Piecewise Exponential Posterior Survival Curves

Why Use Interval-Specific Hazards?

A single exponential survival model assumes the event hazard stays constant for all follow-up. That can be too restrictive when clinical reasoning suggests the event rate may change over time. A piecewise exponential model keeps the hazard constant within each interval but allows it to differ between intervals. It is a useful way to describe changing risk without requiring a smoothly varying hazard curve.

The interval boundaries should be specified before examining the outcome data, or chosen using information independent of those outcomes. The analysis then uses, for each interval, the number of events and the total time participants were at risk there. Right-censored participants contribute exposure up to their censoring times, not an event.

Definition: Let $0=\tau_0<\tau_1<\cdots<\tau_J$ be fixed interval boundaries, and let $\lambda_j$ be the event hazard in interval $j$, measured in events per unit time. The piecewise exponential model assumes $\lambda(t)=\lambda_j$ for $\tau_{j-1}\le t<\tau_j$. For a participant followed from time zero, the survival probability conditional on the hazards is $S(t\mid\lambda_1,\ldots,\lambda_J)=\exp\{-\sum_{j=1}^J\lambda_j u_j(t)\}$, where $u_j(t)=\max\{0,\min(t,\tau_j)-\tau_{j-1}\}$ is the time accumulated in interval $j$ by time $t$.

The event-time likelihood assumes that the hazard is constant within each interval. It also assumes censoring is noninformative for this likelihood: conditional on the modeled information, censoring does not reveal additional information about a participant’s event time. Let $d_j$ be the number of observed events in interval $j$, and $Y_j$ the total person-time at risk in that interval. With these assumptions, the hazard-dependent likelihood is proportional to $\prod_{j=1}^J \lambda_j^{d_j}\exp(-\lambda_jY_j)$. The exposure $Y_j$ is accumulated across participants, so it need not equal the calendar width of the interval.

Gamma Updating Within Each Interval

Use the shape-rate parameterization of the gamma distribution: $\lambda_j\sim\operatorname{Gamma}(a_j,b_j)$ has density proportional to $\lambda_j^{a_j-1}\exp(-b_j\lambda_j)$, with shape $a_j>0$ and rate $b_j>0$. Here $b_j$ has units of time, so that $b_j\lambda_j$ is dimensionless. Suppose the interval hazards have independent gamma priors. Multiplying the prior density by that interval’s likelihood gives the posterior update below.

Posterior update: If $\lambda_j\sim\operatorname{Gamma}(a_j,b_j)$ independently across intervals, and interval $j$ contributes $d_j$ events and $Y_j$ units of person-time, then $\lambda_j\mid\text{data}\sim\operatorname{Gamma}(a_j+d_j,b_j+Y_j)$. The posterior hazards remain independent under this likelihood and prior specification. The posterior mean hazard in interval $j$ is $(a_j+d_j)/(b_j+Y_j)$.

The update follows directly from the density. The prior contributes a factor $\lambda_j^{a_j-1}\exp(-b_j\lambda_j)$, and the likelihood contributes $\lambda_j^{d_j}\exp(-Y_j\lambda_j)$. Their product is proportional to $\lambda_j^{a_j+d_j-1}\exp\{-(b_j+Y_j)\lambda_j\}$, the kernel of a gamma distribution with shape $a_j+d_j$ and rate $b_j+Y_j$. The interval-specific update is therefore the same conjugate calculation as for an exponential hazard, applied separately to each interval.

Worked Example: Updating One Interval’s Hazard

In a hypothetical trial, the first prespecified interval is 0 to 4 months. Suppose its event count is $d_1=3$, and participants contribute $Y_1=6$ patient-months at risk. Use the prior $\lambda_1\sim\operatorname{Gamma}(2,4)$, with the second parameter a rate.

The assumptions for this update are that the hazard is constant during these 4 months, the event and exposure summaries correctly account for observed follow-up and censoring, and the gamma prior is appropriate for this interval’s hazard. The posterior is

\lambda_1\mid\text{data}\sim \operatorname{Gamma}(2+3,\,4+6) = \operatorname{Gamma}(5,10).

Thus the posterior mean hazard is $5/10=0.5000$ events per patient-month. This is a summary of the posterior hazard, not a probability of an event. For reference, the posterior mean is obtained by dividing the shape by the rate.

