Parametric Latency Distributions for Cure Models

Latency Distributions Within a Cure Model

A mixture cure model separates two ideas: whether an individual is susceptible to the event during the period represented by the model, and how long until the event occurs for a susceptible individual. The first is the cure component; the second is the latency distribution. The latency distribution is conditional on susceptibility, not a distribution of observed event times for the whole population.

This distinction matters when choosing a parametric form. A distribution with a heavy tail can imply that susceptible individuals may experience the event after a long wait. The cure fraction, by contrast, represents individuals who never experience the event under the model. A long period without an event can therefore be attributed either to cure or to long latency, and the previous material on identifiability and follow-up explains why those explanations can be difficult to distinguish in finite data.

Definition: Let $c$ be the cure fraction, and let $T^*$ be the event time conditional on being susceptible. Write $S_u(t)=P(T^*>t)$, $f_u(t)$ for its density, and $h_u(t)=f_u(t)/S_u(t)$ for its hazard. The population survival function is $S_{\mathrm{pop}}(t)=c+(1-c)S_u(t)$. The cured individuals have no event time during the modeled horizon; equivalently, their event time is treated as infinite. The cure fraction satisfies $0\leq c<1$ when a susceptible latency distribution is to be learned from event times.

The formulas below assume positive event times measured in a fixed unit, a properly specified continuous distribution for susceptible event times, and a time-constant cure status. For the population survival interpretation, the cure component must represent individuals who will not experience the event, rather than merely individuals who have not experienced it yet. When interpreting observed data, the censoring assumptions used in the cure model must also be appropriate; the latency formulas alone do not make censoring independent.

Three Parametric Choices

Each candidate specifies the distribution of $T^*$. Its parameters describe latency among susceptible individuals, not the cure fraction. In the log-normal model, time is expressed in a chosen unit before taking its logarithm, so the interpretation of $\mu$ depends on that unit.

Definition: In a Weibull latency model with shape $k>0$ and scale $\lambda>0$, the survivor function and hazard are $$ S_u(t)=\exp\left[-(t/\lambda)^k\right], \qquad h_u(t)=\frac{k}{\lambda}(t/\lambda)^{k-1}, \qquad t>0. $$ Here $\lambda$ has units of time, and $k$ determines the hazard pattern: it decreases if $k<1$, is constant if $k=1$, and increases if $k>1$.

Definition: In a log-normal latency model, $\log T^*\sim N(\mu,\sigma^2)$, where $\sigma>0$. If $\phi$ and $\Phi$ are the standard normal density and distribution functions, then $$ S_u(t)=1-\Phi\left(\frac{\log t-\mu}{\sigma}\right), \qquad f_u(t)=\frac{1}{t\sigma}\phi\left(\frac{\log t-\mu}{\sigma}\right), \qquad h_u(t)=\frac{f_u(t)}{S_u(t)}. $$ The log-normal hazard has no simple closed-form expression and eventually decreases toward zero.

Definition: In a log-logistic latency model with scale $\alpha>0$ and shape $\beta>0$, $$ S_u(t)=\frac{1}{1+(t/\alpha)^\beta}, \qquad h_u(t)=\frac{\beta}{\alpha}\frac{(t/\alpha)^{\beta-1}}{1+(t/\alpha)^\beta}, \qquad t>0. $$ The scale $\alpha$ has units of time; $S_u(\alpha)=1/2$. The shape $\beta$ controls the hazard’s pattern and tail.

For the log-normal model, $\exp(\mu)$ is the median latency among susceptible individuals: at $t=\exp(\mu)$, the standardized log time is zero and $S_u(t)=1/2$. In the log-logistic model, $\alpha$ is also the susceptible median. The Weibull scale $\lambda$ has a different interpretation: $S_u(\lambda)=\exp(-1)$, regardless of $k$.

From Susceptible Survival to Population Survival

The cure-model combination follows directly by conditioning on susceptibility. With probability $c$, an individual is cured and survives beyond every finite $t$ under the model. With probability $1-c$, the individual is susceptible and survives beyond $t$ with probability $S_u(t)$. Therefore,

S_{\mathrm{pop}}(t) =P(T>t) =c\cdot 1+(1-c)S_u(t) =c+(1-c)S_u(t).

Differentiating the population survival at finite $t$ gives $f_{\mathrm{pop}}(t)=(1-c)f_u(t)$. Dividing this density by population survival yields the population hazard:

h_{\mathrm{pop}}(t) =\frac{(1-c)f_u(t)}{c+(1-c)S_u(t)} =\frac{(1-c)S_u(t)h_u(t)}{c+(1-c)S_u(t)}.

