Event-Time Density Conditional on Random Effects

From a Conditional Hazard to an Event-Time Density

In a joint model, the event submodel describes how a subject’s event risk depends on the subject’s latent trajectory and shared random effects. Once those random effects are held fixed, the event-time model is subject-specific. For an observed event at time $t_i$, the likelihood needs the density of that event time conditional on the subject’s random effects—not merely the hazard at that instant.

The distinction matters because a hazard is an instantaneous rate among subjects who have remained event-free up to a time, whereas a density accounts for both the instantaneous rate at the event time and the probability of surviving event-free until then. This tutorial derives that relationship and uses the resulting density as the likelihood contribution for an observed event.

Definition: Let $T_i$ be subject $i$'s event time and $b_i$ that subject’s random effects. Conditional on $b_i$ and any covariates or latent trajectory specified by the event submodel, write $S_i(t\mid b_i)=P(T_i>t\mid b_i)$ for the survival function and $\lambda_i(t\mid b_i)$ for the hazard. If the conditional event-time distribution is absolutely continuous, its density is $$ f_i(t\mid b_i)=\lambda_i(t\mid b_i)S_i(t\mid b_i). $$

When the event hazard depends on the latent biomarker trajectory, that trajectory is treated as fixed once the random effects and covariates are given. To keep notation readable, such dependence is suppressed in $f_i(t\mid b_i)$, $S_i(t\mid b_i)$, and $\lambda_i(t\mid b_i)$. The formula applies at times in the subject’s risk period, under the time origin and risk set defined by the model.

Why the Density Is Hazard Times Survival

The hazard at time $t$ describes the event rate at $t$ conditional on surviving just before $t$. In a small interval of width $\Delta t$, the conditional probability of an event is approximately $\lambda_i(t\mid b_i)\Delta t$, when the hazard is locally regular. To have an event in that interval, the subject must also have survived to its start; that probability is approximately $S_i(t\mid b_i)$. Their product gives the probability of an event in the interval to first order, and dividing by its width yields the density.

A direct derivation follows from the definition of hazard. For an absolutely continuous event-time distribution, $\lambda_i(t\mid b_i)=-\frac{d}{dt}\log S_i(t\mid b_i)$. Therefore, $\frac{d}{dt}S_i(t\mid b_i)=-\lambda_i(t\mid b_i)S_i(t\mid b_i)$. The conditional density is the negative derivative of survival, so $f_i(t\mid b_i)=-\frac{d}{dt}S_i(t\mid b_i)$, which gives the stated formula.

\begin{aligned} S_i(t\mid b_i) &=\exp\left\{-\int_0^t\lambda_i(u\mid b_i)\,du\right\},\\ f_i(t\mid b_i) &=-\frac{d}{dt}S_i(t\mid b_i) =\lambda_i(t\mid b_i) \exp\left\{-\int_0^t\lambda_i(u\mid b_i)\,du\right\}. \end{aligned}

The lower limit $0$ represents the chosen time origin; if the model uses another origin, the cumulative hazard must be integrated from the beginning of that subject’s risk period. The integrand in the survival expression is the hazard at every time from the origin through $t$, not just the hazard evaluated at $t$. The density has units of inverse time: a hazard has units of inverse time, while survival is dimensionless.

The assumptions behind this derivation are that the conditional hazard is nonnegative and integrable over finite intervals, the conditional event-time distribution is absolutely continuous over the relevant times, and survival is differentiable in the required sense. These conditions rule out point masses at finite event times. If there is a probability of never experiencing the event, the density over finite times may integrate to less than one; the remaining probability is a mass at infinity, not a missing piece of the finite-time density.

Worked Example: Constant Hazard With a Random-Effects Value

Suppose a hypothetical subject’s hazard, conditional on the random effect $b_i$, is constant in time and has the form $\lambda_i(t\mid b_i)=0.10\exp(0.4b_i)$ per year. For $b_i=1$, find the density of an event at $t=4$ years.

