Prior Predictive Survival Curves Under Parametric Models

Why Check Prior Predictive Survival Curves?

A prior on survival-model parameters can look reasonable when each parameter is considered separately, yet imply implausible patient outcomes when the parameters are translated into survival probabilities over time. For example, a prior might put substantial weight on a median survival of only a few weeks when the trial concerns a chronic condition, or imply that almost no participants experience the event during the planned follow-up. A prior predictive survival curve makes these implications visible on the clinical time scale.

The curve is a model-based prediction before observing trial outcomes. It is not an estimate of the treatment effect and does not show the probability that a particular parameter value is true. Its purpose here is to examine what survival patterns the prior and the chosen parametric model jointly make plausible.

Definition: Let $T$ be a nonnegative event time and let $\theta$ denote the parameters of a survival model. The conditional survival function is $S(t\mid\theta)=P(T>t\mid\theta)$. If $\theta$ has prior distribution $\pi(\theta)$, the prior predictive survival function is $$ S_{\mathrm{prior}}(t)=P(T>t)=\int S(t\mid\theta)\,\pi(\theta)\,d\theta. $$ It averages the model-specific survival probabilities over the prior; it is the survival function for a future event time after accounting for prior uncertainty in $\theta$.

The integration is often done by simulation. Draw parameters $\theta_1,\ldots,\theta_M$ from the prior and plot the corresponding functions $S(t\mid\theta_m)$. These are prior-drawn curves, each representing one possible parameter setting. Their average at a given time estimates $S_{\mathrm{prior}}(t)$. The spread among the curves shows how much the prior allows survival patterns to differ; it is not the same thing as Monte Carlo error in the estimated average.

Parametric Models and Their Assumptions

The exponential model assumes a constant event hazard over time. With rate $\lambda>0$, measured in inverse time units, its survival function is $S(t\mid\lambda)=\exp(-\lambda t)$. Its median event time is $\log(2)/\lambda$. The model is simple and can be useful as a reference, but a constant hazard may not describe a clinical event whose risk changes with time.

A more flexible example is the Weibull model, parameterized here by scale $\eta>0$ and shape $k>0$. The scale has time units; the shape is dimensionless. Its survival function and median are:

S(t\mid\eta,k)=\exp\left[-\left(\frac{t}{\eta}\right)^k\right], \qquad t_{0.5}=\eta(\log 2)^{1/k}.

The Weibull hazard is $h(t)=\frac{k}{\eta}(t/\eta)^{k-1}$. Thus $k=1$ gives a constant hazard, $k>1$ an increasing hazard, and $k<1$ a decreasing hazard. These are assumptions about the event-time distribution, not conclusions learned from the prior check. Both models require a defined event, a time origin, and a meaningful nonnegative time scale. A survival curve must start at 1, remain between 0 and 1, and never increase over time.

A prior predictive check does not require observed event data, and the definition of survival does not depend on a censoring mechanism. If the eventual trial has censoring, the prior for the event-time model and assumptions about observation or censoring need to be considered appropriately in the trial analysis. For this tutorial, the diagnostic target is the event-time survival pattern implied by the parametric model.

From Parameter Draws to a Predictive Curve

In practice, a prior may be sampled using software, or a small set of values can illustrate the calculation. At each clinically relevant time $t$, evaluate the conditional survival probability under each draw and average. If draws are independent from the prior, this average estimates the integral defining $S_{\mathrm{prior}}(t)$.

Worked Example: Exponential Prior and Survival at Trial Landmarks

Suppose, purely hypothetically, that a trial team uses an exponential model with a prior assigning probability 0.25 to $\lambda=0.02$ per month, probability 0.50 to $\lambda=0.05$ per month, and probability 0.25 to $\lambda=0.10$ per month. Calculate the prior predictive survival probability at 6, 12, and 24 months.

For the exponential model, $S(t\mid\lambda)=\exp(-\lambda t)$. At 6 months, the three conditional survival probabilities are $e^{-0.12}=0.8869$, $e^{-0.30}=0.7408$, and $e^{-0.60}=0.5488$. Weighting by their prior probabilities gives:

S_{\mathrm{prior}}(6) =0.25(0.8869)+0.50(0.7408)+0.25(0.5488) =0.7293.

