Converting Odds-Scale Effects to Liability Scale

Why Convert an Effect to the Liability Scale?

For a binary phenotype, a logistic regression coefficient describes a change in log odds per unit of a predictor. A liability model instead represents disease status as the thresholding of a continuous, unobserved quantity. Converting between these scales can make effect estimates easier to compare with results reported on a liability scale, but the conversion depends on the disease prevalence in the population of interest. It is an approximation, not a change of units that can be made without assumptions.

Here the predictor is a genotype dosage \(G\), coded as the number of copies of a specified allele: \(G\in\{0,1,2\}\). A per-allele logistic coefficient is measured in log-odds per additional allele. The target quantity will be the corresponding shift in latent liability, measured in standard deviations of liability per additional allele.

Liability Threshold and Local Conversion

Definition: In a liability-threshold model, an individual has disease when a latent liability \(L\) exceeds a threshold \(t\). For a standardized population liability \(L\sim N(0,1)\) and population disease prevalence \(K\), the threshold is \(t=\Phi^{-1}(1-K)\), where \(\Phi\) is the standard normal cumulative distribution function. A liability effect \(\beta_L\) is the change in the conditional mean of \(L\), in liability standard deviations, for one additional allele.

For the conversion, let \(X=G-E(G)\) be centered dosage. Near the null effect, write the logistic model as \(\operatorname{logit}P(D=1\mid X)=a+\beta_{\mathrm{logit}}X\), where \(D\) is the binary disease indicator. At zero effect, set \(a=\operatorname{logit}(K)\). The conversion compares the change in disease probability per unit of centered dosage implied by this model with the change implied by a liability model.

\beta_L \approx \beta_{\mathrm{logit}}\frac{K(1-K)}{\phi(t)} \quad\text{where}\quad t=\Phi^{-1}(1-K).

Here \(\phi\) is the standard normal density. The coefficient \(\beta_{\mathrm{logit}}\) is the log odds ratio per allele, not the odds ratio itself. Thus an odds ratio \(OR\) must first be transformed as \(\beta_{\mathrm{logit}}=\log(OR)\). The resulting \(\beta_L\) is in liability standard deviations per allele.

Why the Conversion Has This Form

The conversion is a first-order, or local, matching of probability changes at the null effect. It assumes the same population prevalence \(K\) is used to anchor both models. Centering dosage ensures that an allele effect does not create a first-order change in the population-average prevalence.

Differentiate the logistic risk.
At \(\beta_{\mathrm{logit}}=0\), each individual's disease probability is \(K\). For centered dosage \(X\), the derivative of the logistic probability with respect to \(\beta_{\mathrm{logit}}\) is \(K(1-K)X\).

Differentiate the liability risk.
With a small liability shift \(\beta_L X\), the probability is locally \(\Phi(-t+\beta_L X)\). Its derivative with respect to \(\beta_L\), at zero effect, is \(\phi(t)X\).

Match the local changes.
Equating the two derivatives gives \(K(1-K)X\beta_{\mathrm{logit}}=\phi(t)X\beta_L\). For nonzero dosage variation, cancel \(X\) to obtain the stated conversion.

For the liability step, one can define \(L=\beta_L X+\varepsilon\), with normally distributed residual liability and total liability variance standardized to one. The residual variance then depends on \(\beta_L\) and \(\operatorname{Var}(X)\). At zero effect, that residual standard deviation is one; its change contributes no first-order term to the probability derivative. This is why the local derivative is \(\phi(t)X\). The derivation does not establish an exact equality for arbitrary effect sizes.

Worked Example: A Common Outcome

Suppose a hypothetical disease has population prevalence \(K=0.10\), and a hypothetical genetic association has per-allele odds ratio \(1.22\). Convert the effect to the liability scale. Assume the odds ratio is a valid per-allele logistic estimate for the target population, and use the local approximation.

First, take the natural logarithm of the odds ratio: \(\beta_{\mathrm{logit}}=\log(1.22)=0.1989\) log-odds per allele. The threshold and density are \(t=\Phi^{-1}(0.90)=1.2816\) and \(\phi(t)\approx0.1755\). Also, \(K(1-K)=0.10(0.90)=0.0900\).

\beta_L\approx 0.1989\frac{0.0900}{0.1755} =0.1989(0.5128) =0.1020.

The estimated shift is approximately \(0.1020\) liability standard deviations per additional copy of the allele. As a local check, the logistic model implies a probability change of about \(0.0900(0.1989)=0.01790\) per allele at the null. The liability model implies \(\phi(t)\beta_L\approx0.1755(0.1020)=0.01790\). These match by construction; this check does not establish accuracy for a large effect.

Worked Example: A Rare Outcome

Consider a second hypothetical disease with target-population prevalence \(K=0.01\) and per-allele odds ratio \(1.10\). Find the approximate liability effect, assuming the logistic coefficient is small enough for a local conversion to be reasonable.

The log-odds coefficient is \(\beta_{\mathrm{logit}}=\log(1.10)=0.09531\). The threshold is \(t=\Phi^{-1}(0.99)=2.3263\), with \(\phi(t)\approx0.02665\). The prevalence term is \(K(1-K)=0.01(0.99)=0.009900\).

\beta_L\approx0.09531\frac{0.009900}{0.02665} =0.09531(0.3715) =0.03540.

