Piecewise-Constant Baseline Hazard Contributions

Why Interval Contributions Need Care

A piecewise-constant baseline hazard makes the baseline rate simple within each time interval. In a shared-random-effects model, however, the subject-specific hazard need not be constant there: it also depends on the subject’s latent marker trajectory. To calculate survival correctly, integrate the full subject-specific hazard over time, interval by interval.

The result is a conditional contribution: it describes survival for a subject with specified random effects, given the longitudinal trajectory implied by those effects. It is not generally the same as survival after averaging over the population distribution of random effects.

Definition: Let the interval cutpoints be \(0=\kappa_0<\kappa_1<\cdots<\kappa_J\). A piecewise-constant baseline hazard has rate \(\lambda_j>0\) on interval \((\kappa_{j-1},\kappa_j]\), so \(h_0(t)=\lambda_j\) for \(t\) in that interval. In the current-value shared-random-effects model considered here, the subject-specific hazard conditional on random effects \(b_i\) is \(h_i(t\mid b_i)=h_0(t)\exp\{\alpha m_i(t)\}\), where \(m_i(t)\) is the latent marker value and \(\alpha\) is its association parameter.

The interval rates are constant, but \(m_i(t)\) may vary. The baseline intervals and their cutpoints must cover the follow-up times being analyzed. The rates are nonnegative (typically positive), and the latent trajectory and association parameter must be specified well enough that the hazard can be integrated. At a cutpoint the choice of which adjoining interval includes that single time has no effect on a cumulative hazard, because a single point has zero duration.

Deriving the Cumulative-Hazard Sum

By definition, the conditional cumulative hazard through time \(t\) is the integral of the subject-specific hazard from time zero to \(t\). Divide that integral at every cutpoint before \(t\). For each interval, the baseline rate can be taken outside the integral, but the latent-marker multiplier generally cannot.

H_i(t\mid b_i) =\int_0^t h_i(s\mid b_i)\,ds =\sum_{j:\,\kappa_{j-1}<t}\lambda_j \int_{\kappa_{j-1}}^{\min(t,\kappa_j)} \exp\{\alpha m_i(s)\}\,ds,

In this expression, intervals beginning at or after \(t\) contribute zero; equivalently, sum only over intervals with \(\kappa_{j-1}<t\). If the latent marker is constant at \(m_{ij}\) within interval \(j\), the integral reduces to interval length multiplied by the constant multiplier. For the final, possibly partial interval, its length is \(t-\kappa_{j-1}\), not the full interval width.

H_i(t\mid b_i) =\sum_{j:\,\kappa_{j-1}<t} \lambda_j \exp(\alpha m_{ij}) \left\{\min(t,\kappa_j)-\kappa_{j-1}\right\}.

Conditional survival follows from the standard hazard-survival relationship: \(S_i(t\mid b_i)=\exp\{-H_i(t\mid b_i)\}\). Thus, interval contributions add on the cumulative-hazard scale and multiply on the survival scale.

Key result: A constant baseline rate on an interval does not, by itself, make the subject-specific hazard constant on that interval. Use \(\lambda_j\exp\{\alpha m_i(s)\}\) inside the integral unless the latent marker is also constant there.

Worked Calculations

Worked Example: A Constant Latent Marker Within Each Interval

Suppose a hypothetical subject has cutpoints at 0, 1, 2, and 4 time units, with baseline rates \(0.08\), \(0.12\), and \(0.20\) per time unit, respectively. For this calculation, suppose the latent marker is constant within each interval, taking values \(0.5\), \(0.8\), and \(0.6\). Let \(\alpha=0.4\), and calculate conditional survival through \(t=3.2\).