At 12 months, the conditional probabilities are $e^{-0.24}=0.7866$, $e^{-0.60}=0.5488$, and $e^{-1.20}=0.3010$, so:

S_{\mathrm{prior}}(12) =0.25(0.7866)+0.50(0.5488)+0.25(0.3010) =0.5463.

At 24 months, they are $e^{-0.48}=0.6188$, $e^{-1.20}=0.3010$, and $e^{-2.40}=0.09072$, giving $S_{\mathrm{prior}}(24)=0.3279$. In this hypothetical model, the prior predictive probabilities of remaining event-free are about 72.9% at 6 months, 54.6% at 12 months, and 32.8% at 24 months. Whether those values are plausible depends on the disease, event definition, and intended trial population; the calculation alone does not establish clinical plausibility.

The prior-specific medians are $\log(2)/\lambda$: 34.66, 13.86, and 6.931 months, respectively. These are medians conditional on particular parameter values. The median of the prior predictive event-time distribution is instead the time at which the weighted survival curve reaches 0.5; it need not equal the prior-weighted average of these three medians.

That distinction follows directly from the definition. A median of the mixture solves $S_{\mathrm{prior}}(t)=0.5$, whereas an average of conditional medians averages the separate solutions to $S(t\mid\theta)=0.5$. Averaging solutions to an equation is not generally the same as solving the equation after averaging its left-hand side.

S_{\mathrm{prior}}(t_{0.5,\mathrm{prior}})=0.5, \qquad t_{0.5,\mathrm{prior}}\ \ne\ \int t_{0.5}(\theta)\pi(\theta)\,d\theta \quad\text{in general.}

Checking Weibull Curve Shape and Event-Time Implications

Worked Example: A Prior on Weibull Scale

Consider a hypothetical Weibull model with fixed shape $k=2$. Let the prior assign probability 0.50 to $\eta=6$ months and probability 0.50 to $\eta=12$ months. Calculate the prior predictive survival at 6 and 12 months, and interpret the curve pattern.

At 6 months, $S(6\mid6,2)=\exp[-(6/6)^2]=e^{-1}=0.3679$, while $S(6\mid12,2)=\exp[-(6/12)^2]=e^{-0.25}=0.7788$. Therefore:

S_{\mathrm{prior}}(6)=0.50(0.3679)+0.50(0.7788)=0.5734.

At 12 months, the conditional probabilities are $\exp[-(12/6)^2]=e^{-4}=0.01832$ and $\exp[-(12/12)^2]=e^{-1}=0.3679$. Thus $S_{\mathrm{prior}}(12)=0.50(0.01832)+0.50(0.3679)=0.1931$. The prior therefore implies that about 19.3% of event times exceed 12 months, while half of the parameter-specific curves have only about 1.83% survival at that landmark.

The conditional medians are $6\sqrt{\log 2}=4.995$ months and $12\sqrt{\log 2}=9.991$ months. Since $k=2$, both conditional hazards increase over time. The team should ask whether this accelerated event pattern and the substantial prior weight on a 5-month median fit the clinical setting. If not, changing only the scale prior may not be sufficient: the fixed shape assumption also determines the time pattern of risk.

Prior Predictive Treatment-Arm Curves

For a two-arm illustration, suppose an exponential proportional-hazards model has control rate $\lambda_0$ and treatment log hazard ratio $\beta$. With treatment indicator $a$ equal to 0 for control or 1 for treatment, the arm-specific survival curve is $S_a(t)=\exp[-\lambda_0\exp(\beta a)t]$. This construction assumes a constant hazard within each arm and a constant hazard ratio over time. Priors on both parameters induce prior uncertainty in both curves.

Worked Example: Prior-Derived Curves in Two Arms

Suppose, hypothetically, that $\lambda_0=0.04$ per month is fixed and the prior gives equal probability to $\beta=\log(0.5)$, $0$, and $\log(2)$. These imply treatment-to-control hazard ratios of 0.5, 1, and 2. Calculate treatment-arm survival at 6 and 12 months, averaging over the prior on $\beta$.

At 12 months, the control survival probability is $e^{-0.04(12)}=e^{-0.48}=0.6188$. The three possible treatment rates are $0.02$, $0.04$, and $0.08$ per month. Their 12-month survival probabilities are $e^{-0.24}=0.7866$, $e^{-0.48}=0.6188$, and $e^{-0.96}=0.3829$. The prior predictive treatment-arm probability is:

S_{\mathrm{prior},1}(12) =\frac{0.7866+0.6188+0.3829}{3} =0.5961.