If the latency distribution is proper, $S_u(t)\to0$ as $t\to\infty$, so $S_{\mathrm{pop}}(t)\to c$. If $c>0$, then for each of the three specified latency families, whose densities tend to zero, the population hazard tends to zero as the surviving population becomes increasingly enriched with cured individuals. This does not mean the susceptible hazard must tend to zero: for a Weibull distribution with $k>1$, for example, the susceptible hazard increases without bound, even though the population hazard tends to zero when $c>0$.

Worked Example: Weibull Latency and a Nonzero Cure Fraction

Suppose, hypothetically, that the cure fraction is $c=0.25$, and susceptible event times follow a Weibull distribution with $k=0.7$ and $\lambda=8$ months. Find susceptible and population survival at 16 months, and calculate the susceptible and population hazards at that time.

The assumptions for these calculations are that the stated Weibull distribution applies to susceptible individuals, the cure fraction is constant, and time is measured in months. First, $16/8=2$, so $$ S_u(16)=\exp(-2^{0.7})=\exp(-1.6245)\approx0.1970. $$ The population survival is $$ S_{\mathrm{pop}}(16)=0.25+0.75(0.1970)\approx0.3978. $$ Thus, the model assigns about 39.8% population survival beyond 16 months, including the 25% cured fraction.

The susceptible hazard is $$ h_u(16)=\frac{0.7}{8}(16/8)^{-0.3} =(0.0875)(2^{-0.3}) =(0.0875)(0.8122524) \approx0.07107 $$ per month. The population hazard is $$ h_{\mathrm{pop}}(16)= \frac{(0.75)(0.1970)(0.07107)}{0.3978} \approx0.02640 $$ per month. These are instantaneous rates at 16 months, not probabilities of an event during the entire month.

Because $k=0.7<1$, the susceptible Weibull hazard decreases with time. The population hazard is lower at 16 months because it is averaged over the surviving population, which includes cured individuals.

Worked Example: Log-Normal Latency at Its Median

In a hypothetical model, let $\log T^*\sim N(\log 12,0.6^2)$, with time in months, and let $c=0.20$. Calculate susceptible survival, the susceptible hazard, and population survival and hazard at 12 months.

Since $t=\exp(\mu)=12$ months, the standardized log time is $(\log 12-\log 12)/0.6=0$. Therefore $S_u(12)=1-\Phi(0)=0.5$. The susceptible density and hazard are $$ f_u(12)=\frac{\phi(0)}{12(0.6)} =\frac{0.398942}{7.2} \approx0.05541\ \text{per month}, \qquad h_u(12)=\frac{0.05541}{0.5} \approx0.1108\ \text{per month}. $$

The population survival at 12 months is $$ S_{\mathrm{pop}}(12)=0.20+0.80(0.5)=0.60. $$ The population hazard is $$ h_{\mathrm{pop}}(12)=\frac{0.80(0.05541)}{0.60} \approx0.07388\ \text{per month}. $$ The median latency of 12 months refers only to susceptible individuals; it does not say that half the entire population experiences the event by 12 months.

Worked Example: Log-Logistic Latency and Its Tail

Consider a hypothetical log-logistic latency distribution with $\alpha=10$ months and $\beta=2$, and a cure fraction $c=0.30$. Find susceptible and population survival at 20 months and the susceptible hazard there.

Here $20/10=2$. The susceptible survival is $$ S_u(20)=\frac{1}{1+2^2}=\frac{1}{5}=0.20. $$ The population survival is $$ S_{\mathrm{pop}}(20)=0.30+0.70(0.20)=0.44. $$ The susceptible hazard is $$ h_u(20)=\frac{2}{10}\frac{2^{2-1}}{1+2^2} =0.2\frac{2}{5} =0.0800 $$ per month. The model therefore predicts 44% population survival beyond 20 months, of which 30 percentage points are the cured fraction and 14 percentage points are susceptible individuals whose event time exceeds 20 months.

Tail Behavior and Model Implications

All three latency distributions have $S_u(t)\to0$ as $t\to\infty$, so each produces a population survival plateau at $c$ when combined with a cure fraction. They differ in how quickly susceptible survival approaches zero. For fixed positive Weibull shape $k$, its tail decreases exponentially in $t^k$. Log-normal survival decreases more slowly than any such Weibull tail, but faster than any power tail. The log-logistic tail decreases as a power of time, and is therefore the heaviest of these three tails.

For the log-logistic distribution, when $t$ is large relative to $\alpha$, $$ S_u(t)=\frac{1}{1+(t/\alpha)^\beta} \sim(\alpha/t)^\beta. $$ Thus, some susceptible individuals can have very long latencies. Its mean latency is finite only when $\beta>1$; for $\beta\leq1$, the model has an infinite mean. The mean is finite when $\beta>1$, in which case it equals $\alpha(\pi/\beta)/\sin(\pi/\beta)$, where $\pi$ in this expression is the mathematical constant. This tail behavior can be important when interpreting long-term averages or extrapolating far beyond observed follow-up.