Survival at a Time Within or Across Intervals

Given fixed hazards, survival through several intervals is the exponential of the negative cumulative hazard. Under posterior uncertainty, survival is itself a random quantity because it depends on the unknown hazards. Its posterior mean is found by averaging conditional survival over the posterior distribution; it is not generally equal to conditional survival evaluated at posterior mean hazards.

Posterior expected survival: If $\lambda_j\mid\text{data}\sim\operatorname{Gamma}(\alpha_j,\beta_j)$ independently, where $\alpha_j=a_j+d_j$ and $\beta_j=b_j+Y_j$, then $$ E\{S(t)\mid\text{data}\} = \prod_{j=1}^J \left(\frac{\beta_j}{\beta_j+u_j(t)}\right)^{\alpha_j}. $$ This is the posterior mean of the survival probability at time $t$.

To justify the formula, write $S(t)=\prod_j\exp\{-u_j(t)\lambda_j\}$. Independence of the posterior hazards makes the expectation of this product equal the product of the expectations. For a gamma random variable with shape $\alpha$ and rate $\beta$, integrating its density gives

E\{\exp(-u\lambda)\} = \frac{\beta^\alpha}{\Gamma(\alpha)} \int_0^\infty \lambda^{\alpha-1}\exp\{-(\beta+u)\lambda\}\,d\lambda = \left(\frac{\beta}{\beta+u}\right)^\alpha.

The last equality follows from the gamma integral, $\int_0^\infty x^{\alpha-1}e^{-cx}\,dx=\Gamma(\alpha)/c^\alpha$ for $c>0$. Applying it in each interval gives the stated posterior expected survival.

Worked Example: Survival at a Time Crossing an Interval Boundary

Consider a separate hypothetical analysis with intervals 0 to 3 months and 3 to 7 months. In the first interval, the prior is $\operatorname{Gamma}(2,4)$, with $d_1=3$ events and $Y_1=6$ patient-months. In the second, the prior is $\operatorname{Gamma}(1.5,3)$, with $d_2=2$ events and $Y_2=5$ patient-months. The piecewise-constant hazard and noninformative-censoring assumptions apply in both intervals, and the interval priors are independent.

Updating gives $\lambda_1\mid\text{data}\sim\operatorname{Gamma}(5,10)$ and $\lambda_2\mid\text{data}\sim\operatorname{Gamma}(3.5,8)$. At month 5, a participant has accumulated 3 months in the first interval and 2 in the second, so $u_1(5)=3$ and $u_2(5)=2$. Therefore,

E\{S(5)\mid\text{data}\} = \left(\frac{10}{10+3}\right)^5 \left(\frac{8}{8+2}\right)^{3.5} \approx 0.2693\times0.4579 \approx 0.1233.

The posterior mean hazards are $5/10=0.5000$ and $3.5/8=0.4375$ events per patient-month. Substituting those means into the conditional survival formula instead gives $\exp\{-(3)(0.5000)-(2)(0.4375)\}=\exp(-2.375)\approx0.0930$. This is survival evaluated at the posterior mean hazards, not posterior expected survival. The distinction matters because survival is a nonlinear function of the hazards.

Worked Example: Checking the Calculation at the End of Two Intervals

In another hypothetical trial, the prespecified intervals are 0 to 6 months and 6 to 9 months. Suppose the first interval has prior $\operatorname{Gamma}(2,44)$, three events, and 6 patient-months of exposure. The second has prior $\operatorname{Gamma}(1,37)$, three events, and 3 patient-months of exposure. These exposure totals and event counts are assumed to come from correctly recorded follow-up under the piecewise exponential and noninformative-censoring assumptions.

The posterior distributions are $\lambda_1\mid\text{data}\sim\operatorname{Gamma}(5,50)$ and $\lambda_2\mid\text{data}\sim\operatorname{Gamma}(4,40)$. At month 9, the interval contributions are 6 and 3 months. Hence

\begin{aligned} E\{S(9)\mid\text{data}\} &= \left(\frac{50}{50+6}\right)^5 \left(\frac{40}{40+3}\right)^4\\ &= \left(\frac{50}{56}\right)^5 \left(\frac{40}{43}\right)^4\\ &\approx 0.5674\times0.7488\\ &\approx 0.4249. \end{aligned}

Thus the posterior mean survival probability at 9 months is approximately 0.4249. This is an average over posterior uncertainty in both hazards. The posterior mean hazards are $5/50=0.1000$ and $4/40=0.1000$ per month, so survival evaluated at those means is $\exp\{-(6)(0.1)-(3)(0.1)\}=\exp(-0.9)\approx0.4066$, a different quantity.

