Why Marginal Hazards Can Lose Proportionality

Why Conditional Proportionality May Not Survive Averaging

In a shared random-effects joint model, a subject’s latent effects can contribute both to the longitudinal trajectory and to the event hazard. Conditional on those effects, the hazard may have a proportional-hazards form with a constant hazard ratio between two exposure groups. But the marginal hazard compares the subjects who remain at risk in each group after averaging over the latent effects. Those risk sets need not have the same distribution of random effects as follow-up proceeds. Consequently, a constant conditional hazard ratio does not generally imply a constant marginal hazard ratio.

The distinction is between comparing subjects with the same latent effect and comparing the two groups after that latent effect has been integrated out. The latter comparison is affected by who has already experienced the event. This is a survivor-selection effect, not a contradiction between two calculations of the same quantity.

Definition: Let $Z\in\{0,1\}$ denote group and $U>0$ a shared latent random effect. A conditionally proportional hazard has the form $$ \lambda(t\mid Z,U)=U\lambda_0(t)\exp(\beta Z), $$ where $\lambda_0(t)$ is a baseline hazard and $\exp(\beta)$ is the conditional hazard ratio comparing $Z=1$ with $Z=0$ at the same $U$. The marginal hazard $\lambda(t\mid Z)$ averages over the distribution of $U$ among subjects in group $Z$ who remain event-free at time $t$.

For the derivations and examples below, assume that $U$ has the same distribution in both groups at baseline, is independent of $Z$, and is positive. Assume also that the conditional hazard model is correctly specified, that the baseline hazard is common across groups, and that censoring does not alter the event-risk distribution being described. The marginal calculations use event survival; they do not require that $U$ be observed. In an actual joint model, $U$ can be a function of a latent effect that also enters the longitudinal model.

From Conditional Survival to the Marginal Hazard

Write $\Lambda_0(t)=\int_0^t\lambda_0(s)\,ds$ for the cumulative baseline hazard. Conditional on $U$ and $Z=z$, the survival probability through time $t$ is the exponential of minus the cumulative hazard. Averaging that survival probability over the baseline distribution of $U$ gives the group-specific marginal survival:

S(t\mid Z=z) = E\left[\exp\left\{-U\exp(\beta z)\Lambda_0(t)\right\}\right].

The marginal hazard is the negative time derivative of the log marginal survival. Differentiating the expectation (when the relevant expectation is finite) gives:

\lambda(t\mid Z=z) = \lambda_0(t)\exp(\beta z) \frac{ E\left[U\exp\{-U\exp(\beta z)\Lambda_0(t)\}\right] }{ E\left[\exp\{-U\exp(\beta z)\Lambda_0(t)\}\right] }.

The ratio of expectations is the mean of $U$ among event-free subjects at time $t$, in group $z$. To see this, the conditional density of $U$ among survivors is proportional to its baseline distribution multiplied by $\exp\{-U\exp(\beta z)\Lambda_0(t)\}$. Thus the marginal hazard equals the baseline hazard times the group’s exposure multiplier times its current survivor-average random effect. The survivor-average random effect can differ across groups even though their baseline distributions were identical.

At time zero, before events have changed the composition of either risk set, the marginal hazard ratio is $\exp(\beta)$, provided the mean of $U$ exists and is the same in both groups. Later, the groups can have different survivor-average values of $U$. This is why the ratio of their marginal hazards need not remain constant.

A Gamma Example Gives an Explicit Formula

A useful derivation takes $U$ to have a gamma distribution with shape $k$ and rate $k$, so that $E(U)=1$ and $\operatorname{Var}(U)=1/k$. This is an illustrative positive random-effect distribution; it is not an assumption that every joint model must use. The Laplace transform is $E[\exp(-qU)]=(k/(k+q))^k$ for $q\geq 0$. Substituting $q=\exp(\beta z)\Lambda_0(t)$ yields:

S(t\mid Z=z) = \left( \frac{k}{k+\exp(\beta z)\Lambda_0(t)} \right)^k.

Differentiating the log of this expression gives the marginal hazard:

\lambda(t\mid Z=z) = \lambda_0(t)\exp(\beta z) \frac{k}{k+\exp(\beta z)\Lambda_0(t)}.

