Why a Binary-Trait Estimate Needs a Scale
For a binary trait, the recorded outcome is usually coded 0 for controls and 1 for cases. Its observed variance depends on the fraction of cases in the data. If cases are deliberately sampled more often than they occur in the population, that variance reflects the sample composition, not population prevalence. A heritability estimate on this observed 0–1 scale therefore cannot be interpreted directly as the fraction of liability variance explained by the SNP set.
The liability-threshold model represents disease status as the result of an unobserved, continuous liability crossing a threshold. The conversion uses two distinct quantities: the population prevalence, which locates the threshold, and the sample case fraction, which determines the observed binary phenotype variance in the analyzed data. Confusing these quantities can substantially change the converted estimate, especially for uncommon traits.
The Threshold-Model Conversion
Let \(G\) be the centered additive genetic value attributable to the SNP set, and let \(E\) collect the remaining liability components. For the standard derivation, assume \(G\) and \(E\) are independent, centered normal variables and \(L=G+E\) is standardized to have mean 0 and variance 1. Thus, \(\operatorname{Var}(G)=h^2_{SNP,L}\), the SNP heritability on the liability scale. The threshold is chosen so that \(P(L>t)=K\).
The binary phenotype has observed-scale variance \(\hat{p}(1-\hat{p})\) in a sample with case fraction \(\hat{p}\). In the usual threshold-model conversion, the genetic variance on this scale is approximated by \(z^2\operatorname{Var}(G)\). The resulting relationship and its rearrangement are:
Here \(h^2_{SNP,obs}\) is the observed-scale estimate, defined relative to the binary phenotype variance in the analyzed sample; \(h^2_{SNP,L}\) is the liability-scale SNP heritability, a proportion of the unit liability variance. All quantities in the conversion factor are dimensionless. The formula is often called the observed-to-liability scale conversion.
The displayed expression uses \(K^2(1-K)^2\) rather than simply \(K(1-K)\) because \(z\), the density at the population threshold, is related to prevalence through the local change in case probability near that threshold. The formula follows from the threshold model and the standard linear or small-effect approximation for translating liability genetic variance to the binary scale. It is not an exact equality for every possible distribution of genetic effects or arbitrarily large effects.
Deriving the Density-at-the-Threshold Term
A short derivation explains why the standard normal density \(z\) enters the formula. Since \(G\) and \(L\) are centered and jointly normal, with \(\operatorname{Var}(L)=1\) and \(\operatorname{Cov}(G,L)=\operatorname{Var}(G)\), their conditional expectation is:
Because \(G\) is centered, \(\operatorname{Cov}(Y,G)=E[(Y-K)G]=E[G\,1(L>t)]\). Conditioning on \(L\) and integrating over the liability tail gives:
For the last equality, \(\phi'(l)=-l\phi(l)\), so \(\int_t^\infty l\phi(l)\,dl=\phi(t)\). The slope of the best linear prediction of \(Y\) from \(G\) is therefore \(\operatorname{Cov}(Y,G)/\operatorname{Var}(G)=z\). Its explained variance is \(z^2\operatorname{Var}(G)\), which motivates the observed-scale relationship above. The conversion additionally uses the binary variance appropriate to the sample and the prevalence-defined threshold.
Worked Conversions
Worked Example: Uncommon Trait in a Balanced Case-Control Sample
Suppose a hypothetical study estimates observed-scale SNP heritability as \(h^2_{SNP,obs}=0.080\). The trait prevalence is \(K=0.010\), while half of the analyzed sample consists of cases, so \(\hat{p}=0.50\). Assume the estimate is on the observed scale defined above and the threshold-model conversion assumptions are reasonable.
First find the threshold and density. Since \(1-K=0.990\), \(t=\Phi^{-1}(0.990)\approx2.326\), and \(z=\phi(2.326)\approx0.02665\). Substitute these values into the conversion factor:
Thus,
The estimated SNP set accounts for about 4.42% of unit liability variance under this model. The result is lower than the observed-scale estimate because a balanced case-control sample has a large binary variance even though the trait is uncommon in the population.