At 6 months, the three treatment probabilities are $e^{-0.12}=0.8869$, $e^{-0.24}=0.7866$, and $e^{-0.48}=0.6188$, so $S_{\mathrm{prior},1}(6)=0.7641$. These are averages over the specified prior; they do not describe a single hazard ratio or imply that treatment has a 59.6% probability of being beneficial. The prior places equal mass on benefit, no difference, and harm. A clinically useful check would assess whether all three implied patterns, including the harm pattern, are credible for the decision context.

Simulation, Diagnostics, and Limitations

For a continuous prior, simulation approximates the defining integral. With prior draws $\theta_m$, calculate $S(t\mid\theta_m)$ at a grid of clinically meaningful times and average across $m=1,\ldots,M$. An estimate of Monte Carlo error at a fixed time is the sample standard deviation of those $M$ values divided by $\sqrt{M}$. This quantifies numerical uncertainty in the estimated average, not uncertainty represented by the prior itself.

Useful checks include whether the curves start at 1 and decrease, whether they remain within $[0,1]$, and whether survival at clinically important landmarks is plausible. Examine implied medians and whether the curves imply events concentrated implausibly early or late in follow-up. In a treatment model, inspect each arm and the range of treatment-to-control patterns allowed by the prior. Check also that units align: a rate specified per year cannot be multiplied directly by time in months without conversion.

A common mistake is to inspect only the prior mean of a parameter. For example, the average of several exponential rates does not generally yield the prior predictive survival curve: the correct calculation averages $\exp(-\lambda t)$, not $\lambda$ first and then applies the exponential function. In general, averaging a nonlinear function of a parameter is not the same as evaluating the function at the average parameter.

E_{\pi}\!\left[\exp(-\lambda t)\right] \ \ne\ \exp\!\left[-E_{\pi}(\lambda)t\right] \quad\text{in general.}

Another failure is to judge a curve without specifying a meaningful endpoint, time origin, or follow-up horizon. A curve can be mathematically valid yet clinically irrelevant if these definitions do not match the trial question. Parametric checks are also model-dependent: a Weibull prior predictive curve cannot reveal that the Weibull family itself is inadequate for the clinical event process. Compare the generated patterns with clinical knowledge and relevant benchmarks, and treat disagreement as a reason to revisit assumptions or prior choices—not as evidence that the observed trial data are wrong.

The main alternative within prior predictive checking is to generate replicated event times or summaries, such as event counts by a follow-up landmark, rather than inspecting the survival functions directly. Such summaries can be useful but may hide whether implausibility comes from early, late, or changing event risk. Curves retain that time pattern, while remaining conditional on the chosen parametric family and prior. Neither representation replaces clinical judgment or validates the model.

Specify the event and time scale.
State what event ends follow-up, define time zero, and choose clinically relevant landmarks.

Draw from the prior.
Generate parameter values using the proposed joint prior, preserving its specified dependence structure.

Calculate the curves.
Evaluate the parametric survival function for each draw and each time on the chosen grid.

Assess clinical implications.
Review survival probabilities, medians, and event-time patterns, then revise assumptions or priors if implications are not credible.

Key takeaway: A prior predictive survival curve averages model-specific survival probabilities over the parameter prior. Check both the average curve and the range of plausible curves, in clinical time units, while remembering that every implication depends on the chosen parametric model.

Check Your Understanding

Answer each question, then compare with the worked solutions below.

In one sentence, distinguish a prior-drawn survival curve from the prior predictive survival function.
Under an exponential model with $\lambda=0.05$ per month, calculate survival at 10 months and the conditional median.
A prior assigns equal probability to exponential rates of 0.02 and 0.08 per month. Calculate the prior predictive survival probability at 10 months.
Why is the prior predictive median generally not the average of the parameter-specific medians?

Solutions. (1) A prior-drawn curve is conditional on one parameter draw; the prior predictive survival function is the average of conditional survival probabilities over the prior. (2) $S(10\mid\lambda)=e^{-0.5}=0.6065$, and the median is $\log(2)/0.05=13.86$ months. (3) $S_{\mathrm{prior}}(10)=0.50e^{-0.2}+0.50e^{-0.8}=0.50(0.8187)+0.50(0.4493)=0.6340$. (4) The prior predictive median solves the equation that the averaged survival probability equals 0.5; averaging the conditional medians generally does not solve that equation.