Beta Posteriors for Discrete-Time Survival Hazards

From Interval Hazards to Survival

A discrete-time survival analysis divides follow-up into ordered intervals and describes the event process by a conditional probability in each interval. This is useful when event times are recorded in visits, weeks, or other fixed periods rather than precisely, or when the model is intended to estimate interval-specific rather than continuous-time hazards. The central calculation is simple: for each interval, a beta prior combined with the observed events and non-events gives a beta posterior. Survival through several intervals is then the product of the corresponding conditional survival probabilities.

Definition: For interval \(j\), let \(n_j\) be the number of individuals at risk and observed for the interval, and let \(d_j\) be the number of those individuals who experience the event during it. The discrete-time hazard \(h_j\) is the probability of the event in interval \(j\), conditional on being event-free at its start. Thus \(d_j\) events and \(n_j-d_j\) non-events are observed in that interval. Survival through the first \(k\) intervals, conditional on the hazards, is \(S_k=\prod_{j=1}^k(1-h_j)\).

The hazard is conditional: it is not the proportion of the original trial cohort that has experienced the event by interval \(j\). Someone enters an interval’s risk set only if they have not had the event before that interval. A hazard of 0.10 means a 10% event probability during that interval among those event-free at its start, not a 10% probability of surviving the entire study.

Likelihood and Beta Posterior Update

Suppose that, conditional on the risk set size \(n_j\), each individual’s event status in interval \(j\) is a Bernoulli outcome with event probability \(h_j\). The interval outcomes are conditionally independent given \(h_j\). The event count therefore has a binomial likelihood. For clarity, suppose initially that the hazards in different intervals have independent beta priors, \(h_j\sim\operatorname{Beta}(a_j,b_j)\), with \(a_j>0\) and \(b_j>0\).

d_j\mid h_j,n_j\sim\operatorname{Binomial}(n_j,h_j),\qquad L(h_j)\propto h_j^{d_j}(1-h_j)^{n_j-d_j}.

The beta prior has density proportional to \(h_j^{a_j-1}(1-h_j)^{b_j-1}\) on \(0<h_j<1\). Multiplying prior density by likelihood gives the posterior kernel:

\pi(h_j\mid d_j,n_j)\propto h_j^{a_j-1+d_j}(1-h_j)^{b_j-1+n_j-d_j}.

This is the kernel of a beta distribution, so the posterior update is conjugate. Under independent beta priors across intervals, and the interval-specific binomial likelihoods, the joint posterior also factorizes across intervals. Although the risk set in a later interval depends on earlier outcomes, the observed risk counts are treated as data in this likelihood; the likelihood contribution for each hazard uses its own interval’s events and non-events.

Posterior update: If \(h_j\sim\operatorname{Beta}(a_j,b_j)\) before observing the interval data, then \(h_j\mid\text{data}\sim\operatorname{Beta}(a_j+d_j,\ b_j+n_j-d_j)\). The posterior mean is \((a_j+d_j)/(a_j+b_j+n_j)\). The posterior mean is an estimate of the interval event probability among those at risk, not the probability of the event by the end of follow-up.

Worked Examples

Worked Example: Updating One Interval Hazard

In a hypothetical trial interval, 12 participants enter event-free and are observed through the interval. Three experience the event and nine do not. Use a \(\operatorname{Beta}(2,3)\) prior for the interval hazard.

The assumptions for this update are that the 12 interval outcomes are conditionally independent given the common hazard, and that all 12 have an ascertainable event or non-event status for this interval. The likelihood is proportional to \(h^3(1-h)^9\). Multiplying by the prior kernel \(h^{2-1}(1-h)^{3-1}\) gives:

\pi(h\mid\text{data})\propto h^{2-1+3}(1-h)^{3-1+9}, \qquad h\mid\text{data}\sim\operatorname{Beta}(5,12).

