Shared Effects in Continuous-Time Dropout Hazards

From a Shared Effect to a Continuous-Time Hazard

A discrete-time dropout model assigns risk across time intervals. In continuous time, the model instead describes an instantaneous dropout rate: the hazard. This distinction matters when dropout can occur at any time and the observed data include an event time or a censoring time. The shared-effect formulation links that hazard to an outcome random effect, so individuals with different latent outcome profiles can have different dropout risks even when their observed histories are similar.

Let $b_i$ denote a scalar random effect for individual $i$ in the outcome model. Consider a proportional-hazards dropout model in which this same effect enters the dropout hazard. Let $D_i$ be the latent dropout time, and let $\lambda_0(t)$ be a nonnegative baseline hazard. The baseline cumulative hazard is $\Lambda_0(t)=\int_0^t \lambda_0(u)\,du$.

Definition: Conditional on the shared outcome random effect $b_i$, the proportional-hazards dropout model is $$ \lambda_i(t\mid b_i)=\lambda_0(t)\exp(\alpha b_i). $$ Here $\lambda_i(t\mid b_i)$ is the instantaneous dropout hazard at time $t$, conditional on still being observed just before $t$ and on $b_i$. The parameter $\alpha$ controls the association between the shared effect and dropout risk. The baseline hazard describes the time pattern for an individual with $b_i=0$.

The model assumes that the shared effect acts multiplicatively on the hazard and has a constant proportional effect over time. In particular, the hazard ratio comparing effects $b_i$ and $b_i'$, at a time when the baseline hazard is positive, is $\exp\{\alpha(b_i-b_i')\}$. A positive $\alpha$ means larger values of $b_i$ are associated with higher conditional dropout hazards; a negative $\alpha$ means they are associated with lower hazards. This is an association induced by the model, not a causal effect of changing the random effect.

The exponential form guarantees a nonnegative hazard. The baseline cumulative hazard must be finite over the follow-up interval for the survival probability to be positive. The model is conditional on $b_i$; integrating over the random-effect distribution is a separate operation. The joint-model setup uses the shared effect to connect the outcome and dropout processes, with conditional independence of those processes given the shared effects as the key structural assumption.

Deriving Conditional Survival and Event-Time Contributions

Write $S_i(t\mid b_i)=P(D_i>t\mid b_i)$ for the probability of remaining free of dropout through time $t$, conditional on the shared effect. The definition of the hazard implies that the survival function satisfies a differential equation. Integrating that equation yields survival, and multiplying the hazard by survival yields the event-time density.

Relate hazard to survival.
For a continuous event time, $\lambda_i(t\mid b_i)=-\frac{d}{dt}\log S_i(t\mid b_i)$, wherever the derivative exists.

Integrate over time.
Since $S_i(0\mid b_i)=1$, integrating from 0 to $t$ gives $-\log S_i(t\mid b_i)=\int_0^t\lambda_i(u\mid b_i)\,du$.

Substitute the proportional-hazards form.
The multiplier $\exp(\alpha b_i)$ is constant over time conditional on $b_i$, so the integrated hazard is $\exp(\alpha b_i)\Lambda_0(t)$.

S_i(t\mid b_i)=\exp\{-\exp(\alpha b_i)\Lambda_0(t)\}, \qquad f_i(t\mid b_i)=\lambda_0(t)\exp(\alpha b_i) \exp\{-\exp(\alpha b_i)\Lambda_0(t)\}.

The density $f_i(t\mid b_i)=\lambda_i(t\mid b_i)S_i(t\mid b_i)$ describes the conditional density of a dropout at the exact time $t$. Because it is a density, its units are inverse time; its value is not the probability of dropout at one exact instant.

Suppose dropout time is observed only up to a right-censoring time. Let $T_i=\min(D_i,C_i)$, where $C_i$ is the censoring time, and let $\Delta_i=1$ if dropout is observed at $T_i$ and $\Delta_i=0$ if follow-up is right-censored there. If censoring is independent of dropout conditional on the modeled information, the dropout-process contribution is the event density when $\Delta_i=1$ and the survival probability when $\Delta_i=0$. These cases combine as follows.