Latency distribution	Susceptible hazard pattern	Tail implication
Weibull	Decreasing, constant, or increasing according as $k<1$, $k=1$, or $k>1$	Exponential-type tail in $t^k$
Log-normal	Eventually decreases toward zero	Heavier than Weibull tails, lighter than any power tail
Log-logistic	Decreases eventually; can rise and then fall if $\beta>1$	Power tail, with $S_u(t)\sim(\alpha/t)^\beta$

These comparisons concern the susceptible latency distribution. For any of them, adding a positive cure fraction creates a population survival plateau and makes the population hazard tend to zero. A visually flattening population survival curve alone therefore cannot tell whether a chosen latency tail is appropriate or whether the data provide strong evidence for a cure fraction.

Specification Checks and Common Pitfalls

A parametric latency model is a substantive assumption about event times among susceptible individuals. It is not simply a convenient curve-fitting choice. Before interpreting its predictions, check that its parameterization is understood, that the implied hazard pattern is plausible for the application, and that its tail is not driving conclusions far beyond the observed follow-up. When the observed event times show patterns inconsistent with the model’s implied hazard, or when small changes in tail assumptions produce large changes in long-term survival estimates, the specification deserves scrutiny. With limited follow-up, the data may not distinguish a long latency tail from a larger cure fraction.

A useful discipline is to examine the susceptible distribution and the population mixture separately: plot or calculate $S_u(t)$ and $S_{\mathrm{pop}}(t)$ over the times that matter, and verify that the latter approaches $c$. Compare the implied hazard over that same range. A model’s fitted behavior over observed times does not by itself validate extrapolation into unobserved years.

Confusing cure with long latency. A person who has not had the event by the end of follow-up is not thereby known to be cured; the person may be susceptible with a long event time.
Applying the latency distribution to everyone. $S_u(t)$ is conditional on susceptibility. Population survival is $c+(1-c)S_u(t)$, not $S_u(t)$ alone.
Reading parameters across models as interchangeable. The Weibull scale is the time at which susceptible survival is $\exp(-1)$; the log-normal and log-logistic scales above are medians. Equal numerical values do not give the same distribution.
Assuming all tail assumptions are equally mild. The log-logistic power tail can imply a very long latency and even an infinite mean, while the Weibull tail decreases much faster.
Interpreting the population hazard as the susceptible hazard. They differ whenever $c>0$; the population hazard is diluted by cured individuals among those still event-free.

The main alternative is not one particular parametric distribution but a less restrictive latency specification. A more flexible form can reduce dependence on a specific parametric tail, but it does not remove the need to distinguish cure from long latency or solve limited-follow-up identifiability. Parametric forms can be useful for concise interpretation and extrapolation, provided those extrapolations are presented as consequences of the assumed distribution rather than as direct observations.

Key takeaway: Choose a latency distribution for what it says about susceptible event times, and interpret it separately from the cure fraction. The Weibull, log-normal, and log-logistic forms can produce the same population survival plateau but imply markedly different susceptible tails.

Check Your Understanding

Use the definitions above. Each answer includes a concise solution.

A Weibull model has $k=1$. What is its susceptible hazard, and does it increase or decrease with time?
Solution: Substituting $k=1$ gives $h_u(t)=1/\lambda$, a constant hazard of $1/\lambda$ per unit time.
A log-normal latency model has $\mu=\log 20$. What is the susceptible median event time?
Solution: The median is $\exp(\mu)=\exp(\log 20)=20$ units of the chosen time scale. This is conditional on susceptibility.
At a given time, $c=0.40$ and $S_u(t)=0.25$. Find population survival.
Solution: $S_{\mathrm{pop}}(t)=0.40+(1-0.40)(0.25)=0.40+0.15=0.55$, or 55%.
Which of the three specified latency distributions has a power tail?
Solution: The log-logistic distribution. For large $t$, $S_u(t)\sim(\alpha/t)^\beta$.
Why is a zero or small estimated population hazard at late times not, by itself, evidence that the susceptible hazard is zero?
Solution: When $c>0$, cured individuals remain in the population survival denominator. The population hazard is $(1-c)S_u(t)h_u(t)/S_{\mathrm{pop}}(t)$ and, for the three specified latency distributions, tends to zero because their densities tend to zero, even if the susceptible hazard does not tend to zero.

Latency distribution	Susceptible hazard pattern	Tail implication
Weibull	Decreasing, constant, or increasing according as \(k<1\), \(k=1\), or \(k>1\)	Exponential-type tail in \(t^k\)
Log-normal	Eventually decreases toward zero	Heavier than Weibull tails, lighter than any power tail
Log-logistic	Decreases eventually; can rise and then fall if \(\beta>1\)	Power tail, with \(S_u(t)\sim(\alpha/t)^\beta\)