The posterior mean hazard is \(5/(5+12)=5/17=0.2941\). This summarizes the posterior distribution for the event probability in this interval among participants entering it event-free. It is not the observed event proportion \(3/12=0.25\), and neither number is the probability of surviving beyond this interval.

Worked Example: Updating Successive Risk Sets

Consider a hypothetical trial with three discrete follow-up intervals. At the start, 30 participants are at risk. In interval 1, four have the event and three are censored at the interval boundary. Thus 23 enter interval 2. In interval 2, three have the event and two are censored at its boundary, leaving 18 to enter interval 3. In interval 3, two have the event. Use independent \(\operatorname{Beta}(1,1)\) priors for all three hazards.

Censoring at an interval boundary means a participant contributes a known non-event outcome for the interval just completed, then contributes no further interval outcomes. This gives the interval counts \(n_1=30,d_1=4\); \(n_2=23,d_2=3\); and \(n_3=18,d_3=2\). The posterior distributions are:

\begin{aligned} h_1\mid\text{data}&\sim\operatorname{Beta}(1+4,\ 1+30-4)=\operatorname{Beta}(5,27),\\ h_2\mid\text{data}&\sim\operatorname{Beta}(1+3,\ 1+23-3)=\operatorname{Beta}(4,21),\\ h_3\mid\text{data}&\sim\operatorname{Beta}(1+2,\ 1+18-2)=\operatorname{Beta}(3,17). \end{aligned}

In particular, the second posterior’s non-event shape parameter is \(1+23-3=21\). The posterior mean hazards are \(5/32=0.1563\), \(4/25=0.1600\), and \(3/20=0.1500\), respectively.

For survival through all three intervals, use \(S_3=(1-h_1)(1-h_2)(1-h_3)\). Because the posterior hazards are independent, the posterior expectation of this product is the product of their expectations. For a \(\operatorname{Beta}(a,b)\) variable, the mean of \(1-h\) is \(b/(a+b)\). Therefore:

E(S_3\mid\text{data}) =\frac{27}{32}\frac{21}{25}\frac{17}{20} =0.84375(0.84)(0.85) =0.6024.

The posterior mean probability of remaining event-free through all three intervals is about 0.602, or 60.2%. This is an expectation over the posterior distributions of the hazards. A posterior interval for \(S_3\) would additionally describe its uncertainty; the mean alone does not.

Worked Example: How the Prior Affects an Interval Estimate

Suppose a hypothetical interval has 10 participants at risk and one event. Compare a \(\operatorname{Beta}(1,1)\) prior with a \(\operatorname{Beta}(2,8)\) prior. Assume the 10 interval outcomes are observed and satisfy the same binomial assumptions as above.

With the first prior, the posterior is \(\operatorname{Beta}(1+1,1+10-1)=\operatorname{Beta}(2,10)\), whose mean is \(2/12=0.1667\). With the second prior, the posterior is \(\operatorname{Beta}(2+1,8+10-1)=\operatorname{Beta}(3,17)\), whose mean is \(3/20=0.1500\). The data are identical, but the prior distributions differ, so the posterior distributions differ.

This comparison does not establish which prior is appropriate. It shows why the prior must be stated and why sensitivity to plausible prior choices can matter, especially when an interval contains few participants or few events. The beta parameters determine both the center and spread of the prior and posterior distributions.

Reconstructing Survival and Checking the Model

At the end of interval \(k\), survival conditional on the hazards is the product of the probabilities of avoiding the event in each interval so far. For independent beta posteriors, the posterior mean at that endpoint is the product of the posterior means of the interval survival probabilities. However, the full posterior distribution of \(S_k\) is generally not a beta distribution. To describe its uncertainty, one can draw each \(h_j\) from its beta posterior, calculate \(S_k=\prod_{j=1}^k(1-h_j)\) for each set of draws, and summarize the resulting values with posterior quantiles or probabilities.