Therefore the marginal hazard ratio is:

\operatorname{HR}_{\mathrm{marg}}(t) = \exp(\beta) \frac{k+\Lambda_0(t)} {k+\exp(\beta)\Lambda_0(t)}.

This expression depends on follow-up through $\Lambda_0(t)$. If $\beta>0$, it starts at $\exp(\beta)$ and decreases toward 1 as cumulative baseline hazard increases. If $\beta<0$, it starts below 1 and moves toward 1. This direction is a feature of this gamma example, not a universal rule for every distribution of random effects.

Worked Example: A Conditional Hazard Ratio of Two

Suppose the baseline hazard is constant at $0.10$ per year, $U$ has a gamma distribution with shape and rate $k=1$, and $\exp(\beta)=2$. These assumptions give a common baseline hazard, identical baseline random-effect distributions in both groups, and a finite positive random effect with mean 1. The conditional hazard ratio is 2 for any fixed $U$.

Here $\Lambda_0(t)=0.10t$ years in the hazard calculation. The formula for the marginal hazard ratio becomes:

\operatorname{HR}_{\mathrm{marg}}(t) = 2\frac{1+\Lambda_0(t)}{1+2\Lambda_0(t)}.

At $t=5$ years, $\Lambda_0(5)=0.50$ and the marginal hazards are:

\lambda(5\mid Z=0) = 0.10\frac{1}{1+0.50} = 0.06667\text{ per year}, \qquad \lambda(5\mid Z=1) = 0.10(2)\frac{1}{1+1.00} = 0.1000\text{ per year}.

Their ratio is $0.1000/0.06667=1.500$, not 2. At $t=10$ years, $\Lambda_0(10)=1.00$, giving:

\lambda(10\mid Z=0)=0.10\frac{1}{2}=0.05000, \qquad \lambda(10\mid Z=1)=0.20\frac{1}{3}=0.06667, \qquad \operatorname{HR}_{\mathrm{marg}}(10)=1.333.

The conditional hazard ratio remains 2, but the marginal hazard ratio has changed from 1.500 at year 5 to 1.333 at year 10. These are hypothetical calculations under the stated model, not estimates from a study.

Why the Risk Sets Become Different

The mechanism can be seen without a gamma assumption. Subjects with larger $U$ have higher conditional hazards in both groups. As time passes, they are more likely to have experienced the event and leave the risk set. This depletion occurs more quickly in the group with the larger exposure multiplier. As a result, the mean $U$ among survivors can become lower in that group. The marginal hazard ratio is then affected by both the exposure multiplier and the changing random-effect composition.

Worked Example: Survivor Composition With Two Random-Effect Values

Suppose $U$ equals 0.5 or 2 with probability 0.5 each in both groups at baseline. Let $\exp(\beta)=1.5$, and suppose $\Lambda_0(t)=1.00$ at the time of interest. These values satisfy the shared-distribution assumption, and the conditional hazard ratio is 1.5 at either value of $U$.

For group $z$, the probability of surviving conditional on $U=u$ is $\exp\{-u\exp(\beta z)\Lambda_0(t)\}$. In group 0, the survivor weights for $U=0.5$ and $U=2$ are $\exp(-0.5)=0.6065$ and $\exp(-2)=0.1353$. The common baseline probability 0.5 cancels when forming the survivor proportion. Thus:

P(U=2\mid T\geq t,Z=0) = \frac{0.1353}{0.6065+0.1353} = 0.1824.

The mean random effect among group 0 survivors is therefore $0.5(1-0.1824)+2(0.1824)=0.7736$. In group 1, the survivor weights are $\exp(-0.75)=0.4724$ and $\exp(-3)=0.04979$, so:

P(U=2\mid T\geq t,Z=1) = \frac{0.04979}{0.4724+0.04979} = 0.09535, \qquad E(U\mid T\geq t,Z=1)=0.5(1-0.09535)+2(0.09535)=0.6430.

The exposed group’s survivors have the lower mean $U$, because high-$U$ subjects have been depleted more strongly. The marginal hazards are $\lambda_0(t)(0.7736)$ for group 0 and $\lambda_0(t)(1.5)(0.6430)$ for group 1. Their ratio is $1.5(0.6430)/0.7736=1.247$, below the conditional ratio 1.5. The calculation uses survivor composition explicitly rather than assuming that the original 50:50 mixture still describes those at risk.

