Dependence and Variance in Family-Based Scores

Why Dependence Matters for a Family-Based Score

A family-based association test combines evidence from transmissions within families. The score may be written as a sum of offspring contributions, but that does not make the contributions independent. Relatives can share transmission events or uncertainty about the pedigree’s inherited alleles. When this creates covariance between score contributions, the variance of their sum includes more than the individual variances.

The distinction is important for inference: a score is judged relative to its null variance. If positive covariance is omitted, the calculated variance is too small and the standardized score can appear more extreme than it should. If covariance is negative, omitting it can instead overstate the variance. The required calculation is at the level of the joint conditional transmission distribution, not by assuming that every child contributes an independent observation.

For example, in a transmission test for an affected child, a weight can represent that child’s fixed affection status relative to the null mean. The weight and the conditional expected transmission are specified by the test; the point here is how to calculate the variance after defining the score. All expectations and covariances below are conditional on the family information \(C_f\). This conditioning matters: the relevant dependence is dependence that remains in the transmission distribution after conditioning, not a vague claim that relatives are “similar.”

The Family-Level Variance Formula

The variance of a sum contains a variance term for each contribution and a covariance term for each distinct pair. For a family with \(r_f\) contributing offspring, the conditional score variance is

Since the weights are fixed given \(C_f\), this can also be written directly in terms of the genotype or transmission quantities:

The factor of 2 appears because expanding the square of a sum includes both the \(j,k\) and \(k,j\) terms. A positive covariance raises the variance when the two weights have the same sign; a negative covariance lowers it. If the weights have opposite signs, the effect reverses. Thus covariance cannot be treated as automatically positive or automatically an inflation.

How Shared Transmissions Change the Calculation

In an extended pedigree, multiple score contributions can involve a common ancestral transmission or a common uncertain inherited allele. Conditional on the observed pedigree information, the corresponding offspring quantities may then be dependent. The covariance is calculated from their joint conditional distribution. A useful equivalent expression for a pair is

Knowing the two marginal transmission probabilities is not enough to calculate this covariance; the joint probability is also needed. If the same uncertain transmission enters two score terms, counting its marginal variance twice as though it arose from two independent events will generally give the wrong result.

Dependence need not be perfect or positive. For a pair of binary transmission indicators, suppose a hypothetical conditional joint distribution has probabilities \(P(X_1=1,X_2=1\mid C)=0.40\), \(P(1,0\mid C)=0.10\), \(P(0,1\mid C)=0.10\), and \(P(0,0\mid C)=0.40\). Each marginal probability of 1 is 0.50. Thus each variance is 0.25, while the covariance is \(0.40-(0.50)(0.50)=0.15\). With unit weights, the variance of the centered sum is \(0.25+0.25+2(0.15)=0.80\). The arithmetic shows why marginal variances alone cannot recover the score variance.

Combining Families and Standardizing the Score

Suppose the total score is \(U=\sum_{f=1}^{F}S_f\). If distinct families are conditionally independent under the null, then cross-family covariances are zero, and

Within each family, however, the variance in this sum must include all relevant within-family covariances. This is the central distinction: independent families can be combined by adding their family variances, while contributions inside a family must be combined using their joint conditional distribution.

A standardized score uses the square root of the null variance as its standard error. When an asymptotic normal calibration is appropriate, the statistic is \(U/\sqrt{\operatorname{Var}(U\mid C)}\). The variance formula does not by itself guarantee that this statistic is well approximated by a standard normal distribution; that calibration requires its own justification. For a small or highly dependent set of transmission events, the conditional distribution may be discrete, and a variance calculation should not be mistaken for a proof of normality.

Derivation and Practical Calculation

The family-level formula follows directly from the definition of conditional variance. Because each contribution is conditionally centered, \(E(S_f\mid C_f)=0\), so \(\operatorname{Var}(S_f\mid C_f)=E(S_f^2\mid C_f)\). Expand the square and take conditional expectations:

The last equality uses \(E(T_{fj}\mid C_f)=E(T_{fk}\mid C_f)=0\), so each pair’s conditional product expectation equals its covariance. If contributions are not centered under the conditioning used, the variance must instead be computed around the conditional mean of the total score; the displayed simplification cannot be applied without that condition.

In practice, the calculation can be organized as a family-level sequence:

Common Mistake: Assuming Every Child Is Independent

A tempting shortcut is to calculate each child’s contribution variance and add them, on the grounds that the score is a sum over children. That reasoning confuses the algebraic form of the score with the probability model for its terms. The shortcut is correct only if the relevant conditional covariances are zero. As the independent-meiosis example showed, that condition can hold for siblings when parental genotypes are known and their meioses are independent. But it need not hold for contributions involving shared transmissions or uncertainty in an extended pedigree.

The correction is not to add a generic “family correction” of arbitrary size. It is to calculate the covariances implied by the actual conditional transmission distribution and the score definition. Likewise, sibling status alone does not prove that covariance is nonzero. The transmission structure and conditioning information determine the answer.