First compute the conditional hazard: $\lambda_i=0.10\exp(0.4)=0.1491825$ per year. Because it is constant, the cumulative hazard through year 4 is $4\lambda_i=0.5967299$, and the conditional survival probability is $S_i(4\mid 1)=\exp(-0.5967299)\approx0.550609$.

The event-time density is the hazard at year 4 times survival through year 4:

\begin{aligned} f_i(4\mid 1) &=0.10\exp(0.4)\exp\{-4(0.10\exp(0.4))\}\\ &\approx (0.1491825)(0.550609)\\ &\approx 0.0821412\ \text{per year}. \end{aligned}

The result is a density, not a probability of exactly 0.0821412 that the event occurs at year 4. For example, over a sufficiently short interval of width 0.01 year near year 4, the event probability is approximately $0.0821412(0.01)=0.0008214$, with the approximation improving as the interval becomes narrower.

The Observed-Event Likelihood Contribution

If subject $i$ is observed to experience the event at time $t_i$, the contribution of that observed event time to the likelihood, conditional on the random effects, is its density $f_i(t_i\mid b_i)$. This is a density contribution because $T_i$ is continuous. For conditionally independent subjects, the likelihood for a set of observed event times is the product of their individual densities, conditional on their random effects.

L_{\text{event}\mid b} =\prod_{i:\,\text{event observed}} f_i(t_i\mid b_i) =\prod_{i:\,\text{event observed}} \lambda_i(t_i\mid b_i)S_i(t_i\mid b_i).

This expression identifies only the contribution for an observed event. It is conditional on the random effects; it is not yet a likelihood integrated over their distribution. The likelihood is a function of model parameters, with the observed event times held fixed. A density can exceed one for some time units or parameter values; that does not make it a probability greater than one. Changing the time unit changes the numerical value of a density while leaving probabilities over corresponding intervals unchanged.

Worked Example: Two Observed Events Conditional on Random Effects

Consider two hypothetical, conditionally independent subjects with constant hazards. At their specified random-effects values, suppose subject A has hazard $0.10$ per year and an observed event time of 2 years, while subject B has hazard $0.20$ per year and an observed event time of 1 year. Compute their conditional event contribution to the likelihood.

For subject A, the survival probability to 2 years is $\exp(-0.10\times2)=\exp(-0.2)$. Thus $f_A(2\mid b_A)=0.10\exp(-0.2)\approx0.0818731$ per year.

For subject B, the survival probability to 1 year is also $\exp(-0.20\times1)=\exp(-0.2)$. Thus $f_B(1\mid b_B)=0.20\exp(-0.2)\approx0.1637462$ per year.

Conditional independence makes the joint event contribution the product:

\begin{aligned} L_{\text{event}\mid b} &=f_A(2\mid b_A)f_B(1\mid b_B)\\ &\approx (0.0818731)(0.1637462)\\ &\approx 0.0134064\ \text{per year}^2. \end{aligned}

The units are inverse years squared because this is a product of two time densities. The calculation is conditional on the stated random-effects values and hazards; it does not average over uncertainty in the random effects.

Time-Varying Hazards and Common Checks

With a time-varying hazard, it is especially important not to replace the cumulative hazard in the survival function by the hazard at the event time multiplied by time. The hazard at $t$ determines the local event rate; the integrated hazard up to $t$ determines the probability of having remained event-free until then.

Worked Example: A Time-Varying Hazard

For a hypothetical subject with $b_i=0$, let the hazard be $\lambda_i(t\mid b_i)=0.05(1+0.5t)$ per year, where $t$ is measured in years. Find the density at year 2.