Conditional dropout contribution: For an observed time $T_i$ and event indicator $\Delta_i$, $$ L_{Di}(b_i)= \left\{\lambda_0(T_i)\exp(\alpha b_i)\right\}^{\Delta_i} \exp\{-\exp(\alpha b_i)\Lambda_0(T_i)\}. $$ This is a contribution conditional on $b_i$, not the full marginal likelihood for the observed outcomes and dropout data.

For an event, the first factor supplies the hazard at the event time and the exponential factor supplies survival up to that time. For censoring, $\Delta_i=0$, so only the probability of remaining event-free through $T_i$ remains. Any likelihood contribution for the outcome process must also be included in a joint model; it is not part of $L_{Di}(b_i)$.

Worked Example: Conditional Survival Under a Constant Hazard

Suppose the baseline hazard is constant at $0.04$ per month. For an individual with $b_i=1$, let $\alpha=\log(2)$. Find the conditional hazard and the probability of remaining free of dropout for six months.

The shared-effect multiplier is $\exp(\alpha b_i)=\exp\{\log(2)\times 1\}=2$. Thus the conditional hazard is $0.04\times 2=0.08$ per month. A constant baseline hazard gives $\Lambda_0(6)=0.04\times 6=0.24$.

Substituting into the survival formula, $$ S_i(6\mid b_i)=\exp\{-2(0.24)\}=\exp(-0.48)=0.6188. $$

Conditional on $b_i=1$, the probability of no dropout through six months is about $0.6188$, or 61.88%. The corresponding probability of dropout by six months is $1-0.6188=0.3812$, or 38.12%. These are conditional probabilities for this specified random-effect value and hazard model.

Exact Dropout Times, Censoring, and Interval Probabilities

The cumulative baseline hazard need not be linear in time. With a piecewise-constant baseline hazard, it is calculated by summing the hazard rate multiplied by the duration in each interval. At an exact event time, the instantaneous hazard uses the baseline hazard at that time, while survival up to the event time uses the cumulative hazard over the entire preceding period. Keeping those roles distinct is essential.

Worked Example: Event and Censoring Contributions With a Piecewise Baseline

Suppose $\lambda_0(t)=0.02$ per month from month 0 through month 5 and $0.05$ per month after month 5. Let $b_i=-0.4$ and $\alpha=1.5$. Calculate the dropout-process contribution for (a) an observed dropout at month 8 and (b) a subject censored at month 8.

The shared-effect multiplier is $\exp(1.5\times -0.4)=\exp(-0.6)=0.5488$. The cumulative baseline hazard through month 8 is $$ \Lambda_0(8)=(0.02)(5)+(0.05)(3)=0.10+0.15=0.25. $$

The conditional survival probability through month 8 is therefore $$ S_i(8\mid b_i)=\exp\{-(0.5488)(0.25)\} =\exp(-0.1372)=0.8718. $$

(a) At month 8, the baseline hazard is $0.05$ per month, so the conditional hazard is $(0.05)(0.5488)=0.02744$ per month. The event-time contribution is $$ L_{Di}(b_i)=0.02744\exp(-0.1372)=0.02392 $$ per month. This is a density contribution, not a probability of an event exactly at month 8.

(b) For censoring at month 8, $\Delta_i=0$, so the contribution is just $S_i(8\mid b_i)=0.8718$. The two cases use the same survival through month 8; only the observed event indicator determines whether the event-time hazard factor is included.

Probabilities for an interval can be obtained without treating an exact event time as having positive probability. For $0\leq a<b$, the conditional probability of dropout after $a$ but by $b$ is $S_i(a\mid b_i)-S_i(b\mid b_i)$: the subject must survive to $a$, then fail before or at $b$.

Worked Example: Probability of Dropout During a Follow-Up Window

Use the constant baseline hazard $0.04$ per month and the multiplier $2$ from the first example. Find the conditional probability of dropout between months 4 and 7.

The conditional hazard is $0.08$ per month. Thus the cumulative conditional hazards by months 4 and 7 are $0.08(4)=0.32$ and $0.08(7)=0.56$. The interval probability is $$ P(4<D_i\leq 7\mid b_i) =S_i(4\mid b_i)-S_i(7\mid b_i) =\exp(-0.32)-\exp(-0.56) =0.7261-0.5712=0.1549. $$

Under the specified model and conditional on this random-effect value, the probability of dropout in that three-month window is 0.1549, or 15.49%. It is not the probability of dropout at month 4 or month 7.

