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The Basic Joint Model · Tutorial 164 of 697

Event Probability Over a Follow-Up Interval

Learn to obtain interval event probabilities from conditional survival, account for the interval’s starting point, and check calculations against the hazard model.

Advanced 11 min read

What You'll Learn

  • Define conditional survival and cumulative hazard given a subject’s random effects.
  • Calculate the probability of an event between two follow-up times.
  • Distinguish an interval probability from the probability conditional on surviving to the interval start.
  • Apply the calculation to constant and piecewise-constant conditional hazards.
  • Check interval probabilities for time-ordering, units, and consistency with survival.
  • Recognize when conditioning on random effects is not the same as averaging over them.

From Conditional Survival to an Interval Probability

A hazard describes the instantaneous event rate among subjects who have remained event-free up to a given time. It is not itself the probability of an event over a follow-up interval. To answer an interval question, first obtain survival probabilities at the interval’s endpoints, conditional on the subject’s random effects, and then take their difference.

This distinction matters in a shared random-effects model because the event hazard, and therefore survival, depends on the subject’s random effects. Holding those effects fixed gives a subject-specific probability. It does not give a population-average probability over the distribution of random effects.

Definition: Let \(T_i\) be subject \(i\)’s event time and \(b_i\) the subject’s random effects. Write \(\lambda_i(t\mid b_i)\) for the event hazard at time \(t\), conditional on \(b_i\) and any included covariates. The conditional cumulative hazard and conditional survival function are $$ H_i(t\mid b_i)=\int_0^t \lambda_i(u\mid b_i)\,du,\qquad S_i(t\mid b_i)=P(T_i>t\mid b_i)=\exp\{-H_i(t\mid b_i)\}. $$ For \(0\leq a<b\), the probability of an event in \((a,b]\), conditional on \(b_i\), is \(S_i(a\mid b_i)-S_i(b\mid b_i)\). If the probability is instead conditional on the subject having survived to \(a\), it is \(1-S_i(b\mid b_i)/S_i(a\mid b_i)\), provided \(S_i(a\mid b_i)>0\).

These are two different probability statements. The first is conditional on the random effects alone; it includes both subjects who fail before \(a\) (which are not counted in the interval) and subjects who survive to \(a\). The second additionally conditions on being event-free at \(a\). Make that distinction explicit whenever reporting an interval probability.

Why the Difference of Survival Probabilities Works

Suppose the conditional event-time distribution is continuous, as it is under a finite, integrable hazard model without point masses. The event \(\{a<T_i\leq b\}\) is the part of \(\{T_i>a\}\) that is removed by time \(b\). Consequently,

$$ P(a<T_i\leq b\mid b_i) =P(T_i>a\mid b_i)-P(T_i>b\mid b_i) =S_i(a\mid b_i)-S_i(b\mid b_i). $$

The same result follows by integrating the conditional event-time density. For a continuous hazard, that density is \(f_i(t\mid b_i)=\lambda_i(t\mid b_i)S_i(t\mid b_i)\). Since \(dS_i(t\mid b_i)/dt=-f_i(t\mid b_i)\), integration gives

$$ \int_a^b \lambda_i(t\mid b_i)S_i(t\mid b_i)\,dt =\int_a^b f_i(t\mid b_i)\,dt =S_i(a\mid b_i)-S_i(b\mid b_i). $$

For the probability conditional on surviving to \(a\), divide the first interval probability by \(P(T_i>a\mid b_i)=S_i(a\mid b_i)\). This gives \(P(a<T_i\leq b\mid T_i>a,b_i)=1-S_i(b\mid b_i)/S_i(a\mid b_i)\). This conditional probability is often useful for a follow-up question that starts at a landmark time, but it is not interchangeable with the probability from time zero.

The calculation assumes the conditional hazard specifies the event process over the entire interval, and that its cumulative hazard is well-defined. Random effects are held fixed throughout. If the observed data are censored, censoring affects what is observed about \(T_i\), not these probability identities; estimating the model from data requires additional assumptions about the observation and censoring processes.

