Posterior Survival Curves From Piecewise Hazards

From Interval Hazards to a Posterior Curve

A survival curve describes the probability of remaining event-free beyond each follow-up time. With a piecewise-constant hazard model, the hazard is allowed to differ across specified time intervals but is constant within each interval. This gives a flexible way to describe changing event rates without requiring the hazard to follow a single exponential pattern throughout follow-up.

The earlier material on piecewise exponential posterior survival curves introduced this model. Here the emphasis is on turning joint posterior draws of the interval hazards into a curve, then summarizing uncertainty at multiple times. A curve is not one extra parameter: it is a collection of survival probabilities whose values are linked through the same hazards.

Definition: Let the intervals be \((t_{j-1},t_j]\), for \(j=1,\ldots,K\), with \(0=t_0<t_1<\cdots<t_K\). In a piecewise-constant hazard model, the hazard on interval \(j\) is \(\lambda_j\geq 0\). For a time \(t\), define the exposure to interval \(j\) as \(d_j(t)=\max\{0,\min(t,t_j)-t_{j-1}\}\). The survival probability conditional on the hazards is \(S(t\mid\boldsymbol{\lambda})=\exp\{-\sum_{j=1}^K \lambda_j d_j(t)\}\).

The quantity \(d_j(t)\) is the amount of time from interval \(j\) that has elapsed by time \(t\). It is zero for intervals not yet reached, the full interval width for completed intervals, and a partial width for the interval containing \(t\). Hazards have units of inverse time; each product \(\lambda_j d_j(t)\) is dimensionless, as required inside the exponential.

For posterior draw \(m\), use its entire hazard vector \(\boldsymbol{\lambda}^{(m)}\) to calculate \(S^{(m)}(t)=S(t\mid\boldsymbol{\lambda}^{(m)})\). Repeating this calculation for \(M\) joint posterior draws gives \(M\) survival values at every time of interest. Posterior means, medians, or quantiles at a specified time are then computed from those values. Keeping the hazard values from each draw together is essential: the hazards across intervals may be dependent in the posterior.

Why the Formula Accumulates Exposure

Within an interval of constant hazard \(\lambda_j\), the probability of surviving an additional duration \(d\), conditional on having survived to the start of that duration, is \(\exp(-\lambda_j d)\). Survival to time \(t\) requires surviving each elapsed interval portion. Multiplying these conditional survival probabilities gives the piecewise formula.

S(t\mid\boldsymbol{\lambda}) =\prod_{j=1}^K \exp\{-\lambda_j d_j(t)\} =\exp\left\{-\sum_{j=1}^K\lambda_jd_j(t)\right\}.

This also shows why uncertainty accumulates across intervals: each draw’s cumulative hazard is a sum of interval-specific hazard contributions. Earlier interval hazards affect survival at all later times, while a later interval hazard has no effect on survival before that interval begins. The posterior uncertainty in the sum depends not only on uncertainty in each hazard but also on their posterior dependence.

Worked Example: Constructing Several Points on a Curve

Suppose follow-up is divided into three intervals: 0–3, 3–6, and 6–12 months. Consider three illustrative, equally weighted posterior draws of the hazards, in inverse months: \((0.04,0.08,0.05)\), \((0.06,0.04,0.07)\), and \((0.03,0.10,0.04)\). These small hypothetical draws make the calculation visible; they are not enough for precise posterior interval estimation.

At 3 months, only the first interval contributes. For the first draw, \(S^{(1)}(3)=\exp[-(0.04)(3)]=\exp(-0.12)=0.8869\). For the second and third draws, the calculations are \(\exp(-0.18)=0.8353\) and \(\exp(-0.09)=0.9139\).

At 6 months, the first two intervals contribute. The first draw gives \(S^{(1)}(6)=\exp[-(0.04)(3)-(0.08)(3)]=\exp(-0.36)=0.6977\). The other draws give \(\exp[-(0.06)(3)-(0.04)(3)]=\exp(-0.30)=0.7408\), and \(\exp[-(0.03)(3)-(0.10)(3)]=\exp(-0.39)=0.6771\).

At 12 months, the third interval contributes its full six months. For the first draw, \(S^{(1)}(12)=\exp[-(0.04)(3)-(0.08)(3)-(0.05)(6)]=\exp(-0.66)=0.5169\). The corresponding values for the second and third draws are \(\exp(-0.72)=0.4868\) and \(\exp(-0.63)=0.5326\).

The posterior means across these three illustrative draws are \(0.8787\) at 3 months, \(0.7052\) at 6 months, and \(0.5121\) at 12 months. The medians are \(0.8869\), \(0.6977\), and \(0.5169\), respectively. For a real analysis, many more posterior draws would be used to estimate summaries and quantiles reliably.

Summarizing Posterior Uncertainty

At each chosen time \(t\), the posterior distribution of \(S(t)\) is represented by the \(M\) values \(S^{(1)}(t),\ldots,S^{(M)}(t)\). Their average estimates the posterior mean survival probability; their median estimates its posterior median. A central pointwise credible interval can be formed from the appropriate lower and upper posterior quantiles at that time. For example, the 95% pointwise interval uses the 2.5th and 97.5th percentiles of the draws of \(S(t)\).

The interpretation is conditional on the model, prior, and observed data: a 95% posterior credible interval at 6 months assigns 95% posterior probability to the interval containing the 6-month survival probability. It is a statement about that time point. A set of such intervals at many time points is not automatically a 95% simultaneous band for the whole curve.

The survival values from a single hazard draw must be used together across time. In each draw, survival starts at one, cannot increase, and is continuous under this piecewise-exponential construction. The hazard may jump at a boundary, so the curve’s slope may change there; the survival probability itself does not jump.

