Predicting Survival and Cure Probability for Individuals

From Fitted Cure Models to Individual Predictions

A cure model separates two questions: how likely is a person to belong to the cured group, and, if that person is susceptible, how long until the event? Individual prediction combines the answers for a specified covariate profile and time horizon. The distinction matters because a person who has not had the event by a chosen date may still be susceptible and experience it later.

A prediction is conditional on the covariates supplied to the fitted model. It is not a declaration that a particular person is known to be cured or susceptible. Cure status is latent: for someone who remains event-free during follow-up, the data usually cannot directly reveal whether the person is cured or simply has a long latency time.

Definition: Let $C=1$ indicate that an individual is cured of the modeled event, and $C=0$ indicate susceptibility. For covariates $x$, define the cure probability as $\pi(x)=P(C=1\mid X=x)$. Let $S_u(t\mid x)=P(T_u>t\mid C=0,X=x)$ be the survival function for time to event among susceptible individuals. If a cured individual never experiences the event, the marginal survival is $$ S_m(t\mid x)=P(T>t\mid X=x)=\pi(x)+\{1-\pi(x)\}S_u(t\mid x). $$ The finite-time event risk is $F_m(t\mid x)=1-S_m(t\mid x)=\{1-\pi(x)\}\{1-S_u(t\mid x)\}$.

Here $T_u$ is the event time in the susceptible group, while $T$ represents the event time in the full population, including individuals whose event time is effectively infinite. The two expressions for $S_m$ and $F_m$ follow by splitting the population according to latent cure status. The first term of $S_m$ is the cured fraction, which remains event-free; the second is the susceptible fraction still event-free at time $t$.

Assumptions and Prediction Workflow

These predictions rely on the structure of the fitted mixture cure model. The cure probability must be appropriately modeled as a function of the incidence covariates, and the susceptible survival function must be appropriate for the latency covariates. The model treats cured individuals as never experiencing the modeled event, while susceptible individuals have a proper event-time distribution whose survival tends to zero as time tends to infinity. Thus, under this model, $S_m(t\mid x)$ tends to $\pi(x)$.

A fitted model may provide, for example, a logistic regression for $\pi(x)$ and a parametric or semiparametric model for $S_u(t\mid x)$. To make a prediction, calculate both components at the person’s covariates, evaluate susceptible survival at the chosen horizon, and combine them. The model must have been estimated under suitable assumptions, including censoring that is independent of the event process conditional on the modeled information. As in the earlier material on identifiability and follow-up, a plausible fitted curve does not by itself ensure that the data support long-term extrapolation.

Specify the profile and horizon.
Record the covariate values for the person and the time $t$ of interest, with time units.

Predict the cure probability.
Evaluate the incidence model at the profile to obtain $\hat\pi(x)$.

Predict susceptible survival.
Evaluate the latency model at the same profile and horizon to obtain $\hat S_u(t\mid x)$.

Combine and interpret.
Compute $\hat S_m(t\mid x)=\hat\pi(x)+\{1-\hat\pi(x)\}\hat S_u(t\mid x)$, then report survival or its complement, the event risk, in context.

Worked Predictions

Worked Example: Combining Logistic Cure and Weibull Latency Predictions

Suppose a hypothetical fitted model uses a binary covariate $x$, with $x=1$ for a specified profile. Its cure model is $\operatorname{logit}\{\pi(x)\}=-1+0.8x$. For susceptible individuals, the fitted Weibull survival is $S_u(t\mid x)=\exp\{-(t/5)^{1.5}\}$, with time measured in years. Find marginal survival and event risk at three years for a person with $x=1$.

First calculate the cure probability. The logit is $-1+0.8(1)=-0.2$, so

\hat\pi(1)=\frac{\exp(-0.2)}{1+\exp(-0.2)}\approx 0.4502.

The susceptible survival at three years is

\hat S_u(3\mid 1)=\exp\{-(3/5)^{1.5}\} =\exp(-0.4648)\approx 0.6283.

Therefore, marginal survival is

\hat S_m(3\mid 1)=0.4502+(1-0.4502)(0.6283) =0.4502+(0.5498)(0.6283) \approx 0.7956.

The predicted probability of experiencing the event by three years is $1-0.7956=0.2044$, or about 20.44%. The long-term cure probability is about 45.02%, not 20.44%. The latter is the finite-time event risk; it is smaller than the total susceptible fraction because some susceptible individuals remain event-free at three years.

Worked Example: Comparing Two Covariate Profiles

Consider another hypothetical fitted model. The cure probability follows $\operatorname{logit}\{\pi(x)\}=-1+0.5x$. Susceptible survival follows a proportional hazards model $S_u(t\mid z)=\exp\{-H_0(t)\exp(0.7z)\}$, where $H_0(2)=0.2$. Compare marginal survival and two-year event risk for profile A, $(x,z)=(0,0)$, and profile B, $(x,z)=(1,1)$. Time is measured in years.

For profile A, $\hat\pi(0)=\operatorname{logit}^{-1}(-1)\approx0.2689$, so the susceptible fraction is $0.7311$. Its susceptible survival is

\hat S_u(2\mid 0)=\exp\{-0.2\exp(0)\}=\exp(-0.2)\approx0.8187.

Thus $\hat S_m(2\mid 0)=0.2689+(0.7311)(0.8187)\approx0.8675$, and the predicted two-year event risk is $1-0.8675=0.1325$, or about 13.25%.

For profile B, $\hat\pi(1)=\operatorname{logit}^{-1}(-0.5)\approx0.3775$, leaving a susceptible fraction of $0.6225$. The latency-model prediction is

\hat S_u(2\mid 1)=\exp\{-0.2\exp(0.7)\} =\exp(-0.4028)\approx0.6685.