Worked Example: Sample Case Fraction Matches Prevalence
Consider a hypothetical trait with population prevalence \(K=0.10\), sampled at its population case fraction, \(\hat{p}=0.10\). Let the observed-scale SNP heritability estimate be \(0.050\). The threshold is \(t=\Phi^{-1}(0.90)\approx1.282\), giving \(z=\phi(1.282)\approx0.1755\), or \(z^2\approx0.03080\).
The conversion factor is:
Therefore,
The liability-scale estimate is about 14.6%. It is larger than the observed-scale estimate, which is entirely possible: heritability is scale-dependent, and the conversion factor need not be less than one.
Worked Example: A Different Case Fraction Changes the Conversion
Suppose a second hypothetical study concerns a trait with the same population prevalence, \(K=0.050\), but cases are oversampled so that \(\hat{p}=0.25\). Its observed-scale estimate is \(h^2_{SNP,obs}=0.040\). The population threshold is \(t=\Phi^{-1}(0.95)\approx1.645\), and \(z=\phi(1.645)\approx0.1031\), so \(z^2\approx0.01063\).
Using the population prevalence in the numerator and the sample case fraction in the denominator:
The converted estimate is:
The estimate is about 4.53% of liability variance. The calculation uses \(K=0.050\) to set the threshold, but \(\hat{p}=0.25\) to represent the variance of the observed binary phenotype in the analyzed sample. Substituting the sample case fraction for prevalence in both places would answer a different, incorrectly specified calculation.
How to Apply and Interpret the Result
A reliable calculation keeps the inputs separate and in context. Use the population prevalence to determine where the liability threshold lies; use the analyzed sample’s case fraction for the observed binary variance. Then multiply the observed-scale estimate by the conversion factor. The resulting liability-scale estimate refers to the SNP set used for the original estimate, not to all possible genetic influences.
Record \(K\), the population prevalence for the relevant population, and \(\hat{p}\), the fraction of cases in the analyzed sample. Do not treat them as interchangeable.
Calculate \(t=\Phi^{-1}(1-K)\), then evaluate \(z=\phi(t)\).
Multiply \(h^2_{SNP,obs}\) by \(K^2(1-K)^2/\{\hat{p}(1-\hat{p})z^2\}\).
Report the converted result as a proportion of liability variance, and note that it relies on the liability-threshold model and its approximation.
The result is an estimate, not a probability that a particular person will develop the trait. It does not mean that a stated percentage of an individual’s liability is genetic. It describes the estimated fraction of liability variance attributable to the specified SNP set in the population and under the model. The estimate is also specific to the population and prevalence used: changing the prevalence changes the threshold and therefore the conversion.
Common Mistakes and Model Limits
- Using the sample case fraction as the population prevalence. The threshold is set by population prevalence \(K\); the observed binary variance uses sample fraction \(\hat{p}\). They coincide only when sampling reflects population prevalence.
- Using the wrong binary variance. For an analyzed sample with case fraction \(\hat{p}\), the observed variance is \(\hat{p}(1-\hat{p})\), not automatically \(K(1-K)\).
- Reversing the conversion. The liability estimate is the observed estimate multiplied by \(K^2(1-K)^2/\{\hat{p}(1-\hat{p})z^2\}\). Inverting this factor converts in the opposite direction.
- Treating the relationship as exact in every model. The standard conversion uses a linear or small-effect approximation. Strong departures from the threshold-model assumptions, or effects too large for the approximation, can make it less reliable.
- Leaving the prevalence undefined. Prevalence must refer to the population relevant to the trait and analysis. A mismatched prevalence produces a mismatched threshold and conversion.
Check Your Understanding
Use the definitions and conversion logic above to answer the following questions.
- In the threshold model, which quantity determines \(t\): population prevalence \(K\) or sample case fraction \(\hat{p}\)?
- A sample has 40% cases, but the population prevalence is 5%. Which proportion belongs in \(\hat{p}(1-\hat{p})\), and which belongs in the threshold calculation?
- For a trait with \(K=0.10\), \(t\) is approximately 1.282. What is \(z\), and what does it represent?
- Why does the derivation require a centered genetic value \(G\) when writing \(\operatorname{Cov}(Y,G)=E\{G1(L>t)\}\)?
- Does a liability-scale SNP heritability of 0.12 mean that there is a 12% probability that an individual’s trait is caused by the SNPs? Explain what the estimate means instead.