Worked Example: Constant Conditional Hazard Over an Interval

Suppose a hypothetical subject has a constant conditional event hazard of \(0.07\) per year, given their random effects. Find the probability of an event between year 1 and year 4, conditional on the random effects. Then find the probability conditional on surviving to year 1.

With a constant hazard, \(H_i(t\mid b_i)=0.07t\), so \(S_i(t\mid b_i)=\exp(-0.07t)\). The requested interval probability, from time zero as the conditioning origin, is

$$ P(1<T_i\leq4\mid b_i) =e^{-0.07(1)}-e^{-0.07(4)} =e^{-0.07}-e^{-0.28} \approx 0.9323938-0.7557837 =0.1766101 \approx 0.1766. $$

Thus, conditional on the random effects, the probability of an event from year 1 through year 4 is about \(0.1766\), or 17.66%. To condition additionally on being event-free at year 1, divide by survival to year 1:

$$ P(1<T_i\leq4\mid T_i>1,b_i) =1-\frac{S_i(4\mid b_i)}{S_i(1\mid b_i)} =1-e^{-0.07(4-1)} =1-e^{-0.21} \approx 0.1894. $$

This second answer, about 18.94%, uses the surviving subjects at year 1 as its reference group. It is larger than the first interval probability because the two probabilities have different conditioning events.

Using Cumulative Hazard When the Hazard Changes

A hazard need not be constant over time. In that case, the same survival-difference calculation applies, but the cumulative hazard must account for each part of the interval. Integrating the hazard is essential; multiplying the hazard at one time by the interval length is only valid when the hazard is constant over that interval.

Worked Example: A Piecewise-Constant Conditional Hazard

Consider a hypothetical subject whose conditional hazard is \(0.04\) per year from year 0 through year 2, and \(0.10\) per year after year 2. Calculate the probability of an event between years 1 and 3, conditional on the random effects.

The cumulative hazard at year 1 is \(H_i(1\mid b_i)=0.04(1)=0.04\). At year 3, the hazard acts for 2 years at \(0.04\) and 1 year at \(0.10\), giving \(H_i(3\mid b_i)=0.04(2)+0.10(1)=0.18\). Therefore,

$$ P(1<T_i\leq3\mid b_i) =e^{-H_i(1\mid b_i)}-e^{-H_i(3\mid b_i)} =e^{-0.04}-e^{-0.18} \approx 0.9607894-0.8352702 =0.1255192 \approx 0.1255. $$

The interval probability is about 12.55%. If instead the question conditions on survival to year 1, then the cumulative hazard from year 1 to year 3 is \(0.04(1)+0.10(1)=0.14\), and

$$ P(1<T_i\leq3\mid T_i>1,b_i) =1-\exp(-0.14) \approx 0.1306. $$

The two results agree through the identity that the unconditional interval probability is survival to year 1 multiplied by the conditional interval probability: \(e^{-0.04}(1-e^{-0.14})\approx0.1255\).

Substituting a Subject’s Random Effects

In a shared random-effects model, the hazard may depend on a latent longitudinal trajectory, which in turn depends on \(b_i\). Once the conditional hazard is specified, the interval calculation itself is unchanged: evaluate the cumulative hazard at both endpoints with the same \(b_i\), convert to survival, and subtract. The next example isolates this substitution step; it does not require treating noisy observed marker values as the hazard covariate.

Worked Example: A Random Effect Changes the Conditional Hazard

Suppose a hypothetical model has conditional hazard \(\lambda_i(t\mid b_i)=0.05\exp(0.8b_i)\) per year, constant over the interval. For a subject with \(b_i=\log(2)/0.8\), calculate the probability of an event between years 2 and 5, conditional on that random effect.