The instantaneous hazard at year 2 is $\lambda_i(2\mid 0)=0.05(1+0.5\times2)=0.10$ per year. The cumulative hazard, however, integrates the changing rate over the entire two years:

\begin{aligned} \int_0^2 \lambda_i(u\mid 0)\,du &=\int_0^2 0.05(1+0.5u)\,du\\ &=0.05(2+0.25\times 2^2)=0.15. \end{aligned}

Therefore, survival through year 2 is $\exp(-0.15)\approx0.860708$, and the event-time density is $f_i(2\mid0)=0.10\exp(-0.15)\approx0.0860708$ per year. Using the instantaneous hazard $0.10$ as though it were constant over the whole period would instead give survival $\exp(-0.20)$, which is not correct for this time-varying hazard.

Several checks help detect mistakes in deriving or evaluating the density:

Check the time units. If time is in years, both the hazard and density are measured per year; survival is dimensionless.
Check the two time roles. The hazard multiplier is evaluated at $t_i$, while survival uses the integral of the hazard over the full interval from the time origin to $t_i$.
Check the sign. A nonnegative hazard gives a nonnegative cumulative hazard, so the exponent in survival is nonpositive and $0\leq S_i(t\mid b_i)\leq1$.
Check normalization. For a proper finite event-time distribution, the density integrates to one over its support. If survival approaches a positive limit, the finite-time density integrates to less than one and the remaining probability must be accounted for as a possible never-event outcome.

The hazard and density are related but answer different questions. A hazard is a rate conditional on remaining event-free; a density describes the unconditional timing distribution conditional on the random effects. The cumulative hazard is an integrated rate, and exponentiating its negative yields survival. One may specify an event-time distribution directly through a density, but in a joint model the hazard representation is often useful because it expresses event risk as a function of time and the latent trajectory. Regardless of the representation, the density used for an observed event must be compatible with the implied survival distribution.

Common Pitfalls and Scope

A frequent error is to use only $\lambda_i(t_i\mid b_i)$ as the likelihood contribution for an observed event. That omits the probability of reaching $t_i$ without an earlier event. The correction is to use $\lambda_i(t_i\mid b_i)S_i(t_i\mid b_i)$. A related error is to describe $f_i(t_i\mid b_i)$ as the probability of an event at exactly $t_i$. For a continuous event time, the probability at one exact time is zero; the density instead approximates probability per unit time near that time.

This conditional event contribution does not by itself complete the joint-model likelihood. It leaves random-effects integration, censoring contributions, and the longitudinal-data contribution outside its scope. It also relies on the conditional event-time distribution being represented by a hazard with the stated regularity; if events can occur with positive probability at discrete times, the distribution may include point masses and a density-only expression is not sufficient.

Key takeaway: For an observed continuous event at $t_i$, conditional on random effects, the event likelihood contribution is $f_i(t_i\mid b_i)=\lambda_i(t_i\mid b_i)S_i(t_i\mid b_i)$. The hazard is evaluated at the event time, and survival accounts for the entire preceding risk period.

Check Your Understanding

Work through each question, then compare your reasoning with the solutions below.

A hypothetical subject has a constant conditional hazard of $0.25$ per month. Write the event-time density at month 3.
In words, why does an event-time density include survival through the event time as well as the hazard at that time?
A hazard varies over time. Which part of the density uses the hazard evaluated at the event time, and which part uses its history from the time origin?
What are the units of the product of three event-time densities when time is recorded in days?
Can a density value greater than one be valid? Explain.

Solutions. (1) The survival probability is $\exp(-0.25\times3)=\exp(-0.75)$, so $f(3)=0.25\exp(-0.75)\approx0.1181$ per month. (2) An event at that time requires both remaining event-free beforehand and experiencing the event locally at that time; survival represents the first condition and the hazard the second. (3) The hazard factor is $\lambda_i(t_i\mid b_i)$; survival uses $\exp\{-\int_0^{t_i}\lambda_i(u\mid b_i)\,du\}$, which accumulates the hazard history. (4) Each density has units of inverse days, so the product has units of inverse days cubed. (5) Yes. A density is probability per unit time, not a probability at one point; probabilities are obtained by integrating it over intervals.