Worked Example: Using Only the Elapsed Part of an Interval

Use intervals 0–2, 2–5, and 5–8 months. At 4 months, the first interval contributes 2 months, the second contributes only 2 months, and the third contributes none. For one hypothetical posterior draw with hazards \((0.04,0.06,0.10)\) per month, \(S(4)=\exp[-(0.04)(2)-(0.06)(2)]=\exp(-0.20)=0.8187\). Using the full 3-month width of the second interval would incorrectly calculate survival at 5 months rather than at 4 months.

At 8 months, all three intervals are complete, so the same draw gives \(S(8)=\exp[-(0.04)(2)-(0.06)(3)-(0.10)(3)]=\exp(-0.56)=0.5712\). For two additional hypothetical draws, \((0.06,0.04,0.08)\) and \((0.03,0.08,0.06)\), the 8-month values are \(\exp(-0.48)=0.6188\) and \(\exp(-0.48)=0.6188\). The mean of these three survival draws at 8 months is \((0.5712+0.6188+0.6188)/3=0.6029\).

The decrease from 4 to 8 months is calculated within each draw by adding the hazard exposure from the remaining part of the second interval and the full third interval. It is not obtained by treating the 8-month survival probability as a new, unrelated estimate.

Dependence Across Intervals Matters

A frequent error is to summarize each interval hazard separately and combine those summaries as if the hazards were independent. Posterior draws generally carry their joint dependence, and the uncertainty of cumulative hazard depends on it. The following deliberately simple example isolates that issue.

Worked Example: Preserving Joint Posterior Draws

Consider two intervals, each two months long, and four equally weighted hypothetical posterior draws of the hazards (in inverse months): \((0.02,0.10)\), \((0.04,0.08)\), \((0.06,0.06)\), and \((0.08,0.04)\). At month 4, the cumulative hazard in every paired draw is \(2\lambda_1+2\lambda_2=0.24\). Thus every draw gives \(S(4)=\exp(-0.24)=0.7866\), and the posterior survival distribution in this illustrative set has no spread at month 4.

If the two marginal hazard samples were incorrectly recombined independently, possible pairs would include \((0.02,0.04)\) and \((0.08,0.10)\). Those produce cumulative hazards \(0.12\) and \(0.36\), and survival values \(\exp(-0.12)=0.8869\) and \(\exp(-0.36)=0.6977\). The extra spread is an artifact of discarding the negative association in the original paired draws. In an actual analysis, calculate each survival value from the hazards in the same joint posterior draw.

Diagnostics, Alternatives, and Limits

Several checks can reveal a calculation or interpretation problem:

Check the endpoints and direction. Each draw should satisfy \(S^{(m)}(0)=1\) and should be non-increasing over time. Values outside \([0,1]\), an increasing curve, or a discontinuity at an interval boundary indicate an error in the exposure calculation or implementation.
Check units and interval widths. Hazards expressed per month must be multiplied by durations in months. Inspect partial intervals carefully, especially when requested times do not match the interval boundaries.
Check the joint posterior computation. If posterior draws are used, assess whether the sampling algorithm has mixed adequately and whether the retained draws are sufficient for stable curve summaries. Instability in tail quantiles can make credible intervals unreliable even when the point estimate looks stable.
Check late follow-up uncertainty. Later survival values accumulate contributions from more intervals. If data provide little information about late hazards, posterior summaries there may be sensitive to the model or prior; a smooth-looking curve does not itself establish precision.

A plug-in curve computed from posterior mean hazards is one alternative summary, but it is generally not the posterior mean survival curve. In general, \(\exp\{-E[H(t)\mid\text{data}]\}\neq E[\exp\{-H(t)\}\mid\text{data}]\), where \(H(t)=\sum_j\lambda_jd_j(t)\) is cumulative hazard. Transforming each draw first and then summarizing preserves the posterior distribution of survival. A frequentist Kaplan–Meier curve is another familiar descriptive approach, but it does not provide this model-based posterior distribution. These approaches answer related but distinct inferential questions.

Piecewise hazards also impose a modeling choice: within each interval the hazard is constant, and its behavior can change only at chosen boundaries. Results may depend on the boundary placement and on the prior and likelihood used to generate the hazard draws. Posterior summaries describe uncertainty under those assumptions; they do not remove the need to assess whether the model is adequate for the trial’s follow-up and event pattern.

Key takeaway: For every joint posterior hazard draw, accumulate the correct interval-specific exposure, transform cumulative hazard to survival, and only then summarize across draws at each time. This yields coherent pointwise posterior summaries while retaining the uncertainty and dependence encoded by the model.

Check Your Understanding

Work through the questions, then compare with the solutions.

For intervals 0–2 and 2–5 months, with hazards \(0.05\) and \(0.10\) per month in one draw, calculate survival at 4 months.
What does a 95% pointwise posterior credible interval at month 6 describe, and what does it not automatically describe?
Why should hazards from the same posterior draw be kept together when calculating survival?
A calculated draw-specific survival curve rises between months 3 and 4. Name a likely calculation issue.

Solutions. (1) The exposures are 2 months in the first interval and 2 months in the second, so \(S(4)=\exp[-(0.05)(2)-(0.10)(2)]=\exp(-0.30)=0.7408\). (2) It describes a posterior interval for survival at month 6, conditional on the model, prior, and data; it is not automatically a 95% simultaneous band covering the entire curve. (3) Joint draws preserve posterior dependence among interval hazards, which affects cumulative-hazard and survival uncertainty. Independently recombining hazards can create a distribution that was not represented by the posterior draws. (4) Survival must be non-increasing in this model. Check for an incorrectly signed cumulative hazard, incorrect interval exposure, or inconsistent ordering of calculated times.