The first two intervals are complete, while only \(3.2-2=1.2\) time units of the third interval have elapsed. The interval-specific effective rates are \(0.08e^{0.4(0.5)}\), \(0.12e^{0.4(0.8)}\), and \(0.20e^{0.4(0.6)}\). Therefore,

\begin{aligned} H_i(3.2\mid b_i) &=1(0.08e^{0.2})+1(0.12e^{0.32})+1.2(0.20e^{0.24})\\ &=0.0977122+0.1652553+0.3050998\\ &=0.5680673. \end{aligned}

The conditional survival is \(S_i(3.2\mid b_i)=e^{-0.5680673}\approx0.5666\). For this specified latent profile and random effects, the model assigns about 0.5666 conditional probability of remaining event-free through 3.2 time units.

Worked Example: A Linear Latent Trajectory Across Two Intervals

Consider a hypothetical latent trajectory \(m_i(s)=0.2+0.15s\), corresponding to specified random intercept and slope values. Let \(\alpha=0.5\); let \(h_0(s)=0.10\) on \((0,2]\) and \(h_0(s)=0.18\) on \((2,4]\). Find conditional survival through \(t=3\).

Here \(\alpha m_i(s)=0.1+0.075s\), so the multiplier changes continuously with time. Integrate it over each interval rather than multiplying by its value at an endpoint or midpoint:

\begin{aligned} H_i(3\mid b_i) &=0.10\int_0^2 e^{0.1+0.075s}\,ds +0.18\int_2^3 e^{0.1+0.075s}\,ds\\ &=\frac{0.10}{0.075}(e^{0.25}-e^{0.1}) +\frac{0.18}{0.075}(e^{0.325}-e^{0.25})\\ &=0.2384736+0.2400126\\ &=0.4784862. \end{aligned}

The integration uses \(\int e^{0.1+0.075s}\,ds=e^{0.1+0.075s}/0.075\). The second interval ends at 3, so only its portion from 2 to 3 contributes. Conditional survival is \(e^{-0.4784862}\approx0.6197\). This example shows why a piecewise-constant baseline does not justify treating the full subject-specific hazard as constant.

Worked Example: Ending Partway Through an Interval

Suppose a hypothetical subject has baseline rate \(0.05\) per time unit on \((0,2]\) and \(0.15\) per time unit on \((2,5]\). Let the latent marker be constant at \(0.4\), with \(\alpha=0.5\), and suppose follow-up ends at \(t=3.5\). Compute the conditional survival contribution through that time.

The multiplier is \(e^{0.5(0.4)}=e^{0.2}\approx1.221403\). The first interval contributes two time units, and the second contributes only \(3.5-2=1.5\) time units:

\begin{aligned} H_i(3.5\mid b_i) &=2(0.05e^{0.2})+1.5(0.15e^{0.2})\\ &=e^{0.2}(0.100+0.225)\\ &=1.221403(0.325)=0.3969559. \end{aligned}

Thus \(S_i(3.5\mid b_i)=e^{-0.3969559}\approx0.6724\). If an event is observed exactly at 3.5, the hazard at that time is \(0.15e^{0.2}\approx0.1832\) per time unit, and the conditional event-time density contribution is \(h_i(3.5\mid b_i)S_i(3.5\mid b_i)\approx0.1232\) per time unit. If follow-up is right-censored at 3.5 instead, the event part is absent and the survival contribution is \(0.6724\), under the usual conditional noninformative-censoring assumption.

Justification and Practical Evaluation

The sum formula follows by partitioning the integral, not by assuming that hazards add as probabilities. For example, if follow-up crosses two intervals, the cumulative hazard is the integral over the first interval plus the integral over the second. Exponentiating its negative gives the product of the corresponding survival factors.

Find the intervals reached.
List each interval that overlaps the time from zero to \(t\), and identify the length of any partial final interval.

Write the full subject-specific hazard.
On interval \(j\), use \(\lambda_j\exp\{\alpha m_i(s)\}\). Keep the time dependence in \(m_i(s)\) if the trajectory changes.

Integrate and add.
Calculate each interval’s integral, multiply by its baseline rate, and sum the contributions to obtain \(H_i(t\mid b_i)\).