Using and Checking the Curve

The formula gives posterior expected survival at any chosen time by using the portion of each interval elapsed by that time. Before the first boundary, only the first hazard contributes. At a boundary, the completed interval contributes its full width, while the next interval contributes zero. After the final modeled boundary, the calculation requires an explicit model for what happens next; the piecewise model alone does not determine survival beyond its specified intervals.

A posterior interval for survival can be obtained by drawing hazards from their independent gamma posteriors, calculating $S(t)$ for each draw, and taking posterior quantiles of those survival values. Such an interval describes posterior uncertainty about survival at that time. The posterior mean formula above gives a point summary, not an uncertainty interval.

Several checks help identify problems before interpreting a result:

Exposure accounting: $Y_j$ must sum participants’ at-risk time in interval $j$, including appropriate contributions from censored participants. Check interval boundaries and units; mixing years with months changes the hazard scale.
Event allocation: an event belongs to the interval containing its event time. Check that the event counts and exposure summaries use the same prespecified boundaries.
Survival behavior: conditional survival cannot increase with time because cumulative hazard is nondecreasing. A calculation that increases across intervals usually has a sign, boundary, or indexing error.
Model adequacy: if event rates appear to change substantially within an interval, the constant-hazard assumption may be too crude. Sparse data in an interval can also leave its posterior strongly influenced by its prior.

Model Choice and Common Pitfalls

Compared with a single exponential model, the piecewise exponential model allows different hazards across intervals; the single exponential model is the special case in which all interval hazards are constrained to be equal. Compared with a Weibull model with fixed shape, a piecewise model can represent abrupt changes at chosen boundaries rather than requiring a smooth hazard pattern. In return, its estimates can depend on the placement of those boundaries, and adding intervals can leave little exposure or few events in some intervals. The model is most useful when interval boundaries have a clinical rationale and the data support estimating interval-specific hazards.

A common mistake is to calculate posterior mean survival as $\exp\{-\sum_j u_j(t)E(\lambda_j\mid\text{data})\}$. That is survival at the posterior mean hazards, not the posterior expectation of survival. The correct calculation averages $\exp\{-\sum_j u_j(t)\lambda_j\}$ over the joint posterior, producing the gamma-transform formula above. Another error is to treat a censored participant as having an event at the censoring time; censoring contributes time at risk but not an event count.

Key takeaway: Update each interval’s gamma prior with that interval’s event count and person-time, then combine the independent posterior hazards through cumulative hazard. To report posterior expected survival, average survival over the hazard posterior rather than simply substituting posterior mean hazards.

Check Your Understanding

Answer each question, then compare with the solutions.

A hazard has prior $\operatorname{Gamma}(1,5)$. The interval contributes 2 events and 4 patient-months. What is the posterior distribution and posterior mean hazard?
For posterior hazard $\operatorname{Gamma}(3,12)$, calculate posterior expected survival through 2 months within its interval.
At time $t$, a participant has spent 4 months in interval 1 and 1 month in interval 2. The posterior hazards are independent $\operatorname{Gamma}(2,10)$ and $\operatorname{Gamma}(3,20)$. Write the expression for posterior expected survival at $t$.
Why does a censored participant contribute to $Y_j$ but not to $d_j$?

Solutions. 1. The posterior is $\operatorname{Gamma}(1+2,5+4)=\operatorname{Gamma}(3,9)$; its mean is $3/9=0.3333$ events per patient-month. 2. The expectation is $(12/(12+2))^3=(6/7)^3\approx0.6297$. 3. It is $(10/(10+4))^2(20/(20+1))^3\approx0.5102\times0.8638\approx0.4407$. 4. Censoring establishes that the participant remained at risk until the censoring time, so that time contributes exposure; it does not establish that the event occurred, so it adds nothing to the event count.