A Weibull Baseline Hazard Contribution

From a Weibull Baseline to a Conditional Contribution

A Weibull baseline gives the event process a specific time pattern before the latent marker is taken into account. In a joint model with a current-value link, the hazard is then multiplied by a function of the subject’s latent marker value at each time. The resulting survival contribution depends on the marker’s whole trajectory up to the time in question, not just its value at that endpoint.

The calculations below are conditional on a subject’s random effects. In the shared-random-effects model, those effects determine the latent marker trajectory; conditioning on them makes that trajectory known. The contribution can then be calculated from the hazard. This is distinct from a marginal contribution, which averages over the random-effects distribution.

Definition: Use the Weibull baseline hazard parameterization $h_0(t)=\lambda\rho t^{\rho-1}$, for $t>0$, with $\lambda>0$ and $\rho>0$. Its cumulative baseline hazard is $H_0(t)=\lambda t^\rho$. Given subject $i$'s random effects $b_i$, let $m_i(t)$ be the latent marker value and define the current-value hazard by $$ h_i(t\mid b_i)=h_0(t)\exp\{\alpha m_i(t)\} =\lambda\rho t^{\rho-1}\exp\{\alpha m_i(t)\}. $$ Here $\alpha$ is the association parameter; the product $\alpha m_i(t)$ must be dimensionless. The conditional survival function is $S_i(t\mid b_i)=\exp\{-\int_0^t h_i(u\mid b_i)\,du\}$, and the conditional event-time density is $f_i(t\mid b_i)=h_i(t\mid b_i)S_i(t\mid b_i)$.

The units matter. If time is measured in days, $h_i(t\mid b_i)$ has units of inverse days and $H_i(t\mid b_i)=\int_0^t h_i(u\mid b_i)\,du$ is dimensionless. In this parameterization, $\lambda$ has units of time raised to the power $-\rho$. The shape $\rho$ controls the baseline time pattern: the baseline hazard is increasing for $\rho>1$, constant for $\rho=1$, and decreasing for $0<\rho<1$. These statements describe the baseline only; the marker multiplier can also change the subject-specific hazard over time.

Deriving Survival and Density

For a continuous event time, the conditional hazard is the instantaneous event rate among subjects who have not yet experienced the event, conditional here on $b_i$. The defining relation between hazard and survival is $h_i(t\mid b_i)=-\frac{d}{dt}\log S_i(t\mid b_i)$. Integrating from 0 to $t$, and using $S_i(0\mid b_i)=1$, gives the cumulative-hazard expression for survival:

\begin{aligned} -\log S_i(t\mid b_i) &=\int_0^t h_i(u\mid b_i)\,du\\ &=\lambda\rho\int_0^t u^{\rho-1}\exp\{\alpha m_i(u)\}\,du,\\ S_i(t\mid b_i) &=\exp\left\{-\lambda\rho\int_0^t u^{\rho-1}\exp\{\alpha m_i(u)\}\,du\right\}. \end{aligned}

To obtain the event-time density, differentiate the survival function. Since its exponent is minus the cumulative hazard, the chain rule gives $-\frac{d}{dt}S_i(t\mid b_i)=h_i(t\mid b_i)S_i(t\mid b_i)$. Thus the density contribution at an observed event time $t$ is the instantaneous hazard at that time multiplied by the probability of surviving up to that time.

f_i(t\mid b_i) =\lambda\rho t^{\rho-1}\exp\{\alpha m_i(t)\} \exp\left\{-\lambda\rho\int_0^t u^{\rho-1}\exp\{\alpha m_i(u)\}\,du\right\}.

These formulas assume the stated Weibull baseline and current-value link, a positive event time, and a marker trajectory defined over the interval from time 0 to $t$. Conditional on the random effects, the trajectory is treated as specified. For a subject censored at $c$, the event-process contribution is $S_i(c\mid b_i)$; for an observed event at $t$, it is $f_i(t\mid b_i)$. These are the conditional contributions, before any integration over random effects. The usual interpretation also presumes that censoring can be handled as noninformative conditional on the modeled information.

When the Marker Trajectory Is Constant

If $m_i(u)=m_i$ throughout the interval, the marker multiplier is constant and can be taken outside the integral. In this special case, the integral has a closed form:

H_i(t\mid b_i) =\lambda\rho\exp(\alpha m_i)\int_0^t u^{\rho-1}\,du =\lambda t^\rho\exp(\alpha m_i), \qquad S_i(t\mid b_i)=\exp\{-\lambda t^\rho\exp(\alpha m_i)\}.

