Recalibrating Absolute Risk Without Changing Rank

Why Recalibrate Absolute Risk?

A polygenic score can order people by predicted risk appropriately while its reported absolute risks are too high or too low in a target population. One reason is a difference in baseline incidence: even if the score’s association with the outcome has the same strength, the target population may experience the outcome more or less often over the relevant time horizon. An intercept adjustment can address this kind of mismatch while leaving the score coefficient unchanged.

The adjustment has a specific aim: match the mean predicted risk to the incidence in a clearly defined target population. It does not, by itself, establish that risks are correct at every score level. In particular, if the relationship between score and outcome is steeper or flatter in the target population, changing only the intercept cannot repair that difference.

Model and Definition

Let $Y=1$ indicate that a person experiences a binary outcome within a specified time horizon, and let $S$ be that person’s polygenic score. Suppose the existing score-based model uses a logistic link:

\widehat{P}(Y=1\mid S)=\operatorname{expit}(\hat a+\hat b S), \qquad \operatorname{expit}(u)=\frac{1}{1+e^{-u}}.

Here, $\hat a$ is the fitted intercept and $\hat b$ is the fitted score coefficient. The model treats the log odds of the outcome as a linear function of $S$. For the basic recalibration considered here, the score, outcome definition, time horizon, and score coefficient are held fixed; only the intercept changes. The target incidence must refer to the same outcome and time horizon in the target population.

Definition: Intercept-only absolute-risk recalibration replaces $\hat a$ with a new intercept $\hat a_{\mathrm{T}}$, while retaining $\hat b$: $$ \widehat{P}_{\mathrm{T}}(Y=1\mid S) =\operatorname{expit}(\hat a_{\mathrm{T}}+\hat b S). $$ The new intercept is chosen so the mean predicted risk in the target population equals its specified incidence $q_{\mathrm{T}}$. This matches average risk; it does not necessarily make predictions correct conditional on every score value.

If the target population is represented by $n$ individuals with scores $S_1,\ldots,S_n$, an empirical version of the calibration condition is

\frac{1}{n}\sum_{i=1}^{n} \operatorname{expit}(\hat a_{\mathrm{T}}+\hat b S_i) =q_{\mathrm{T}}.

This equation uses the target population’s score distribution. If individuals have unequal population weights $w_i$, normalized so that $\sum_i w_i=1$, the weighted mean replaces the unweighted mean. A case-control sample’s raw case fraction is generally not the target incidence, so it cannot be substituted for $q_{\mathrm{T}}$ without an appropriate incidence source or sampling adjustment.

What an Intercept Shift Does

Write the adjustment as $\delta=\hat a_{\mathrm{T}}-\hat a$. For any fixed score, the new linear predictor is the old one plus $\delta$. On the odds scale, this has a particularly simple interpretation: the recalibrated odds equal the original odds multiplied by $e^\delta$. The probabilities themselves are not multiplied by a common constant.

\frac{\widehat{P}_{\mathrm{T}}(Y=1\mid S)} {1-\widehat{P}_{\mathrm{T}}(Y=1\mid S)} = e^\delta \frac{\widehat{P}(Y=1\mid S)} {1-\widehat{P}(Y=1\mid S)}.

For $\hat b>0$, predicted risk increases with score both before and after recalibration, so the ordering of individuals by predicted risk is unchanged. More generally, for any two scores $S_i$ and $S_j$, the difference between their recalibrated linear predictors is $\hat b(S_i-S_j)$, exactly as before. If $\hat b<0$, the ordering is still unchanged from the original model, though higher scores correspond to lower predicted risks. An intercept shift changes neither that direction nor the relative ordering. It can, however, change whether a person crosses a fixed clinical risk threshold.

Existence and Uniqueness of the Calibrating Intercept

For a finite target sample with positive weights, an intercept solving the mean-risk equation exists and is unique whenever $0<q_{\mathrm{T}}<1$. To see why, define the weighted mean predicted risk as a function of a candidate intercept $a$:

g(a)=\sum_{i=1}^{n}w_i\operatorname{expit}(a+\hat b S_i), \qquad \sum_{i=1}^{n}w_i=1.

The expit function is continuous, and its derivative is $\operatorname{expit}(u)\{1-\operatorname{expit}(u)\}$. Therefore,

g'(a)=\sum_{i=1}^{n}w_i \operatorname{expit}(a+\hat b S_i) \left\{1-\operatorname{expit}(a+\hat b S_i)\right\}>0.

Each term is positive for a finite $a$, so $g(a)$ is strictly increasing. Also, $g(a)$ approaches 0 as $a$ approaches $-\infty$, and approaches 1 as $a$ approaches $+\infty$. By continuity, it takes each value strictly between 0 and 1, and strict increase means it takes that value only once. In practice, a numerical root-finding method can find the intercept. The existence of a solution guarantees a match to the mean; it does not guarantee that the fixed-slope risk model is otherwise appropriate.

Worked Calculations

Worked Example: Moving One Risk to a New Incidence

Suppose a hypothetical person has score-based linear predictor $-2$, giving original risk $\operatorname{expit}(-2)$. For a target population in which everyone has this same score, the specified incidence is $0.10$. Find the intercept adjustment and the recalibrated risk.

The original risk and odds are

\operatorname{expit}(-2)=\frac{1}{1+e^2}\approx 0.1192, \qquad \text{original odds}=e^{-2}\approx 0.1353.

