Risk Differences and Absolute Risks from Genotype Models

From Genotype Effects to Risks

A logistic association model summarizes genotype effects on the odds of a binary outcome. Yet a reader often wants a probability: for example, the probability of disease among people with zero copies of an allele, compared with the probability among people with one or two copies. Those probabilities and their differences are useful for describing absolute risk, but they are not read directly from a logistic slope. They depend on the model intercept as well as the genotype coefficient.

This distinction matters especially in case-control studies. Sampling people according to outcome status can preserve genotype odds-ratio slopes, as in the earlier material on case-control sampling, while shifting the intercept. Without further information, that shifted intercept does not identify population risks. The calculations below distinguish model-based risks from the odds-ratio contrasts that case-control sampling can estimate.

Definition: For a binary outcome \(Y\), define the risk at genotype \(g\) as \(p_g=P(Y=1\mid G=g)\), for the stated population and outcome window. The risk difference comparing genotypes \(g_1\) and \(g_0\) is \(p_{g_1}-p_{g_0}\). It is an absolute difference in probabilities, often reported in percentage points. A logistic model specifies the log odds of the outcome; its predicted probability is obtained by applying the inverse-logit function.

Calculating Genotype-Specific Probabilities

For an additive dosage model, let \(G\) be the number of copies of the allele of interest, coded \(0,1,\) or \(2\). A simple logistic model is

\operatorname{logit}\{P(Y=1\mid G=g)\}=\alpha+\beta g, \qquad \operatorname{logit}(p)=\log\!\left(\frac{p}{1-p}\right).

Here \(\alpha\) is the log odds at dosage zero and \(\beta\) is the change in log odds for each additional allele copy. Inverting the logit gives the genotype-specific risk:

p_g=\frac{\exp(\alpha+\beta g)}{1+\exp(\alpha+\beta g)} =\frac{1}{1+\exp\{-(\alpha+\beta g)\}}.

The model’s predicted risk is a probability between zero and one. A risk difference is calculated by subtracting two such probabilities, not by subtracting logistic coefficients or odds. For a two-degree-of-freedom genotypic model, use separate indicators for the nonreference genotypes; each genotype’s linear predictor is then calculated with its corresponding coefficient.

Worked Example: Risks Under an Additive Dosage Model

Suppose a hypothetical fitted model for a binary outcome has \(\alpha=-2.00\) and \(\beta=0.70\), with \(G=0,1,2\). Calculate the three model-predicted risks and the risk differences.

For \(G=0\), the linear predictor is \(-2.00+0.70(0)=-2.00\), so \(p_0=1/(1+\exp(2.00))=0.1192\), or 11.92%.

For \(G=1\), the linear predictor is \(-2.00+0.70(1)=-1.30\), so \(p_1=1/(1+\exp(1.30))=0.2142\), or 21.42%.

For \(G=2\), the linear predictor is \(-2.00+0.70(2)=-0.60\), so \(p_2=1/(1+\exp(0.60))=0.3543\), or 35.43%.

The risk difference comparing dosage 1 with dosage 0 is \(0.2142-0.1192=0.0950\), a difference of 9.50 percentage points. Comparing dosage 2 with dosage 0 gives \(0.3543-0.1192=0.2351\), or 23.51 percentage points. These are model-based differences in predicted outcome probability; they are not claims that genotype causes the outcome.

Odds Ratios and Risk Differences Are Different Scales

For the additive model, the odds ratio comparing dosage \(g+1\) with dosage \(g\) is \(\exp(\beta)\). This ratio is constant across dosage steps under the model. The risk difference, however, depends on the baseline linear predictor and is generally not constant across those steps.

The dependence can be seen algebraically. Let \(\eta=\alpha+\beta g\), so \(p_g=\exp(\eta)/(1+\exp(\eta))\). The next dosage has linear predictor \(\eta+\beta\). Subtracting the two probabilities and combining fractions gives

\begin{aligned} p_{g+1}-p_g &=\frac{\exp(\eta+\beta)}{1+\exp(\eta+\beta)} -\frac{\exp(\eta)}{1+\exp(\eta)}\\ &=\frac{\exp(\eta)\{\exp(\beta)-1\}} {[1+\exp(\eta)][1+\exp(\eta+\beta)]}. \end{aligned}

The numerator contains the odds-ratio contrast through \(\exp(\beta)\), but the denominator also depends on the starting risk through \(\eta\). Thus the same odds ratio can correspond to different risk differences at different baseline risks. In particular, an odds ratio alone cannot provide an absolute risk or risk difference.

Worked Example: A Genotypic Model With Separate Contrasts

Consider a hypothetical model with genotype 0 as the reference and separate indicators for genotypes 1 and 2: \(\operatorname{logit}\{P(Y=1\mid G)\}=-2.40+0.50I(G=1)+1.10I(G=2)\). Calculate the predicted risks and compare each nonreference genotype with genotype 0.

For genotype 0, both indicators equal zero. Thus \(p_0=1/(1+\exp(2.40))=0.08317\), or 8.317%.

For genotype 1, the linear predictor is \(-2.40+0.50=-1.90\). Hence \(p_1=1/(1+\exp(1.90))=0.1301\), or 13.01%. Using the full-precision model probabilities, the risk difference is approximately \(0.04694\), or 4.694 percentage points.

For genotype 2, the linear predictor is \(-2.40+1.10=-1.30\). Hence \(p_2=1/(1+\exp(1.30))=0.2142\), or 21.42%. The risk difference is \(0.2142-0.08317=0.1310\), approximately 13.10 percentage points.

The coefficients correspond to log odds ratios against genotype 0, not percentage-point changes. The predicted probabilities and differences require converting each linear predictor through the inverse logit.

