Why a Family Likelihood Is Multivariate
In a pedigree, relatives’ quantitative phenotypes are generally correlated. A likelihood that treats each relative’s phenotype as an independent observation would therefore discard part of the model: the dependence encoded by the pedigree covariance matrix. The Gaussian likelihood puts the family’s phenotype vector, its modeled mean, and its covariance together in one probability density.
This tutorial derives that density and its log form. It assumes that the covariance matrices for the pedigree have already been specified, as in the earlier material on building pedigree covariance matrices. The task here is to translate a chosen mean and covariance model into a likelihood, not to estimate its parameters or derive a particular covariance matrix from genetic data.
The mean vector specifies the expected phenotype for each person. A common fixed-effects model is \(\mu=X\beta\), where \(X\) is a known design matrix containing an intercept and any covariates, and \(\beta\) is a vector of unknown regression coefficients. A variance-components model expresses the covariance as a weighted sum of known pedigree matrices. For example,
Here \(A_k\) is a specified covariance-pattern matrix for component \(k\), \(\tau_k\) is its variance-component coefficient, \(I_n\) is the \(n\)-dimensional identity matrix, and \(\sigma_e^2\) is the residual variance. The parameter collection \(\theta\) includes the mean parameters and covariance parameters. The matrices and coefficients must yield a positive-definite \(V(\theta)\) for the ordinary nonsingular Gaussian density to apply. In particular, the residual term represents independent residual variation in this model; it does not mean the phenotypes themselves are independent.
The Multivariate Gaussian Likelihood
For an observed vector \(y\), define its residual vector relative to the modeled mean as \(r=y-\mu\). The multivariate normal density, and thus the likelihood as a function of the parameters, is
The determinant \(|V|\) describes the overall volume and scale of the modeled joint distribution. The quadratic form measures how far the observed phenotype vector lies from its mean, accounting for the covariance and hence the family relationships. It is a squared distance in covariance-scaled coordinates, not generally the sum of squared residuals divided by a common variance.
Worked Example: A Two-Person Pedigree Likelihood
Consider two related individuals with hypothetical, standardized phenotypes \(y=(2,4)^\mathsf{T}\), and suppose their modeled means are both 3. Let the covariance matrix be
The assumptions for this calculation are that the two-person phenotype vector is jointly Gaussian with this mean and covariance, and that \(V\) is positive definite. Its determinant is \(2(2)-1(1)=3>0\), and its diagonal entries are positive, so this symmetric matrix is positive definite.
The residual vector is \(r=(-1,1)^\mathsf{T}\). The inverse covariance and quadratic form are
Substituting \(n=2\), \(|V|=3\), and the quadratic form into the density gives
This is the joint density at the two observed standardized phenotype values under the stated model. The off-diagonal covariance matters: the calculation uses the full inverse matrix, rather than multiplying two separate univariate densities.
From Density to Log-Likelihood
Taking the natural logarithm of the Gaussian likelihood turns its product and power terms into sums. Because the logarithm is strictly increasing, a parameter value that gives a larger likelihood also gives a larger log-likelihood. Applying the logarithm to each factor yields
The first term depends only on the number of phenotypes. The second penalizes covariance matrices with a large determinant, while the final term penalizes a poor fit of the observed vector to the modeled mean, measured relative to \(V\). Both covariance-dependent terms are essential: retaining only the quadratic form would reward some covariance choices simply for making residuals appear small, without accounting for the distribution’s scale.
The expression follows directly from the density: \(\log(a b)=\log(a)+\log(b)\), \(\log(|V|^{-1/2})=-\tfrac12\log|V|\), and \(\log(\exp(-Q/2))=-Q/2\), where \(Q=r^\mathsf{T}V^{-1}r\). Thus the log-likelihood is not a separate model; it is an equivalent representation of the same Gaussian likelihood.