Since every person has the same score, setting the recalibrated risk to $0.10$ also sets the target mean risk to $0.10$. Its log odds are $\log\{0.10/(1-0.10)\}=\log(1/9)\approx -2.1972$. Thus the required shift is

\delta=-2.1972-(-2)\approx -0.1972, \qquad \operatorname{expit}(-2+\delta) =\operatorname{expit}(-2.1972)\approx 0.1000.

The odds are multiplied by $e^\delta\approx 0.8210$, while the risk changes from about 11.92% to 10.00%. With only one score value there is no ordering to compare, but the calculation illustrates how the intercept sets the absolute risk for that score.

Worked Example: Matching a Target Mean While Preserving Rank

Consider a hypothetical target population with two equally common score groups. Their original linear predictors are $-3$ and $-2$, and the desired target incidence is $0.20$. Find the common intercept shift and check the resulting mean risk.

Let $x=e^\delta$, so the two recalibrated risks are $x/(e^3+x)$ and $x/(e^2+x)$. The target-mean condition is

\frac{1}{2}\left\{\frac{x}{e^3+x}+\frac{x}{e^2+x}\right\}=0.20.

Multiplying through by the positive denominators gives $4x^2+1.5(e^3+e^2)x-e^5=0$. The positive root is $x\approx 2.82605$, so $\delta=\log(x)\approx 1.03896$. The shifted predictors are approximately $-1.96104$ and $-0.96104$. Substitution gives

\operatorname{expit}(-1.96104)\approx 0.12335, \qquad \operatorname{expit}(-0.96104)\approx 0.27665, \qquad \frac{0.12335+0.27665}{2}\approx 0.20000.

Before recalibration, the two risks are $\operatorname{expit}(-3)\approx0.04743$ and $\operatorname{expit}(-2)\approx0.11920$, with mean approximately $0.083314$. After recalibration, the higher-score group still has the higher predicted risk. The difference in probabilities between the groups has changed, but their ordering has not.

Worked Example: Why Matching Incidence May Not Match Score-Specific Risk

Suppose a hypothetical target population has two equally common score values, $S=0$ and $S=1$. A model with fixed score coefficient $\hat b=1$ is recalibrated to the target incidence $0.15$. However, the target’s actual risks in these two groups are $0.05$ and $0.25$, respectively. Determine what the intercept-only model predicts and compare its score gradient with the target’s.

The intercept $a_{\mathrm{T}}$ must satisfy $\{\operatorname{expit}(a_{\mathrm{T}})+\operatorname{expit}(a_{\mathrm{T}}+1)\}/2=0.15$. Numerically solving gives $a_{\mathrm{T}}\approx-2.321$. Substitution yields

\operatorname{expit}(-2.321)\approx0.0894, \qquad \operatorname{expit}(-1.321)\approx0.2106, \qquad \frac{0.0894+0.2106}{2}\approx0.1500.

The recalibrated model matches the target incidence, but its predicted risks at $S=0$ and $S=1$ are about 8.94% and 21.06%, rather than the target values of 5% and 25%. On the log-odds scale, the target difference between the two groups is

\log\left(\frac{0.25}{0.75}\right) - \log\left(\frac{0.05}{0.95}\right) \approx 1.846,

whereas the model’s log-odds difference is $\hat b(1-0)=1$. Changing the intercept moves both predicted logits by the same amount and leaves their difference at 1. Thus the intercept corrects the average incidence but cannot correct this steeper target risk gradient. The example is hypothetical and illustrates a limitation of the method, not evidence about any particular population.

When the Adjustment Is Useful—and When It Is Not Enough

An intercept-only adjustment is appropriate when the main mismatch to address is a different baseline incidence and the existing score gradient is considered suitable for the target population. A matched mean can make the absolute-risk scale more relevant to that population without sacrificing the model’s score ordering. The result depends on the target population definition, its score distribution, and the outcome time horizon used to specify incidence.

The central limitation is that a single intercept supplies only one degree of freedom. If the risk gradient differs across score levels, matching the overall mean can leave systematic overprediction in some parts of the score distribution and underprediction in others. A difference in gradient may arise from a different score–outcome association or from a relationship that is not adequately described by the fixed model form. Average incidence alone cannot determine which conditional risks are correct.

Common Mistake

Wrong reasoning: “The target incidence is twice the source incidence, so I should double every predicted probability.” This generally does not correspond to an intercept change in a logistic model; multiplying probabilities may even produce values above 1.
Correction: Intercept shifts multiply odds by the same factor, not probabilities. Choose the shift by solving the target mean-risk equation, then check both the resulting average and the fixed ranking.
Wrong reasoning: “The mean recalibrated risk equals target incidence, so the risk estimates are calibrated at every score.” A mean match imposes only one population-level constraint.
Correction: If the target score gradient differs, an intercept alone cannot change it. Matching overall incidence is not proof of score-specific accuracy.

Key takeaway: Intercept-only recalibration shifts every model logit by the same amount. It can match the target population’s mean absolute risk while preserving the score ranking, but it cannot repair a difference in the risk gradient across score values.

Check Your Understanding

Answer each question by distinguishing the change in average risk from the change in risk ranking or gradient.

In the logistic score model, which parameter changes during intercept-only recalibration, and which parameter remains fixed?
Why does adding the same intercept shift to every person preserve the original ordering of predicted risks?
A target incidence is specified as 0.12. What equation defines the recalibrated intercept for a target sample with scores $S_1,\ldots,S_n$?
Why is doubling each predicted probability generally not equivalent to intercept recalibration?
A recalibrated model matches target incidence but overpredicts risk at low scores and underpredicts at high scores. What feature of the target relationship might an intercept shift be unable to correct?