Convert to survival.
Calculate \(e^{-H_i(t\mid b_i)}\). If an event is observed at \(t\), combine survival with the hazard at \(t\) for the conditional event-time density.

When \(m_i(s)\) has a form that permits analytic integration, use the corresponding integral as in the linear-trajectory example. Otherwise, numerical quadrature can approximate each interval integral. A useful check is to refine the numerical integration grid and confirm that the result stabilizes. Also check that every interval duration is nonnegative, that the rates and exponential multipliers are positive, and that \(H_i(t\mid b_i)\) does not decrease as \(t\) increases.

Diagnostics, Alternatives, and Limitations

A piecewise-constant baseline is flexible relative to a single parametric shape, but its fit can depend on the chosen cutpoints. Examine the number of events and the amount of follow-up in each interval: intervals with very little information can produce unstable rate estimates. Compare fitted survival patterns with observed event-time patterns across follow-up, and assess sensitivity to reasonable alternative cutpoints. Abrupt changes in the fitted baseline rate at cutpoints are allowed by the model, but may be implausible in some applications.

The Weibull baseline hazard is a parametric alternative with a smooth, structured time pattern; it can be more parsimonious when that pattern is adequate, but can be restrictive when it is not. A piecewise-constant baseline allows local changes but introduces choices about the number and placement of intervals. Neither approach removes the need to model the latent-marker association correctly or to integrate over random effects for population-level likelihood or survival quantities.

A frequent error is to multiply each baseline rate by an interval length and stop, or to substitute an observed marker measurement for the latent marker. The first omits \(\exp\{\alpha m_i(s)\}\); the second changes the specified conditional model. A related error is using the full width of the final interval when follow-up stops early. In every case, the correction is to integrate the model’s full hazard only over the time actually observed.

Finally, conditional survival is not marginal survival. For random effects with population distribution \(p(b_i)\), marginal survival is obtained by averaging the conditional survival over that distribution:

S_i(t)=\int \exp\{-H_i(t\mid b_i)\}\,p(b_i)\,db_i.

In general this is not equal to \(\exp\{-\int H_i(t\mid b_i)p(b_i)\,db_i\}\), because averaging and exponentiating do not commute. The interval calculation here supplies the conditional quantity to be averaged when a marginal quantity is required.

Key takeaway: With a piecewise-constant baseline, split follow-up at the cutpoints and integrate the full hazard \(\lambda_j\exp\{\alpha m_i(s)\}\) within each interval. Use the actual elapsed duration in the final interval, then exponentiate the negative total cumulative hazard to obtain conditional survival.

Check Your Understanding

Try each question, then compare with the solutions below.

A baseline rate is \(0.10\) on \((0,2]\), and the latent marker is constant at \(0\) there with \(\alpha=0.7\). What is the cumulative hazard through time 1.5?
Why is it generally incorrect to use \(\lambda_j\exp\{\alpha m_i(\kappa_j)\}\) times the interval width when \(m_i(s)\) varies within interval \(j\)?
A subject has cumulative-hazard contributions \(0.12\) and \(0.08\) from two successive intervals. What is conditional survival at the end of the second interval?
For a rate of \(0.2\) on \((0,3]\), a constant marker \(m_i=0.5\), and \(\alpha=0.4\), calculate the contribution from time 1 to time 2.

Solutions. (1) The multiplier is \(e^{0.7(0)}=1\), and only 1.5 time units have elapsed, so \(H=0.10(1.5)(1)=0.150\). (2) The multiplier changes with \(s\), so the correct contribution is \(\lambda_j\int \exp\{\alpha m_i(s)\}\,ds\); an endpoint value times width is not generally equal to that integral. (3) The total cumulative hazard is \(0.12+0.08=0.20\), giving conditional survival \(e^{-0.20}\approx0.8187\). (4) The interval length is one time unit and the multiplier is \(e^{0.2}\), so the contribution is \(0.2(1)e^{0.2}\approx0.2443\).