From Survival Predictions to Event-Time Timing
A survival prediction answers how likely a patient is to remain event-free beyond a specified time. A conditional event-time density answers a different question: among patients with this observed history who have remained event-free to the landmark, how is the probability of an event distributed over later times? The density describes the relative concentration of predicted event timing. It is not itself a probability at one exact time.
Let \(T_i\) be patient \(i\)'s event time, let \(t_L\) be the landmark, and let \(H_i(t_L)\) denote the patient's observed longitudinal history through that landmark. We consider a patient known to be event-free at \(t_L\). The target is the distribution of \(T_i\) conditional on both \(H_i(t_L)\) and \(T_i>t_L\). All time units must be consistent: if time is measured in months, a density has units of inverse months.
The density is useful when the timing distribution matters, not only the risk by a horizon. For example, two predictions can have the same probability of an event within a year while placing that probability at different points in the year. A density curve makes that difference visible.
Conditioning on History and Landmark Survival
Write \(b_i\) for the patient's latent random effects and \(x_i\) for observed covariates. Conditional on \(b_i\) and the covariates, let \(h_i(t\mid b_i,x_i)\) be the event hazard and \(S_i(t\mid b_i,x_i)\) the event-free survival function from the model's time origin. For a continuous event-time model, \(S_i(t\mid b_i,x_i)=\exp\{-\int_0^t h_i(u\mid b_i,x_i)\,du\}\). The joint model supplies the hazard, including whatever longitudinal association structure has been specified.
First use the observed longitudinal history to obtain a distribution for the latent effects, denoted \(p(b_i\mid H_i(t_L),x_i)\). This distribution reflects uncertainty about the patient's latent trajectory after observing their measurements; it is generally not a point estimate. Survival to the landmark supplies additional information: latent-effect values associated with greater earlier hazard are less compatible with having remained event-free.
This is the survival conditioning developed in the earlier material on conditioning on survival to the landmark: the survival likelihood reweights the latent-risk distribution. Once the weights have been updated, the conditional survival beyond the landmark is \(S_i(t\mid H_i(t_L),T_i>t_L,x_i)=\int [S_i(t\mid b_i,x_i)/S_i(t_L\mid b_i,x_i)]p_L(b_i)\,db_i\), for \(t\ge t_L\). The ratio is the probability of surviving from \(t_L\) to \(t\) conditional on \(b_i\).
Differentiating this conditional survival function gives the event-time density. The density is a weighted average of each latent-effect-specific density from the landmark onward, using the landmark-updated weights.
An equivalent expression is often convenient if the weights before survival conditioning are available:
The denominator is the predicted probability of surviving to \(t_L\), conditional on the observed history and covariates under the pre-survival latent-effect distribution. It normalizes the numerator so that the result conditions on survival rather than treating survival to the landmark as another future event.
Why the Density Formula Holds
For a fixed value of \(b_i\), the probability of surviving from the landmark to \(t\) is \(S_i(t\mid b_i,x_i)/S_i(t_L\mid b_i,x_i)\). Its derivative with respect to \(t\) is the negative of the latent-effect-specific density after the landmark. Since \(dS_i(t\mid b_i,x_i)/dt=-h_i(t\mid b_i,x_i)S_i(t\mid b_i,x_i)\), the density conditional on \(b_i\) is \(h_i(t\mid b_i,x_i)S_i(t\mid b_i,x_i)/S_i(t_L\mid b_i,x_i)\). Averaging that density over \(p_L(b_i)\) gives the displayed formula.
This derivation also explains why simply averaging the hazard over the random-effects distribution is not enough. The marginal density averages hazard multiplied by survival, because a patient must still be event-free at \(t\) to contribute event density there. And the relevant distribution of \(b_i\) is the one updated by survival through the landmark.
Worked Calculations
Worked Example: Survival Reweights Two Latent-Risk Groups
Suppose the posterior distribution based on a patient's biomarker history assigns probability 0.60 to a lower-risk latent group with constant hazard \(0.03\) per month and probability 0.40 to a higher-risk group with constant hazard \(0.10\) per month. The patient has survived to month 8. Find the updated group probabilities.
For an exponential event time with rate \(\lambda\), survival to month 8 is \(e^{-8\lambda}\). The unnormalized weights are therefore \(0.60e^{-0.03(8)}=0.60e^{-0.24}=0.4719767\) and \(0.40e^{-0.10(8)}=0.40e^{-0.8}=0.1797316\). Their sum is \(0.6517083\).
Normalize each weight: \(0.4719767/0.6517083=0.7242\), and \(0.1797316/0.6517083=0.2758\), to four significant figures. Thus, conditional on the observed history and survival to month 8, the lower-risk group has updated weight about 0.7242, and the higher-risk group about 0.2758. Survival has shifted the weights toward the lower-risk group; it has not established which latent group the patient belongs to.
Worked Example: Density Six Months After a Landmark
At a landmark, suppose the survival-updated weights for two latent groups are 0.70 and 0.30. Their constant hazards after the landmark are \(0.04\) and \(0.12\) per month, respectively. Calculate the conditional event-time density six months after the landmark.
For each group, the density \(u\) months after the landmark is \(\lambda e^{-\lambda u}\), where \(u=t-t_L\). The mixture density at \(u=6\) is \(0.70(0.04e^{-0.04(6)})+0.30(0.12e^{-0.12(6)})\).
