The FBAT Statistic and Its Conditional Null

Why Condition on Family Transmissions?

A family-based association test asks whether offspring genotypes are associated with their phenotypes, using the way alleles are transmitted within families. Rather than comparing allele frequencies across potentially different populations, the test compares observed offspring genotypes with the genotypes expected from their parents. The key is to make this comparison conditional on the family information that determines which transmissions were possible.

For a biallelic marker, the FBAT score combines each offspring’s phenotype weight with that offspring’s deviation from its conditional expected genotype dosage. The score is useful only together with its conditional variance: that variance describes how much the score would fluctuate under the null model given the observed conditioning information.

Definition: Let $X_{fi}$ be the number of copies of the tested allele, coded 0, 1, or 2, for offspring $i$ in family $f$. Let $w_{fi}$ be that offspring’s phenotype weight, treated as fixed in the conditional calculation. Let $S_f$ denote the family information being conditioned on, such as the parental genotypes in a fully genotyped trio. The family FBAT score is $$ U_f=\sum_i w_{fi}\{X_{fi}-E(X_{fi}\mid S_f)\}. $$ The combined score across families is $U=\sum_f U_f$.

For a quantitative phenotype, a weight might be the phenotype minus a specified reference mean. For an affected-status phenotype, a weight might be the affected indicator minus a prespecified prevalence. The precise choice of weights is part of the test specification; what matters for the conditional calculation is that the weights are fixed while the possible transmissions are evaluated.

The conditional null is a statement about the score given the conditioning information. In the standard FBAT formulation, the robust null is commonly expressed as no linkage or no association. Under that null, conditional on $S_f$ and the fixed phenotype weights, the expected centered transmission score is zero. Thus, the test compares the observed weighted transmission deviations with the variation expected under the null. A zero expected score does not mean every offspring’s expected genotype equals the observed genotype; it means the conditional expectation of the total score is zero.

Conditional Mean and Variance

Write $m_{fi}=E(X_{fi}\mid S_f)$, and define the centered genotype deviation $D_{fi}=X_{fi}-m_{fi}$. Given $S_f$, the phenotype weights are fixed, so the conditional expectation of the family score follows directly from linearity of expectation.

E(U_f\mid S_f)= E\left\{\sum_i w_{fi}D_{fi}\mid S_f\right\} =\sum_i w_{fi}E(D_{fi}\mid S_f) =\sum_i w_{fi}\{E(X_{fi}\mid S_f)-m_{fi}\} =0.

For the variance, the weights are again constants in the conditional calculation. The general expression includes covariance between offspring genotype deviations. It is important not to omit those covariance terms unless the transmission structure justifies doing so.

\begin{aligned} V_f &=\operatorname{Var}(U_f\mid S_f)\\ &=\operatorname{Var}\left(\sum_i w_{fi}D_{fi}\mid S_f\right)\\ &=\sum_i\sum_j w_{fi}w_{fj} \operatorname{Cov}(X_{fi},X_{fj}\mid S_f). \end{aligned}

For one offspring, this reduces to $V_f=w_{f1}^2\operatorname{Var}(X_{f1}\mid S_f)$. If offspring transmissions are conditionally independent given the parental genotypes, the cross-covariances are zero, and the family variance is the sum of the individual weighted variances. For independent families, the combined conditional variance is $V=\sum_f V_f$. If dependence remains under the conditioning scheme, the relevant covariances must be included in $V$.

Theorem: Under the conditional null, the FBAT score has conditional mean zero and conditional variance $V=\operatorname{Var}(U\mid S)$. When $V>0$, the standardized score is $$ Z_{\mathrm{FBAT}}=\frac{U}{\sqrt{V}}. $$ With many independent informative families and no single family dominating the total variance, this statistic is commonly calibrated approximately by a standard normal distribution. The corresponding two-sided test uses $|Z_{\mathrm{FBAT}}|$; equivalently, $Z_{\mathrm{FBAT}}^2$ is compared approximately with a chi-squared distribution with 1 degree of freedom. For small samples, an exact conditional distribution can be obtained by enumerating the possible transmissions under the null.

The standardization expresses the observed score in units of its conditional standard deviation. It does not turn the score into an estimated change in phenotype per allele. If $V=0$, the conditioning information leaves no transmission variation relevant to the score, so the standardized statistic is undefined and the data provide no information for this test.

Worked Examples

Worked Example: A Single Trio With Two Heterozygous Parents

Suppose, hypothetically, both parents have genotype $Aa$, where $A$ is the tested allele. Their one offspring has dosage $X=2$ and a fixed quantitative-trait weight $w=1.5$. Conditional on these parental genotypes, Mendelian transmission gives offspring dosages 0, 1, and 2 with probabilities $1/4$, $1/2$, and $1/4$, respectively. This assumes correct genotypes and ordinary Mendelian segregation at the marker.

The conditional mean dosage is $$ E(X\mid S)=0(1/4)+1(1/2)+2(1/4)=1. $$ The observed centered dosage is $2-1=1$, so the score is $$ U_f=w\{X-E(X\mid S)\}=1.5(2-1)=1.5. $$

The conditional variance of dosage is $$ \operatorname{Var}(X\mid S) =(0-1)^2(1/4)+(1-1)^2(1/2)+(2-1)^2(1/4)=0.5. $$ Therefore, $$ V_f=w^2\operatorname{Var}(X\mid S)=1.5^2(0.5)=1.125, \qquad Z_{\mathrm{FBAT}}=\frac{1.5}{\sqrt{1.125}}=1.414. $$

The standardized score is 1.414 conditional standard deviations above zero. A standard-normal two-sided approximation would give a p-value of about 0.157. But with just this one trio, the exact conditional score takes values $-1.5,0,1.5$ with probabilities $1/4,1/2,1/4$; the exact two-sided p-value for an absolute score at least 1.5 is 0.5. This illustrates why the normal approximation should not be treated as exact in a tiny sample.

