Why Predictions Need to Be Updated
A dynamic prediction is conditional on the information available at a particular time. As follow-up continues, that information changes: a patient may remain event-free and provide another biomarker measurement. A prediction made using the earlier history is therefore not simply carried forward unchanged. The new survival information and the new measurement both affect which latent patient profiles remain plausible, and hence can change the predicted survival probability.
The key operation is sequential conditioning. At each visit, start with the distribution of the patient’s latent characteristics given the history so far. Update it for survival since the previous visit and for the new biomarker result. Then use the updated distribution to calculate survival beyond the new landmark. This procedure produces a sequence of predictions, each based only on information available at its landmark.
The Recursive Update
Assume that, conditional on \(b_i\) and \(x_i\), the joint model specifies a longitudinal measurement density \(f(y_{ij}\mid b_i,x_i,s_j)\) and a survival function \(S_i(t\mid b_i,x_i)=P(T_i>t\mid b_i,x_i)\). The shared random effects allow the biomarker trajectory and event risk to be associated. The update below uses the joint model’s conditional measurement density and survival function; it does not replace either model.
For this sequential calculation, assume the visit schedule is fixed or otherwise ignorable for the model, that the patient’s biomarker is observed only while the patient is event-free, and that measurement errors are conditionally independent across visits given the latent effects and covariates. Also assume the event and biomarker processes are conditionally independent given \(b_i\) and \(x_i\), as encoded by the specified joint model. If these assumptions do not hold, the update needs to account for the relevant dependence or observation process.
Write \(\pi_{i,k-1}(b_i)\) for the distribution of the latent effects given the patient’s history through \(s_{k-1}\), including survival through \(s_{k-1}\). For a patient who remains event-free to \(s_k\) and has observed \(Y_{ik}=y_{ik}\), the posterior density of the latent effects is proportional to the previous density multiplied by two new pieces of information: survival from \(s_{k-1}\) to \(s_k\), and the new measurement’s likelihood.
The survival factor is a conditional probability: given survival through \(s_{k-1}\) and latent effects \(b_i\), the probability of surviving through \(s_k\) is \(S_i(s_k\mid b_i,x_i)/S_i(s_{k-1}\mid b_i,x_i)\). The denominator normalizes the updated density so its integral is one. If no biomarker result is observed at visit \(s_k\), the measurement-density factor is omitted only when missingness is ignorable under the model; survival information can still update the latent-effects distribution.
Once the distribution has been updated, calculate survival beyond the new landmark by averaging the latent-effect-specific conditional survival probabilities over that distribution:
Why the Recursion Is Valid
The update follows from Bayes’ rule and the conditional-independence assumptions. Given the history through \(s_{k-1}\), the new observation is that the patient survived the next interval and then had measurement \(y_{ik}\). Conditional on \(b_i\) and \(x_i\), the probability contribution from survival is the ratio of survival functions above, and the contribution from the measurement is its conditional density. Multiplying these factors by the existing posterior gives an unnormalized density for \(b_i\); dividing by its integral gives the posterior.
For the prediction, condition first on \(b_i\). Given survival through \(s_k\), the probability of surviving to \(u\) is \(S_i(u\mid b_i,x_i)/S_i(s_k\mid b_i,x_i)\). The patient’s \(b_i\) is not known, so average that probability over its posterior distribution given the complete history and survival to the landmark. This gives the stated integral. It is important that this is an average of conditional survival probabilities, not generally the survival probability evaluated at an average random effect.
Worked Sequential Updates
Worked Example: Two Measurements and Two Successive Predictions
Hypothetical data. Suppose the latent profile is either high-risk \(H\) or low-risk \(L\). Before the first measurement, the probabilities are \(P(H)=0.40\) and \(P(L)=0.60\). A positive first biomarker result has likelihood 0.80 under \(H\) and 0.20 under \(L\). The patient is then followed to visit \(s_1\). Conditional on profile, the probabilities of surviving from time 0 to \(s_1\) are 0.70 for \(H\) and 0.90 for \(L\). If the patient survives, a positive second measurement has likelihood 0.75 under \(H\) and 0.25 under \(L\). The survival probabilities from \(s_1\) to \(s_2\) are 0.65 for \(H\) and 0.88 for \(L\).
First update, using the first measurement. The unnormalized weights for \(H\) and \(L\) are \(0.40(0.80)=0.320\) and \(0.60(0.20)=0.120\). Their sum is \(0.440\), so
The predicted probability of surviving to \(s_1\), given this measurement, is \(0.7273(0.70)+0.2727(0.90)=0.7545\). This prediction averages the two profile-specific survival probabilities using the posterior weights after the first measurement.
Second update, using survival and the new measurement. The patient is observed to survive to \(s_1\), so the profile weights are first multiplied by the survival probabilities for that interval. The positive second measurement then contributes its likelihood:
The total is \(0.4432\), giving \(P(H\mid T_i>s_1,Y_{i0},Y_{i1})=0.3818/0.4432=0.8615\), and the corresponding probability for \(L\) is \(0.1385\). The predicted probability of surviving from \(s_1\) to \(s_2\) is therefore \(0.8615(0.65)+0.1385(0.88)=0.6819\). In this hypothetical example, the second positive result increases the high-risk weight, while survival to \(s_1\) had already updated the weights before that result was used.
Worked Example: Survival and a Measurement Can Pull in Different Directions
Hypothetical data. At a visit \(s_{k-1}\), suppose the current posterior probabilities are \(P(H\mid\text{history})=0.60\) and \(P(L\mid\text{history})=0.40\). The patient survives to \(s_k\), with interval survival probabilities 0.80 for \(H\) and 0.95 for \(L\). The new biomarker result is more likely under \(L\): its likelihood is 0.30 under \(H\) and 0.70 under \(L\). For the next interval, the survival probabilities are 0.70 for \(H\) and 0.90 for \(L\).