For the event density, multiply this survival probability by the hazard at $t$. Even with a constant marker, the density is not just the hazard: it includes survival up to $t$.

Worked Example: Constant Latent Marker

Suppose time is measured in days, $\lambda=0.02$ per day squared, $\rho=2$, the constant latent marker is $m_i=1.5$ units, and $\alpha=0.4$ per marker unit. Find the conditional survival at day 3 and the conditional event density there.

The assumptions for the calculation are the Weibull baseline and current-value link above, with this subject’s marker held constant over time. The association multiplier is $\exp(\alpha m_i)=\exp(0.4\times1.5)=\exp(0.6)\approx1.822$. The cumulative baseline hazard by day 3 is $H_0(3)=0.02(3^2)=0.18$, so

H_i(3\mid b_i)=0.18\exp(0.6)\approx0.3280, \qquad S_i(3\mid b_i)=\exp(-0.3280)\approx0.7204.

The baseline hazard at day 3 is $0.02(2)(3^{1})=0.12$ per day. Including the marker multiplier gives $h_i(3\mid b_i)=0.12\exp(0.6)\approx0.2187$ per day. Therefore,

f_i(3\mid b_i)=0.12\exp(0.6)\exp\{-0.18\exp(0.6)\}\approx0.1575\ \text{per day}.

Conditional on these random effects and model parameters, the probability of surviving beyond day 3 is about 0.7204. The density at day 3 is about 0.1575 per day; it is a density, not the probability of an event occurring at exactly day 3.

When the Marker Changes Over Time

With a changing marker, the current-value multiplier changes inside the cumulative-hazard integral. In general, $H_i(t\mid b_i)$ is not $H_0(t)\exp\{\alpha m_i(t)\}$: that expression uses only the endpoint marker and misses the marker values at earlier times. The integral may require numerical quadrature or a special-function expression. A useful exception is a linear marker trajectory combined with certain Weibull shapes, for which the integral can sometimes be evaluated analytically.

Worked Example: A Linear Marker With a Shape-Two Weibull Baseline

Consider time in years, $\lambda=0.04$ per year squared, $\rho=2$, and $m_i(u)=2+0.3u$ marker units. Let $\alpha=0.5$ per marker unit. Calculate conditional survival and event density at year 2. These parameters satisfy the model assumptions, and the linear trajectory specifies the latent marker throughout the required interval.

The hazard is $0.08u\exp\{0.5(2+0.3u)\}=0.08u\exp(1+0.15u)$ per year. The cumulative hazard is therefore

\begin{aligned} H_i(2\mid b_i) &=0.08e^1\int_0^2 u e^{0.15u}\,du\\ &=0.08e^1 \left[\frac{e^{0.15u}(0.15u-1)+1}{(0.15)^2}\right]_0^2\\ &=0.08e^1 \frac{e^{0.3}(-0.7)+1}{0.0225}\\ &\approx0.5325. \end{aligned}

Thus $S_i(2\mid b_i)=\exp(-0.5325)\approx0.5872$. At year 2, the marker is $m_i(2)=2.6$, so the hazard is $0.16\exp(0.5\times2.6)=0.16e^{1.3}\approx0.5871$ per year. The event density is

f_i(2\mid b_i)=0.5871(0.5872)\approx0.3447\ \text{per year}.

The cumulative hazard used the marker trajectory over the entire two years, whereas the hazard factor in the density used the marker value at year 2. That distinction is essential: earlier marker values affect survival to the endpoint, while the endpoint value determines the instantaneous hazard there.

Observed Event and Censoring Contributions

Let $T_i$ denote the observed follow-up time and let $\delta_i=1$ indicate an observed event, while $\delta_i=0$ indicates right censoring. Conditional on the random effects, the event-process contribution can be written as $f_i(T_i\mid b_i)^{\delta_i}S_i(T_i\mid b_i)^{1-\delta_i}$. This compact form selects the density for an event and survival for censoring. It does not treat a censored subject as having an event at the censoring time.

Worked Example: Event Versus Right Censoring

Use the model from the constant-marker example. At day 3 the conditional survival is approximately 0.7204 and the conditional density is approximately 0.1575 per day. If a subject has an observed event at day 3, the event-process contribution is $f_i(3\mid b_i)\approx0.1575$ per day. If a different subject is censored at day 3, the contribution is $S_i(3\mid b_i)\approx0.7204$. The two contributions differ because an observed event supplies both the fact of surviving until that time and the event at that time, while right censoring supplies only survival through the censoring time.