Assumptions, Diagnostics, and Alternatives

A marginal hazard ratio will generally vary over time when the random effect affects the event hazard and its distribution among survivors changes. But variation is not guaranteed in every special case. For example, if the random effect is constant across subjects, there is no heterogeneity for survival to select on. The mathematical form also depends on the distribution of $U$, the baseline hazard, and whether the distribution of $U$ is the same between groups. If baseline random-effect distributions differ by group, the marginal contrast can reflect that difference from the outset as well as subsequent selection.

A practical diagnostic is to distinguish the estimand before interpreting a proportional-hazards assessment. If the model specifies conditional proportional hazards, inspect whether that conditional model is plausible given its covariates and latent-effect structure. If the target is a population-level marginal comparison, calculate or estimate marginal survival and hazards by integrating over the random-effect distribution, then examine the ratio over time. A single constant ratio may conceal relevant time variation. Censoring and the observation process also matter for estimation: the expressions above describe event survival under the stated model, not a license to ignore informative censoring or misspecified longitudinal and event components.

There are several different summaries, and they answer different questions:

Approach	What it compares	Main implication
Conditional hazard ratio	Groups at the same latent random effect	Can be constant under the conditional model
Marginal hazards from integration	Group-specific survivor populations after averaging over random effects	Can vary as survivor compositions change
One-coefficient marginal proportional-hazards fit	A constant-hazard-ratio summary for the observed comparison	May be a useful summary, but need not equal the conditional coefficient or describe every follow-up time

The conditional coefficient $\beta$ is not automatically a population-average effect. Conversely, a marginal hazard ratio that changes over time does not by itself show that the conditional model is wrong. A marginal Cox fit may compress a time-varying marginal contrast into one estimate; its interpretation depends on the data and follow-up and should not be substituted for the conditional parameter without explanation. When the scientific question concerns absolute event experience, marginal survival probabilities can also be informative, but they are a different summary from a hazard ratio.

Common Pitfall

Wrong reasoning: “The conditional model has a constant hazard ratio, so integrating out the random effects must leave the same constant ratio.” This treats the baseline random-effect distribution as if it remained unchanged among survivors in both groups. It does not: survival weights depend on both $U$ and the group-specific hazard multiplier.

Correction: Average the conditional survival probabilities first, then derive the marginal hazards, or equivalently average $U$ using its distribution among survivors at each time. The time-dependent weighting is essential. Also state whether a reported hazard ratio is conditional on the latent effects or marginal over them; without that distinction, the numerical value can be misinterpreted.

Key takeaway: Conditional proportional hazards do not generally imply marginal proportional hazards. Integrating over a shared random effect changes the composition of each group’s risk set over time, and the resulting marginal hazard ratio can therefore vary with follow-up.

Check Your Understanding

Use the conditional-versus-marginal distinction and the calculations above to answer these questions. Short solutions follow the questions.

In the gamma example with $k=1$, $\exp(\beta)=2$, and $\Lambda_0(t)=0$, what is the marginal hazard ratio?
Why can the marginal hazard ratio differ between groups even when their baseline distributions of $U$ are identical?
For the gamma model, state the marginal hazard ratio as a function of $\Lambda_0(t)$, $k$, and $\exp(\beta)$.
In the two-value example, why does the exposed group have a smaller survivor proportion with $U=2$ at the stated time?

Solutions. (1) Substitution gives $2(1+0)/(1+2(0))=2$; at time zero there has been no survival-based change in composition. (2) The group with the larger hazard multiplier removes high-$U$ subjects from its risk set more quickly, changing its survivor-average $U$. (3) The ratio is $\exp(\beta)\{k+\Lambda_0(t)\}/\{k+\exp(\beta)\Lambda_0(t)\}$. (4) At the same baseline value $U=2$, survival in group 1 is weighted by $\exp(-2(1.5)(1.00))=\exp(-3)$, while survival in group 0 is weighted by $\exp(-2(1.00))=\exp(-2)$. The stronger depletion of high-$U$ subjects in group 1 gives the smaller survivor proportion, $0.09535$ rather than $0.1824$.