Assumptions, Interpretation, and Checks

The formulas above rely on several assumptions that should be made explicit when the model is used. The event time is represented by a continuous-time hazard; the hazard follows the specified proportional form conditional on the shared effect; and the baseline hazard and cumulative hazard are defined over the observed follow-up range. The censoring process must be handled consistently. In the simple contribution shown here, censoring is assumed independent of dropout conditional on the modeled information, so its mechanism does not add a dropout-event factor. If censoring itself carries information about dropout beyond that conditioning set, this contribution alone is not an adequate likelihood.

The shared-effect assumption is also substantive: conditional on the latent effects included in the model, the outcome and dropout processes are assumed independent. The dropout association is then represented through their common dependence on $b_i$. A positive $\alpha$, for example, says that individuals with larger $b_i$ have higher dropout hazards, conditional on the model. It does not establish that a particular outcome value causes dropout.

Several basic checks follow directly from the model equations. For increasing $t$, $\Lambda_0(t)$ must be nondecreasing, and $S_i(t\mid b_i)$ must remain between 0 and 1 and be nonincreasing. The conditional hazard must be nonnegative. For an exact event, the hazard factor must be evaluated at the event time, while the cumulative hazard must include all preceding intervals. A negative cumulative increment, an increasing survival curve, or use of an interval probability as an exact-time density signals a calculation or implementation error.

More substantive warning signs include an implausible time pattern in the baseline hazard, unstable association estimates, or systematic discrepancies between predicted and observed dropout patterns across follow-up. These may indicate a poor baseline specification, a proportional-hazards restriction that is unsuitable, or a shared-effect structure that fails to capture the relevant dependence. Numerical checks cannot verify the conditional independence assumption, which is a modeling assumption rather than a directly testable fact from observed data alone.

The continuous-time model is related to, but not interchangeable with, a discrete-time hazard model. A discrete-time model describes event probabilities over specified intervals; the continuous-time model describes an instantaneous rate and derives survival by integrating that rate. A model with no shared effect would omit the dependence of dropout risk on outcome random effects, while a separate frailty formulation would need its own interpretation and assumptions about the relationship to the outcome process. The shared proportional-hazards specification is attractive for its clear conditional interpretation, but it does not by itself determine the baseline hazard, establish identifiability, or guarantee that the assumed latent association is supported by the data.

Common Pitfall

Common mistake: Treating $\lambda_i(t\mid b_i)$ as the probability of dropout at time $t$, or using only the event-time hazard for a dropout observed at $t$. Correction: A hazard is an instantaneous rate among those still at risk. The exact event-time density is hazard multiplied by survival to that time, $\lambda_i(t\mid b_i)S_i(t\mid b_i)$; a right-censored observation contributes survival alone.

A second frequent error is to use the baseline hazard at the event time as though it also summarized all earlier risk. It does not: the event-time factor uses the instantaneous baseline hazard, and the survival factor uses the integrated baseline hazard from the start of follow-up to the observed time.

Check Your Understanding

Answer each question using the conditional model and formulas in this tutorial. Brief solutions follow the questions.

If $\alpha=0$, what does the model imply about the dropout hazard's dependence on $b_i$?
With constant baseline hazard $0.03$ per month, $\alpha=1$, and $b_i=\log(2)$, calculate conditional survival through 5 months.
For a subject right-censored at time $T_i$, which part of the conditional dropout contribution remains?
Explain why an exact event-time contribution is a density rather than a probability at a single time.

Solutions. (1) Since $\exp(\alpha b_i)=\exp(0)=1$, the hazard is $\lambda_0(t)$ and does not depend on $b_i$. (2) The multiplier is $\exp\{\log(2)\}=2$, and $\Lambda_0(5)=0.03(5)=0.15$. Hence $S_i(5\mid b_i)=\exp\{-2(0.15)\}=\exp(-0.30)=0.7408$. (3) With $\Delta_i=0$, the contribution is $\exp\{-\exp(\alpha b_i)\Lambda_0(T_i)\}=S_i(T_i\mid b_i)$; there is no event-time hazard factor. (4) For a continuous event time, the probability at exactly one time is zero. The hazard multiplied by survival gives the density per unit time, which is used in the likelihood for an observed exact event time.