Then $\hat S_m(2\mid 1)=0.3775+(0.6225)(0.6685)\approx0.7936$. The two-year event risk is about $1-0.7936=0.2064$, or 20.64%. These are model-based predictions for two specified profiles. Because both the cure and latency covariates differ, this contrast alone does not isolate either component’s effect and should not be interpreted as a causal comparison.

Survival After a Landmark Time

A useful extension is to ask about someone who is already known to be event-free at time $t$. Such a person’s conditional probability of being cured is not generally $\pi(x)$. Event-free follow-up updates the probability because susceptible individuals can also remain event-free.

Derivation: A cured individual is necessarily event-free at time $t$. Therefore $P(C=1,T>t\mid X=x)=\pi(x)$, while $P(T>t\mid X=x)=S_m(t\mid x)$. By the definition of conditional probability, $$ P(C=1\mid T>t,X=x)=\frac{\pi(x)}{S_m(t\mid x)}. $$ For $t_2>t_1$, the event risk between those times conditional on being event-free at $t_1$ is $$ P(t_1<T\leq t_2\mid T>t_1,X=x) =\frac{S_m(t_1\mid x)-S_m(t_2\mid x)}{S_m(t_1\mid x)}. $$

Worked Example: Updating the Cure Probability After Event-Free Follow-Up

Suppose a hypothetical profile has cure probability $\pi(x)=0.40$, and its susceptible survival is $S_u(2\mid x)=0.70$ and $S_u(5\mid x)=0.25$. Find the marginal survivals at two and five years, the probability of being cured given survival to two years, and the conditional event risk from year two through year five.

The marginal survival probabilities are

S_m(2\mid x)=0.40+(0.60)(0.70)=0.82, \qquad S_m(5\mid x)=0.40+(0.60)(0.25)=0.55.

Among individuals with this profile who are event-free at two years, the probability of being cured is $0.40/0.82\approx0.4878$, or 48.78%. This exceeds the original 40% cure probability because the event-free group includes all cured individuals, whereas some susceptible individuals have already had the event. The conditional probability of an event after year two and by year five is

\frac{0.82-0.55}{0.82}=\frac{0.27}{0.82}\approx0.3293.

So, among those event-free at two years, the model predicts a 32.93% event risk during the next three years. The probability of eventual event among this same event-free group is $(0.82-0.40)/0.82\approx0.5122$, under the model’s assumption that susceptible survival tends to zero.

Checks, Alternatives, and Limitations

Several basic checks help catch errors in computation or interpretation. For a valid mixture prediction with $0\leq S_u(t\mid x)\leq1$, marginal survival must satisfy $\pi(x)\leq S_m(t\mid x)\leq1$. If susceptible survival decreases with time, marginal survival must also be nonincreasing. The finite-time event risk must lie between zero and $1-\pi(x)$, and it approaches $1-\pi(x)$ only as susceptible survival approaches zero. A result outside these bounds signals a calculation, coding, or model-output problem.

The main alternative is an ordinary survival model without a cure fraction, which does not include a positive long-term survival plateau as part of its structure. Within a cure model, susceptible survival $S_u$ answers a conditional question about susceptible individuals; marginal survival $S_m$ answers the population-level question for everyone with the specified covariates. Reporting the conditional curve as if it were marginal survival omits the cured group and changes the estimand.

The long-term cure probability is especially dependent on assumptions and follow-up. If the study has limited follow-up, a long susceptible latency tail can resemble a cured fraction. Predictions at horizons well beyond the observed data therefore rely heavily on model structure. Individual predictions are also plug-in estimates: they do not, by themselves, show the uncertainty arising from estimating model parameters. Numerical precision should not be mistaken for certainty about a person’s latent status or future.

Common mistake: “This person has a 45% predicted cure probability, so their event risk by three years is 55%.” This treats the complement of the cure probability as a finite-time risk. The correction is to account for susceptible survival at three years: $F_m(3\mid x)=\{1-\pi(x)\}\{1-S_u(3\mid x)\}$. The 55% is the model’s long-run susceptible fraction; only some of that group may experience the event by three years.

Check Your Understanding

Use the mixture formula and distinguish the time horizon in each answer. Solutions are included after each question.

A model gives $\pi(x)=0.30$ and $S_u(4\mid x)=0.60$. What are marginal survival and event risk at four years?
Solution. $S_m=0.30+(0.70)(0.60)=0.72$. The four-year event risk is $1-0.72=0.28$, or 28%.
For $\pi(x)=0.25$, what is the model’s long-term event probability if susceptible survival tends to zero? Why is it not necessarily the event risk at a particular finite time?
Solution. The long-term event probability is $1-\pi(x)=0.75$. At a finite time, some susceptible individuals may remain event-free, so finite-time event risk is $(1-\pi(x))(1-S_u(t\mid x))$, which can be below 0.75.
If $S_m(3\mid x)=0.80$ and $\pi(x)=0.32$, what is the predicted cure probability among those event-free at three years?
Solution. It is $\pi(x)/S_m(3\mid x)=0.32/0.80=0.40$, or 40%.
At two horizons, a fitted model gives $S_m(2\mid x)=0.75$ and $S_m(6\mid x)=0.60$. What is the event risk from year two through year six among those event-free at year two?
Solution. The conditional risk is $\{0.75-0.60\}/0.75=0.20$, or 20%.
A reported marginal survival is below the fitted cure probability at the same profile and time. Is that compatible with the mixture formula?
Solution. No. Since $S_m(t\mid x)=\pi(x)+\{1-\pi(x)\}S_u(t\mid x)$ and $S_u(t\mid x)\geq0$, marginal survival cannot be below $\pi(x)$. Recheck the inputs, computation, or model output.