Why Case-Control Sampling Does Not Give Population Risks

In a case-control study, inclusion in the analysis depends on outcome status: cases and controls are sampled at different rates from the source population. Suppose the probability of being sampled is \(s_1\) for an individual with \(Y=1\) and \(s_0\) for an individual with \(Y=0\), and assume these sampling probabilities do not also depend on genotype. At any fixed genotype \(g\), sampling changes the outcome odds as follows:

\frac{P(Y=1\mid G=g,\text{sampled})} {P(Y=0\mid G=g,\text{sampled})} = \frac{s_1}{s_0} \frac{P(Y=1\mid G=g)} {P(Y=0\mid G=g)}.

Taking logarithms shows the intercept shift directly. If the population model has intercept \(\alpha\) and genotype slope \(\beta\), the sampled log odds are \(\alpha+\log(s_1/s_0)+\beta g\). The sampling factor adds the same constant to the log odds for every genotype. Consequently, the genotype slope, and hence the genotype odds ratio, is preserved under these assumptions; the intercept is not.

Key result: When case and control sampling probabilities depend on outcome but not genotype within outcome group, the sampled logistic intercept equals the population intercept plus \(\log(s_1/s_0)\), while genotype slopes are unchanged. The case-control sample alone does not generally reveal \(s_1/s_0\) or the population intercept, so it does not identify population absolute risks.

Worked Example: The Same Odds Ratio, Different Sampled Risks

Suppose a hypothetical source population has risks \(p_0=0.10\) and \(p_1=0.20\) for two genotype groups. Cases are sampled with probability \(s_1=0.50\), and noncases with probability \(s_0=0.10\). Calculate the population and sampled odds ratios, and the sampled risks.

The population odds are \(0.10/0.90=1/9\) and \(0.20/0.80=1/4\). Their ratio is \((1/4)/(1/9)=9/4=2.25\).

Among sampled people in the first genotype group, the probability of being a case is \((0.50\times0.10)/[(0.50\times0.10)+(0.10\times0.90)]=0.05/0.14=5/14=0.3571\). For the second genotype group it is \((0.50\times0.20)/[(0.50\times0.20)+(0.10\times0.80)]=0.10/0.18=5/9=0.5556\).

Using these exact fractions, the sampled odds are \((5/14)/(9/14)=5/9\) and \((5/9)/(4/9)=5/4\). Their ratio is \((5/4)/(5/9)=9/4=2.25\), the same odds ratio as in the source population. But the sampled risks, 35.71% and 55.56%, are not the source-population risks of 10% and 20%.

This example shows why the sample proportion of cases is not generally an estimate of population prevalence in a case-control design. Knowing the genotype odds ratio does not, by itself, recover either genotype-specific population risk.

Recovering Absolute Risks Requires Calibration

Population risks can be calculated if the logistic intercept is calibrated to the source population. One route is to know the case and control sampling probabilities, allowing the sampled intercept to be adjusted by subtracting \(\log(s_1/s_0)\). Another is to have suitable external information, such as a known population baseline risk at a reference genotype, together with the relevant genotype odds ratios. The calibration information must correspond to the population and outcome window being discussed.

Worked Example: Converting an Odds Ratio Using a Baseline Risk

Suppose external population information gives a risk of 8% for genotype 0, and a fitted association supplies an odds ratio of 2.00 for genotype 1 versus genotype 0. Calculate the genotype 1 risk and the risk difference, assuming the odds ratio applies to this population and comparison.

The genotype 0 odds are \(0.08/(1-0.08)=0.08/0.92=2/23\). Multiplying by the odds ratio gives genotype 1 odds of \(2(2/23)=4/23\). Converting those odds to a probability gives \(p_1=(4/23)/(1+4/23)=4/27=0.1481\), or 14.81%.

Using the unrounded risk, the risk difference is \(4/27-0.0800=0.06815\), approximately 6.815 percentage points. The odds ratio of 2.00 does not mean that risk doubles: here the risk ratio is \((4/27)/0.0800=50/27\approx1.852\). The baseline risk is what makes the absolute-risk calculation possible.

Interpretation and Common Pitfalls

A genotype-specific risk is meaningful only with its population, outcome definition, and time window stated. If the model also contains covariates, its inverse-logit predictions at fixed covariate values are conditional risks. A population-average risk generally requires averaging predictions over an appropriate covariate distribution; it is not necessarily the prediction at an average covariate value.

Common mistake: “The logistic coefficient is 0.70, so each allele adds 70 percentage points of risk.” This confuses a log-odds coefficient with an absolute probability difference. The correction is to exponentiate the coefficient for the per-allele odds ratio, then use the intercept and genotype value in the inverse-logit formula to obtain risks; subtract those risks to obtain a risk difference.

A related mistake is to report the case fraction in a case-control sample as population prevalence and use it to calculate absolute risk. The case fraction reflects the study’s sampling design as well as the source population. Unless the sampling fractions or suitable population calibration information are known, the absolute-risk interpretation is not justified. Report the odds-ratio association supported by the design, and state clearly what additional information would be needed for absolute risks.

Check Your Understanding

Use the definitions and calculations above to answer these questions.

In a dosage logistic model, what quantities are needed to calculate the risk at dosage 2?
Why can a constant per-allele odds ratio coexist with different risk differences at different baseline risks?
Under outcome-dependent sampling that is independent of genotype within outcome group, which part of the logistic model is shifted and which part is preserved?
Why is the fraction of sampled participants who are cases generally not a valid estimate of source-population prevalence in a case-control study?
If the reference-genotype risk and a genotype odds ratio are known, what calculation converts them to a comparison-genotype risk?