The assumptions are the same Weibull/current-value model and right censoring that is noninformative conditional on the modeled information. Neither number is a marginal likelihood contribution: obtaining that requires averaging the conditional contribution over the random-effects distribution.

Interpretation, Checks, and Alternatives

The current-value association has a direct hazard-ratio interpretation at a fixed time and for a fixed baseline: comparing two latent marker values separated by one unit, the hazard ratio is $\exp(\alpha)$, provided their other modeled quantities are the same. If $\alpha>0$, a higher latent marker value corresponds to a higher hazard; if $\alpha<0$, it corresponds to a lower hazard. This describes a conditional association, not a causal effect. A changing marker can make the event hazard vary over time even when the baseline hazard is constant, provided the marker’s time variation affects the hazard—for example, when $\alpha\ne0$. If $\alpha=0$, the marker has no effect on the hazard, even if its trajectory changes.

A useful check is to set $\alpha=0$ in the formulas. The marker term must disappear, leaving $S_i(t)=\exp(-\lambda t^\rho)$ and $f_i(t)=\lambda\rho t^{\rho-1}\exp(-\lambda t^\rho)$. If $m_i(u)$ is constant, the formulas should reduce to the closed forms above. In a numerical implementation, check that the integrated hazard is nonnegative, survival lies between 0 and 1, and the event density is nonnegative. For a changing trajectory, compare numerical integration at progressively finer tolerances; a survival value materially above 1 or a negative cumulative hazard indicates a coding or numerical error.

A Weibull baseline is parsimonious and can be convenient when its time pattern is plausible. Its restriction is that the baseline hazard follows a monotone power pattern, or is constant when $\rho=1$; it cannot represent every irregular baseline shape. The exponential baseline is exactly the special case $\rho=1$. More flexible baseline specifications are alternatives when the Weibull shape is inadequate, but flexibility brings additional modeling and estimation choices. Regardless of baseline choice, a changing latent marker requires integrating its time-varying contribution to the hazard when computing survival.

The integral also limits closed-form calculation. Constant marker trajectories yield a simple Weibull expression, but a general longitudinal trajectory usually requires numerical integration. A further modeling limitation is that the current-value link depends on the marker at each instant, not automatically on its slope or accumulated history. Those are different link choices and should not be silently substituted into this model.

Common pitfall: Replacing the cumulative hazard by $H_0(t)\exp\{\alpha m_i(t)\}$ when the marker changes over time. This incorrectly applies the endpoint marker value to the whole follow-up interval. The correct quantity is $\lambda\rho\int_0^t u^{\rho-1}\exp\{\alpha m_i(u)\}\,du$. Also, a changing marker does not by itself guarantee a changing event hazard: if $\alpha=0$, its time variation has no effect on the hazard.

Check Your Understanding

Answer each question, then compare with the solutions.

For a Weibull baseline with $\lambda=0.1$ per month squared and $\rho=2$, what is the baseline cumulative hazard at 2 months?
If $m_i(u)$ changes over time, why is $H_0(t)\exp\{\alpha m_i(t)\}$ generally not the conditional cumulative hazard?
For a constant marker $m_i$, state the conditional survival function under the Weibull/current-value model.
When $\rho=1$ and $\alpha=0$, with $\lambda=0.3$ per month, what are the conditional survival and event density at 4 months?
For an observed event at $T_i$, which conditional contribution is used, and how does it differ from the contribution for right censoring at that time?

Solutions. 1. $H_0(2)=0.1(2^2)=0.4$, dimensionless. 2. The marker multiplier varies within the interval, so the hazard must be integrated using $m_i(u)$ at each $u$, not just its endpoint value. 3. $S_i(t\mid b_i)=\exp\{-\lambda t^\rho\exp(\alpha m_i)\}$. 4. With $\rho=1$ and $\alpha=0$, $S_i(4)=\exp(-0.3\times4)=\exp(-1.2)\approx0.3010$. The hazard is $0.3$ per month, so $f_i(4)=0.3\exp(-1.2)\approx0.09036$ per month. 5. An observed event contributes $f_i(T_i\mid b_i)=h_i(T_i\mid b_i)S_i(T_i\mid b_i)$; right censoring contributes $S_i(T_i\mid b_i)$, because no event at that time was observed.