship \(1/4\), so their genetic covariance is \(2(1/4)(6)=3\). Because each full-sibling pair shares a household, its total covariance is \(3+1.5=4.5\). Half siblings have kinship \(1/8\), giving genetic covariance \(2(1/8)(6)=1.5\); they live in different households, so there is no household contribution.
Putting these entries in the stated order gives
For example, the \(A,C\) entry is 1.5 because they are half siblings but do not share a household. The \(A,B\) entry is larger, 4.5, because full-sibling genetic covariance is 3 and a shared household adds 1.5. The household contribution is not restricted to genetically related pairs; it is determined by household membership.
Worked Example: First Cousins and a Genetically Unrelated Housemate
Consider hypothetical individuals \(X,Y,M\), all in one household. \(X\) and \(Y\) are first cousins whose parents are full siblings; assume the relevant founders and unrelated spouses are noninbred and unrelated to one another. \(M\) is genetically unrelated to both cousins, such as an unrelated housemate. Let \(\sigma_A^2=8\), \(\sigma_H^2=2\), and \(\sigma_e^2=1\).
For first cousins, \(\phi_{XY}=1/16\), so their genetic covariance is \(2(1/16)(8)=1\). For either cousin and \(M\), the kinship is 0, so the genetic covariance is 0. The cousin pair share a household, as do both cousins with \(M\), so every off-diagonal entry receives household covariance 2. Each diagonal is \(8+2+1=11\).
The resulting matrix is
The \(X,Y\) covariance is 3, the sum of genetic covariance 1 and household covariance 2. Each cousin–\(M\) covariance is 2, entirely due to sharing a household. This example illustrates why a nonzero covariance should not automatically be interpreted as evidence of genetic relatedness.
Reading and Checking a Matrix
A reliable calculation begins with the pedigree relationships and household assignments, not with the variance components alone. For each pair, determine the kinship and whether household membership matches; for each person, remember that all three variance components contribute to the diagonal. A useful sequence is:
Use one consistent order for rows, columns, and all pairwise calculations.
For each pair, identify \(\phi_{ij}\) and whether the two people share a household.
Apply \(2\phi_{ij}\sigma_A^2\), add household covariance when applicable, and add residual variance only on the diagonal.
Mirror pairwise covariances across the diagonal and interpret entries in squared phenotype units.
As a basic check, a noninbred diagonal should equal \(\sigma_A^2+\sigma_H^2+\sigma_e^2\) under this model. An off-diagonal entry should not contain \(\sigma_e^2\). If two unrelated people share a household, their covariance is \(\sigma_H^2\); if related people live in different households, it is their genetic covariance alone. These checks can reveal a common bookkeeping error before the matrix is used in further analysis.
Common Mistake
Wrong reasoning: “The model assumes the individual genetic effects \(g_i\) and \(g_j\) are independent, so their covariance is zero; only the household term can make relatives’ phenotypes covary.” That claim conflicts with the pedigree model. For example, a parent and child have \(\phi_{ij}=1/4\), so their genetic covariance is \(2(1/4)\sigma_A^2=\sigma_A^2/2\), which is nonzero when \(\sigma_A^2>0\).
Correction: The genetic, household, and residual components are mutually independent as types of components, but the genetic effects across individuals have the specified pedigree covariance. Residuals are independent across individuals, and distinct household effects have zero covariance. Keeping these distinctions explicit prevents incorrectly deleting the genetic contribution to relatives’ off-diagonal entries.
Check Your Understanding
Use the covariance formula and distinguish genetic, household, and residual contributions in each answer.
- For two unrelated, noninbred people who share a household, what is their expected phenotype covariance?
- Two noninbred full siblings live in different households. Express their covariance in terms of \(\sigma_A^2\), \(\sigma_H^2\), and \(\sigma_e^2\).
- Why does a diagonal entry include residual variance while an off-diagonal entry does not under this model?
- For a parent and child with \(\sigma_A^2=10\), what is their genetic covariance before any household contribution is added?
- State the distinction between independence of component types and dependence of genetic effects across relatives.