The result is approximately \(0.03540\) liability standard deviations per allele. For the local probability check, \(0.009900(0.09531)=0.0009436\), while \(0.02665(0.03540)=0.0009434\), with the small difference due to rounding. A rare outcome has a high threshold and a low density at that threshold, so the conversion factor depends on both prevalence terms rather than on \(K\) alone.

Prevalence Is Not the Case Fraction in a Case-Control Sample

In a case-control study, investigators may deliberately sample a proportion of cases very different from the disease prevalence in the source population. The sample case fraction describes the study design; it is not generally an estimate of \(K\). Under standard case-control sampling, a logistic slope can be estimated without the population intercept, but a liability conversion still needs the target-population prevalence. Use prevalence appropriate to the population to which the liability scale refers, and state that choice.

Worked Example: Correcting a Case-Control Prevalence Substitution

Suppose a hypothetical case-control sample is half cases, but the disease prevalence in the target population is \(K=0.05\). The estimated per-allele odds ratio is \(1.35\). Compare conversion using the target prevalence with the incorrect substitution of the sample case fraction. Assume the logistic estimate is otherwise suitable and the local approximation is acceptable.

The log-odds coefficient is \(\log(1.35)=0.3001\). For the target prevalence, \(t=\Phi^{-1}(0.95)=1.6449\), \(\phi(t)\approx0.1031\), and \(K(1-K)=0.04750\). Thus:

\beta_L\approx0.3001\frac{0.04750}{0.1031} =0.3001(0.4607) =0.1383.

This is approximately \(0.1383\) liability standard deviations per allele. If one incorrectly used \(K=0.50\), then \(t=0\), \(\phi(t)=0.3989\), and \(K(1-K)=0.25\), giving \(0.3001(0.25/0.3989)=0.1881\). That is about 36% larger than the target-prevalence conversion. The discrepancy arises because the sample case fraction is not the population prevalence defining the liability threshold.

When the Approximation Can Fail

The conversion is most defensible as a local approximation for a modest effect, a correctly specified per-allele log-odds coefficient, and a clearly defined target-population prevalence. Important conditions and failure modes include:

Large effects: the logistic and liability models need not have matching probability changes away from the null. A first-order derivative cannot guarantee an accurate finite-effect conversion.
Uncertain or mismatched prevalence: the multiplier depends on \(K\). A prevalence from a different age range, ancestry group, diagnostic definition, or time period may define the wrong threshold.
Different predictor units: the formula is for a one-allele increase in dosage. If a score or genotype has been standardized, first identify its units; do not interpret the result as per allele without converting the coefficient accordingly.
Model mismatch: non-additive genetic effects, effect modification, or a logistic coefficient with an inappropriate covariate specification can undermine the interpretation as a common per-allele effect.
Unrepresentative association estimates: selection, population structure, or other bias in the estimated log-odds coefficient is not repaired by changing its scale.

A finite-effect alternative is to calculate probabilities under a specified probit model and compare them directly with the logistic predictions. For example, if a reference genotype has risk \(p_0\), and an additional allele changes its risk to \(p_1\), a probit-model liability shift with residual standard deviation one would be \(\Phi^{-1}(p_1)-\Phi^{-1}(p_0)\). This is not automatically an exact conversion of a logistic coefficient: \(p_0\) must be specified, and the assumed probit probabilities must correspond to the population model. It is useful as a sensitivity check, not as a universal replacement for the local formula.

The local conversion is also distinct from converting an entire score's predictive performance or variance explained. A per-allele effect describes a coefficient for a specified dosage unit. Additional assumptions, including allele-frequency and genotype-variance information, are needed for variance-based summaries. Keep the reported scale and unit attached to every estimate.

Common Mistake and Practical Reporting

Common mistake: “The study included 50% cases, so I use \(K=0.50\) in the conversion.” This confuses case-control sampling with population prevalence. Use the prevalence in the population whose liability threshold is being described; report it and its source. If that prevalence is uncertain, calculate the conversion over a plausible range and show how much the result changes.

A transparent report gives the per-allele odds ratio or log-odds coefficient, the target prevalence \(K\), the threshold \(t\), the conversion factor, and the resulting liability effect with units. It should label the result approximate and explain whether prevalence is assumed known. If the estimated effect is not small, compare against a defensible finite-effect sensitivity analysis rather than presenting the local conversion as exact.

Check Your Understanding

Answer each question, then compare with the solutions below.

A hypothetical outcome has prevalence \(K=0.20\) and per-allele odds ratio \(1.15\). What quantities must you calculate before obtaining its local liability effect?
Why does a case-control sample's proportion of cases usually not serve as \(K\) in the conversion?
What does a converted effect of \(0.08\) mean if the predictor is allele dosage?
Give one reason the local conversion could be unreliable even if the arithmetic is correct.

Solutions. (1) Calculate \(\beta_{\mathrm{logit}}=\log(1.15)\), the threshold \(t=\Phi^{-1}(0.80)\), and \(\phi(t)\); then use \(\beta_L\approx\beta_{\mathrm{logit}}[0.20(0.80)]/\phi(t)\). (2) Case-control sampling can set the case fraction by design, while \(K\) is the prevalence in the population defining the liability threshold. (3) It is an estimated shift of \(0.08\) liability standard deviations per additional allele, under the stated model and approximation; it is not an 8% change in disease probability. (4) Examples include a large effect, an incorrect target prevalence, a standardized predictor mistaken for allele dosage, or a misspecified association model.