Worked Example: Components Determine the Covariance and Density
Suppose two hypothetical siblings have a model with a shared familial component and independent residual variation:
The first matrix is taken as given for this illustration; its entries encode the specified familial covariance pattern. Adding the two components gives
This covariance is positive definite because its diagonal entries are positive and its determinant is positive. Let the mean vector be \((0,0)^\mathsf{T}\), and let the hypothetical observed phenotypes be \(y=(1,0)^\mathsf{T}\). Then
Therefore the likelihood is
The nonzero off-diagonal entry comes from the shared component, while the residual component adds variance to each individual’s diagonal. The likelihood uses their sum as one joint covariance matrix.
Mean Models, Pedigree Blocks, and Independent Families
Covariates enter through the mean vector, not by changing the Gaussian formula. For example, if \(x_i\) is an individual’s age, a linear mean model might specify \(\mu_i=\beta_0+\beta_1 x_i\). The likelihood then compares each observed phenotype with its modeled mean while still using the pedigree covariance to account for dependence. A covariate in the mean does not by itself remove familial correlation.
When several pedigrees are sampled independently, their phenotype vectors are independent under the model conditional on their specified covariates and parameters. The overall likelihood is the product of their pedigree-specific likelihoods, and the overall log-likelihood is the sum. Equivalently, the full covariance matrix is block diagonal, with one covariance block per pedigree. Independence is across pedigrees in this setup, not among members of a pedigree.
Worked Example: Summing Log-Likelihoods Across Two Pedigrees
Suppose two independent hypothetical pedigrees each contain two observed phenotypes. In both, the mean vector is \((2,2)^\mathsf{T}\), and the covariance matrix is
This matrix is positive definite because its diagonal entries are positive and its determinant is positive. Let the observed vectors be \(y_1=(2,3)^\mathsf{T}\) and \(y_2=(1,2)^\mathsf{T}\). Their residuals are \(r_1=(0,1)^\mathsf{T}\) and \(r_2=(-1,0)^\mathsf{T}\). The inverse is
For each pedigree, the quadratic form is \(r_j^\mathsf{T}V^{-1}r_j=1/0.9375=1.0667\). The two independent likelihood contributions multiply, so their log-likelihood contributions add:
The term \(-2\log(2\pi)\) occurs because there are four phenotype observations in total. The determinant contribution is positive here because each determinant is less than one, so \(-\tfrac12\log(0.9375)>0\). The two family contributions are added on the log scale; treating all four people as independent would instead use a different covariance matrix and a different likelihood.
Assumptions and a Common Pitfall
A Gaussian pedigree likelihood is appropriate only when its assumptions describe the data and sampling setup well enough for the intended analysis. In particular:
- The phenotype vector is modeled as jointly multivariate normal with the stated mean and covariance. Marginal normality alone does not establish the required joint model.
- The covariance matrix is symmetric, finite, and positive definite for the ordinary density written here. A singular covariance requires a different treatment and cannot be inserted into the displayed inverse-and-determinant formula.
- The mean and covariance are correctly specified for the individuals and covariates included. Within-family dependence must be represented by the covariance rather than ignored.
- For a product of pedigree likelihoods, distinct pedigrees are assumed independent under the model. If observations share modeled dependence across family units, the block-diagonal factorization is not justified.
A frequent mistake is to say, “There are two relatives, so multiply two univariate normal densities.” That reasoning assumes their phenotypes are independent. With covariance \(V\) having nonzero off-diagonal entries, the correct joint density includes \(V^{-1}\) and \(|V|\); it generally cannot be factored into the product of the two marginal densities. Another mistake is to interpret a larger likelihood as the probability that a model is true. A likelihood is a function of parameter values for observed data, not a probability distribution over competing models unless an additional probability framework is specified.
Check Your Understanding
Use the definitions and formulas above to answer the following.
- In a pedigree Gaussian likelihood, what information is represented by the off-diagonal entries of \(V\)?
- Why does the log-likelihood contain both \(\log|V|\) and a covariance-weighted quadratic form?
- For what reason does a nonzero covariance between relatives prevent the joint density from being treated as a product of their marginal densities?
- What additional independence assumption justifies adding log-likelihood contributions from separate pedigrees?
- Explain why a likelihood value is not the probability that the specified Gaussian model is true.