Worked Example: Sibling Weights Can Reverse the Contribution Sign

Consider a hypothetical family with two offspring and parental genotypes $Aa$ and $aa$. Each offspring therefore has dosage $X_i=1$ or 0 with probability $1/2$, so $E(X_i\mid S)=0.5$ and $\operatorname{Var}(X_i\mid S)=0.25$. Suppose the observed dosages are $X_1=1$ and $X_2=0$, with fixed weights $w_1=2$ and $w_2=-1$. Assume the two meioses are conditionally independent given the parents.

The first offspring contributes $$ w_1(X_1-0.5)=2(1-0.5)=1. $$ The second contributes $$ w_2(X_2-0.5)=(-1)(0-0.5)=0.5. $$ Thus $U_f=1+0.5=1.5$. The second contribution is positive even though that offspring received fewer tested alleles than expected: a negative weight paired with a negative genotype deviation gives a positive product.

Conditional independence makes the covariance zero, so $$ V_f=(2^2)(0.25)+((-1)^2)(0.25)=1.25, \qquad Z_{\mathrm{FBAT}}=\frac{1.5}{\sqrt{1.25}}=1.342. $$ The score is a sum of weighted transmission deviations. Its positive total does not, by itself, describe transmissions only among offspring with positive weights; it summarizes the combined contributions. In this small example, the exact null score can be $-1.5,-0.5,0.5,$ or $1.5$, each with probability $1/4$. The exact two-sided p-value for $|U_f|\geq 1.5$ is $0.5$, whereas the normal approximation based on $Z=1.342$ is about 0.180.

Worked Example: Combining Three Independent Trio Scores

Suppose three hypothetical families each have one offspring of parents $Aa$ and $aa$, so each offspring dosage has conditional mean 0.5 and variance 0.25. Let the phenotype be binary, with $Y=1$ for affected and $Y=0$ for unaffected, and use the prespecified weight $w=Y-0.25$. Assume the families are independent conditional on their parental genotypes. The observed outcomes are: family 1 affected with $X=1$; family 2 unaffected with $X=0$; family 3 affected with $X=0$.

The weights are $0.75,-0.25,0.75$. The three score contributions are $$ 0.75(1-0.5)=0.375,\qquad (-0.25)(0-0.5)=0.125,\qquad 0.75(0-0.5)=-0.375. $$ Thus the combined score is $U=0.375+0.125-0.375=0.125$.

Each family’s conditional variance is $w_f^2(0.25)$. Independence permits adding these variances: $$ V=0.25\{0.75^2+(-0.25)^2+0.75^2\} =0.25(0.5625+0.0625+0.5625)=0.296875. $$ Therefore, $$ Z_{\mathrm{FBAT}}=\frac{0.125}{\sqrt{0.296875}}=0.2294. $$ This small standardized score is close to zero. With only three families, a normal-approximation p-value is not a substitute for the exact conditional calculation; the example shows the score and variance arithmetic, not a claim of strong evidence for or against association.

Interpretation and Practical Checks

The sign of a score contribution is determined by the product $w_{fi}\{X_{fi}-E(X_{fi}\mid S_f)\}$. A positive contribution can arise from a positive phenotype weight paired with an allele excess, or from a negative phenotype weight paired with an allele deficit. A negative contribution arises when those signs differ. The sign of the total score summarizes all such products, not one simple count of transmissions in a particular phenotype group.

Under the specified coding, a large positive standardized score means the observed combined weighted deviations are in the positive direction relative to their conditional null variation; a large negative score means they are in the negative direction. A two-sided test assesses departures in either direction. Neither direction establishes that the tested allele causes the phenotype. The result concerns association under the test’s coding, conditioning information, phenotype weights, and assumptions.

The conditional calculation relies on a correctly specified transmission distribution, accurate genotypes and pedigrees, and a valid choice of conditioning information. It also treats phenotype weights as fixed for the calculation. Conditioning on parental genotypes can protect the comparison from some population differences because expected transmission is computed within the family, but it does not correct genotyping errors or make every study design problem disappear.

Common Mistakes

Reading the total sign as a description of one phenotype group. A positive total does not necessarily mean offspring with positive weights received excess tested alleles. Negative-weight offspring with allele deficits also contribute positively.
Using the observed genotype instead of its conditional expectation. The score uses $X-E(X\mid S)$, not $X$ alone. The expectation must reflect the family’s possible Mendelian transmissions.
Adding variances without checking dependence. The sum of individual weighted variances is justified when the relevant transmissions are conditionally independent. Otherwise, include the covariance terms.
Treating $Z_{\mathrm{FBAT}}$ as an effect estimate. It is a standardized score, not a per-allele phenotype change. Effect estimation requires a separate procedure.
Assuming the normal approximation is exact for a few families. The conditional distribution can be discrete in small samples; use exact conditional calibration when appropriate.

Key takeaway: FBAT forms a phenotype-weighted sum of offspring genotype deviations from their conditional expectations. Its null mean is zero, its conditional variance measures transmission variability, and the standardized score is interpreted relative to that variability, with calibration suited to the sample size.

Check Your Understanding

Use the conditional score definition and variance expression to answer these questions.

For one offspring with weight $w$, dosage $X$, and conditional mean dosage $m$, write its score contribution and conditional variance.
Why does the general family variance include covariance terms between offspring?
Can a negative-weight offspring with fewer tested alleles than expected make a positive score contribution? Explain using the product in the FBAT score.
What does a positive standardized FBAT statistic summarize, and what does it not establish?
Why might an exact conditional p-value differ from a standard-normal approximation in a very small sample?