The unnormalized updated weights are \(0.60(0.80)(0.30)=0.144\) for \(H\) and \(0.40(0.95)(0.70)=0.266\) for \(L\). The normalizing total is \(0.410\), so the updated high-risk probability is \(0.144/0.410=0.3512\); the low-risk probability is \(0.6488\). The predicted survival probability over the next interval is
The survival information and the measurement both reduce the high-risk weight in this example. To see why both factors matter, omit survival and update using only the biomarker: \(0.60(0.30)/\{0.60(0.30)+0.40(0.70)\}=0.3913\). That is not the correct weight after the patient has also survived the interval. The correct sequential update uses all new information, not just the measurement.
Worked Example: An Ignorably Missing Measurement
Hypothetical data. At a landmark, suppose the current posterior probabilities are \(0.50\) for \(H\) and \(0.50\) for \(L\). The patient remains event-free through the next scheduled visit, but the biomarker result is missing. Assume the missingness is ignorable and the survival probabilities over that interval are 0.60 for \(H\) and 0.90 for \(L\). For the following interval, the survival probabilities are 0.50 for \(H\) and 0.85 for \(L\).
Because there is no observed measurement likelihood to include, update using survival alone. The unnormalized weights are \(0.50(0.60)=0.300\) and \(0.50(0.90)=0.450\), with total \(0.750\). Thus the updated probabilities are \(P(H\mid\text{updated history})=0.300/0.750=0.4000\) and \(P(L\mid\text{updated history})=0.6000\). The next-interval prediction is \(0.4000(0.50)+0.6000(0.85)=0.7100\).
The missing measurement does not act like a neutral or average-valued result; it contributes no measurement likelihood under the stated ignorable-missingness assumption. If the chance of a result being missing depends on the unobserved result or latent risk, omitting a missingness model may instead produce a biased update.
Diagnostics, Alternatives, and Limitations
Sequential calculation is equivalent to updating from the full history at once when the same model and assumptions are used. In a full-history calculation, the latent-effects density is proportional to the prior density multiplied by every observed measurement likelihood and by the survival probability through the current landmark. In the recursion, those same factors enter one visit at a time. This equivalence is a useful check: the final weights should not depend on whether the analyst computes them recursively or in a single batch.
In practice, the integral over \(b_i\) may be approximated using quadrature or a set of weighted samples. With weighted samples, each particle’s weight is multiplied by its survival probability over the new interval and its measurement likelihood, then the weights are normalized. These are computational ways to implement the same update, not different definitions of the prediction. A simpler plug-in calculation using one estimated latent effect may be easier, but it does not generally equal the average over uncertainty in \(b_i\).
Several checks help identify errors:
- Check the landmark conditioning. A prediction from \(s_k\) is conditional on survival through \(s_k\); use a conditional survival ratio, not an unconditioned survival probability from time zero.
- Check that follow-up is counted once. If the current posterior already includes survival through \(s_{k-1}\), multiply only by survival from \(s_{k-1}\) to \(s_k\). Multiplying again by survival from time zero double-counts earlier follow-up.
- Check the direction of each likelihood. A measurement more likely under one latent profile should increase that profile’s weight, all else equal. A low marker value is not automatically low risk; the measurement model determines its likelihood.
- Check the visit and event ordering. A biomarker result observed at a visit requires the patient to be event-free at that visit. Do not use a result recorded after the event as though it were available at the landmark.
- Check numerical normalization. Updated weights must be nonnegative and sum to one; the predicted conditional survival probability must lie between zero and one.
A landmark approach that models risk from the information observed at each visit is a possible alternative to the joint-model recursion. It can be useful, but it does not automatically use a shared latent trajectory and survival process in the same way. The joint-model calculation is also limited by its assumptions: misspecified biomarker or survival models, poorly estimated association, informative measurement timing or missingness, and sparse measurements can all undermine calibration. Sequential updating organizes the available information; it cannot correct a model that describes that information poorly.
Check Your Understanding
Answer each question, then compare with the brief solutions below.
- A patient’s latent profile is \(H\) or \(L\), with current weights 0.30 and 0.70. The patient survives the next interval with probabilities 0.60 under \(H\) and 0.90 under \(L\). There is no new measurement. What are the updated weights?
- Why is the survival factor for an update from \(s_{k-1}\) to \(s_k\) a ratio of survival probabilities?
- Suppose a new biomarker likelihood is 0.80 under \(H\) and 0.20 under \(L\). Holding the prior weights and survival information fixed, which profile’s posterior weight will the measurement tend to increase?
- What is wrong with updating a posterior that already conditions on survival through \(s_{k-1}\) by multiplying by \(S_i(s_k\mid b_i,x_i)\) rather than the conditional survival probability for the next interval?
Solutions. (1) The unnormalized weights are \(0.30(0.60)=0.18\) and \(0.70(0.90)=0.63\), totaling 0.81. The updated weights are \(0.18/0.81=0.2222\) for \(H\) and \(0.63/0.81=0.7778\) for \(L\). (2) Conditional on having survived through \(s_{k-1}\), the probability of surviving through \(s_k\) is \(P(T_i>s_k\mid T_i>s_{k-1},b_i,x_i)=S_i(s_k\mid b_i,x_i)/S_i(s_{k-1}\mid b_i,x_i)\). (3) The measurement tends to increase the weight for \(H\), because its likelihood is larger under \(H\). (4) The full survival probability includes survival up to \(s_{k-1}\), which is already represented in the current posterior. Using it again counts earlier survival twice; use the ratio for the new interval.