The first component is \(0.04e^{-0.24}\approx0.0314651\) per month, so its weighted contribution is \(0.70(0.0314651)\approx0.0220256\) per month. The second is \(0.12e^{-0.72}\approx0.0584103\) per month, contributing \(0.30(0.0584103)\approx0.0175231\) per month. Therefore the conditional density is approximately \(0.0395487\) per month.
This value describes the density at that time, not a 3.95% event probability. For instance, over a sufficiently short interval of width 0.1 month around that time, the probability is approximately \(0.0395487(0.1)=0.003955\), or about 0.396%, if the density is nearly constant across that interval.
Worked Example: A Single Exponential Model
Suppose a patient's event hazard is constant at \(0.05\) per month, and the patient has survived to month 10. Find the conditional density at month 14 and the conditional probability of surviving from month 10 through month 14.
With one known rate, survival from the landmark to month 14 depends on the four-month elapsed time \(u=14-10=4\). The density is \(0.05e^{-0.05(4)}=0.05e^{-0.2}\approx0.04094\) per month.
The conditional probability of surviving those four months is \(e^{-0.05(4)}=e^{-0.2}\approx0.8187\). These quantities are not interchangeable: the first is a density at month 14; the second is a probability over the full four-month interval. In this special constant-hazard model, the conditional density has the same exponential form after any landmark, but that simplification need not hold for a time-varying hazard or a mixture of latent effects.
Assumptions, Diagnostics, and Alternatives
The formulas are conditional on a correctly specified joint model and its fitted parameters. In particular, the model must provide a valid event hazard and survival function, and its assumptions connecting the longitudinal history and event process through latent effects must be appropriate. Conditional independence of the longitudinal measurements and event process given the random effects and covariates is a common joint-model assumption; if it is implausible, the posterior weights and predictions may be biased. Censoring must also be handled under the model's censoring assumptions. The formulas above describe the density of finite event times: if survival has a positive limit, this density integrates to less than 1, with the remaining probability representing no finite event, as can occur for a cured group with zero hazard. A cure model may be needed to specify that mass or the cure mechanism; competing event types require cause-specific or competing-risk quantities.
In computation, evaluate the density on a time grid using the same integration or approximation for latent effects that is used for survival prediction. Check that values are nonnegative, that the density agrees with the negative numerical derivative of the conditional survival curve, and that integrating the density from \(t_L\) to later times agrees with the corresponding decline in conditional survival. If the event is eventually certain, the density integrates to 1 over the future; if a positive probability of never experiencing the event is allowed, its integral is less than 1 and the remaining mass is the probability of no event.
A coarse time grid can miss a narrow peak, while unstable or inaccurate integration over random effects can produce a jagged curve or a mismatch between density and survival. Recheck the grid resolution and integration accuracy, and compare density-derived interval probabilities with direct differences of survival predictions. The earlier material on computing window risk gives the identity that the probability of an event between \(a\) and \(b\), where \(t_L\le a<b\), is \(S_i(a\mid H_i,T_i>t_L)-S_i(b\mid H_i,T_i>t_L)\). This interval probability is often more stable and easier to communicate than a single density value.
The main alternatives answer different questions. A survival curve gives the probability of remaining event-free through each time; a conditional hazard gives the instantaneous event rate among those still event-free at that time; and a window risk gives the probability of an event over a stated interval. The density describes how event probability is distributed locally over future time. None is universally preferable: use the density when relative timing is the target, and report interval risks when a probability over a clinically meaningful window is needed.
Common Pitfall and Limitations
- Wrong reasoning: “The density is 0.04 per month, so the probability of an event at that exact time is 4%.” A continuous event-time distribution assigns probability zero to any exact time; the density is probability per unit time.
- Correction: Multiply by a short interval width for an approximate probability, or integrate the density across the interval. For an exact interval probability, use the difference in conditional survival at its endpoints.
- Wrong reasoning: Average the latent-effect-specific hazards using weights based only on biomarker history, even though the patient survived to the landmark.
- Correction: Survival to the landmark changes the latent-effect weights. Update them by the survival likelihood before averaging future event-time densities.
A density peak is also not a guarantee that an event will occur near that time. It is conditional on the fitted model, the observed history, and survival to the landmark; it does not account for model misspecification or unrepresented sources of uncertainty. In particular, treating fitted parameters as fixed yields a conditional prediction that does not include parameter-estimation uncertainty.
Check Your Understanding
Try each question, then compare with the solutions below.
- Why is a conditional event-time density value not the probability of an event at that exact time?
- A latent group has history-based weight 0.50 and survival probability 0.80 to the landmark. A second group has weight 0.50 and survival probability 0.40. What are the survival-updated weights?
- If conditional survival is 0.90 at month 5 after the landmark and 0.75 at month 7, what is the probability of an event between those months?
- For a continuous model, how can a computed density curve be checked against its conditional survival curve?
Solutions. (1) A density has units of inverse time; probability over an interval is approximately density times interval width for a short interval, and probability at one exact time is zero. (2) The unnormalized weights are \(0.50(0.80)=0.40\) and \(0.50(0.40)=0.20\), with sum 0.60. The updated weights are \(0.40/0.60=0.6667\) and \(0.20/0.60=0.3333\). (3) The event probability is \(0.90-0.75=0.15\), or 15%. (4) Compare the density with the negative derivative of conditional survival, and check that its integral over an interval equals the survival decline across that interval.