Substitution gives \(\exp(0.8b_i)=\exp(\log 2)=2\), so the conditional hazard is \(0.05(2)=0.10\) per year. The cumulative hazards are \(H_i(2\mid b_i)=0.10(2)=0.20\) and \(H_i(5\mid b_i)=0.10(5)=0.50\). Thus,

$$ P(2<T_i\leq5\mid b_i) =e^{-0.20}-e^{-0.50} \approx 0.8187308-0.6065307 =0.2122001 \approx 0.2122. $$

For this specified random effect, the conditional interval probability is about 21.22%. This is not automatically the probability for a randomly selected subject. A population-level probability generally requires averaging the conditional probability over the relevant distribution of random effects; inserting a typical or estimated \(b_i\) does not perform that averaging.

Checks, Alternatives, and Limitations

A reliable calculation has several quick checks. First, cumulative hazard must be nondecreasing, so \(H_i(b\mid b_i)\geq H_i(a\mid b_i)\); hence survival must satisfy \(S_i(b\mid b_i)\leq S_i(a\mid b_i)\). The difference is therefore nonnegative and no larger than \(S_i(a\mid b_i)\). Second, all hazards and cumulative hazards must use compatible time units: if time is measured in years, a hazard stated per month must be converted or integrated on a consistent scale. Third, the conditional-survival ratio must use the same \(b_i\) at both endpoints.

The survival difference is usually the most direct approach when conditional survival is already available. Integrating \(\lambda_i(t\mid b_i)S_i(t\mid b_i)\) over the interval is an equivalent route and can be useful for numerical integration or checking a derivation. For a discrete-time event model with possible probability mass at particular times, endpoint conventions must be handled according to that model; the continuous-hazard density argument above does not apply unchanged. Under the continuous model used here, an event exactly at a fixed endpoint has probability zero.

A common mistake is to report \(1-S_i(b\mid b_i)\) as the probability of an event in \((a,b]\). That quantity is the probability of an event by \(b\), including events before \(a\). The correction is to subtract survival at the interval’s start: \(S_i(a\mid b_i)-S_i(b\mid b_i)\). A related mistake is to divide by \(S_i(a\mid b_i)\) without saying that the probability is now conditional on survival to \(a\).

The calculation is only as credible as the conditional hazard used to obtain \(H_i\). A misspecified time pattern, an invalid extrapolation of the hazard, or a model that does not represent relevant heterogeneity can produce misleading interval probabilities even when the subtraction is performed correctly. Likewise, an estimated conditional probability has uncertainty from parameter estimation and, when random effects are estimated, from uncertainty about those effects. A plug-in value alone does not express that uncertainty.

Key takeaway: Given random effects, calculate the interval event probability as \(S_i(a\mid b_i)-S_i(b\mid b_i)\). If conditioning on survival to \(a\) as well, use \(1-S_i(b\mid b_i)/S_i(a\mid b_i)\). State which conditioning applies and use cumulative hazards on a consistent time scale.

Check Your Understanding

Answer each question, then compare with the solutions below.

  1. A conditional hazard is \(0.03\) per month and is constant. What is the probability of an event between months 2 and 6, given the random effects?
  2. For the same hazard, what is the probability of an event between months 2 and 6 conditional on surviving to month 2?
  3. A conditional survival model gives \(S_i(3\mid b_i)=0.82\) and \(S_i(7\mid b_i)=0.61\). Find the probability of an event in \((3,7]\) conditional on \(b_i\). What additional quantity is needed to calculate the probability conditional on surviving to month 3?
  4. Why is \(1-S_i(7\mid b_i)\) not the probability of an event between months 3 and 7?

Solutions. (1) The cumulative hazards are \(0.03(2)=0.06\) and \(0.03(6)=0.18\), so the probability is \(e^{-0.06}-e^{-0.18}\approx0.9418-0.8353=0.1065\). (2) Conditional on survival to month 2, it is \(1-e^{-0.03(6-2)}=1-e^{-0.12}\approx0.1131\). (3) The interval probability is \(0.82-0.61=0.21\). The conditional-on-survival probability is \(0.21/S_i(3\mid b_i)=0.21/0.82\approx0.2561\). (4) \(1-S_i(7\mid b_i)\) includes events at any time before month 7, including events before month 3; subtract \(S_i(7\mid b_i)\) from \(S_i(3\mid b_i)\) to isolate the interval.

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