A short justification for the posterior mean formula is the factorization of expectation. If the hazards are independent under the posterior, then their transformed values \(1-h_j\) are independent too. Repeatedly applying the expectation rule for independent variables gives:

E(S_k\mid\text{data}) =E\left[\prod_{j=1}^{k}(1-h_j)\,\middle|\,\text{data}\right] =\prod_{j=1}^{k}E(1-h_j\mid\text{data}) =\prod_{j=1}^{k}\frac{b_j+n_j-d_j}{a_j+b_j+n_j}.

Here the final expression uses posterior parameters \(a_j+d_j\) and \(b_j+n_j-d_j\). If hazards have a dependent prior, the product-of-expectations step does not generally hold; the joint posterior must be used to calculate survival summaries.

The binomial update requires a clear definition of who was observed for each interval. The examples assume censoring occurs at interval boundaries, so participants censored at the end of an interval are known to be event-free during that interval. If someone is censored partway through an interval, their event status for the whole interval is not known. Treating them automatically as a non-event through the interval can bias the update. Use intervals and likelihood contributions that correctly represent their observed time, or use an appropriate model for censoring and within-interval event times.

Other checks follow directly from the model. Every interval must satisfy \(0\leq d_j\leq n_j\), and risk counts should reconcile with earlier events and censoring. For example, if all losses occur at interval boundaries, then the next risk count equals the current risk count minus the events and boundary censorings. Large changes in estimated hazards across adjacent intervals may reflect a genuine pattern, but can also arise from sparse risk sets. A zero-event interval is valid and yields a posterior, but it does not prove the hazard is zero. A very small risk set makes the posterior especially sensitive to the prior.

Comparison, Limitations, and Common Pitfalls

A discrete-time hazard model estimates a separate conditional event probability for each interval. A piecewise exponential model instead represents event timing with a continuous-time hazard that is constant within specified periods and uses time-at-risk information. The discrete model is natural when the event is only known to have occurred within a period and all interval outcomes are represented appropriately. The piecewise exponential model retains a continuous-time interpretation and can make use of exposure time within intervals. Neither approach is automatically preferable: the observation process and scientific question determine which is suitable.

The beta-binomial update is computationally convenient, but the independence assumptions may be restrictive. It treats participants’ interval outcomes as conditionally independent and uses one common hazard for all individuals in a given interval. It does not by itself adjust for covariates, changing treatment exposure, clustering, competing events, or unobserved heterogeneity. It also estimates hazards only at the chosen interval resolution; changing interval boundaries can change the model and its interpretation.

Common mistake: It is incorrect to use the original cohort size as the denominator in every interval, or to subtract only events when calculating the next risk set. Later hazards are conditional on remaining event-free and under observation at the start of their intervals. For example, in the three-interval example, \(n_2=30-4-3=23\), not 30, and \(n_3=23-3-2=18\). Use each interval’s actual risk set, count its non-events as \(n_j-d_j\), and apply the beta update to those interval-specific counts.

Check Your Understanding

Answer each question using the interval-specific definitions and assumptions in this tutorial.

With a \(\operatorname{Beta}(2,4)\) prior, 15 participants at risk, and 5 events, what is the posterior distribution and posterior mean hazard?
At the start of an interval, 40 participants are at risk; 6 have the event and 4 are censored at the interval boundary. How many enter the next interval?
If the posterior hazards for two intervals are independent \(\operatorname{Beta}(2,8)\) and \(\operatorname{Beta}(3,7)\), what is the posterior mean survival through both intervals?
Why is a participant censored partway through an interval not automatically a known non-event for that full interval?

Solutions. 1. The posterior is \(\operatorname{Beta}(2+5,4+15-5)=\operatorname{Beta}(7,14)\); its mean is \(7/21=0.3333\). 2. There are \(40-6-4=30\) participants entering the next interval. 3. The mean survival is \((8/10)(7/10)=0.56\), by independence and the beta means of \(1-h_j\). 4. They are known to be event-free only up to their censoring time; without observing them through the interval, their event status for the full interval is unknown.