Computing Window Risk From Two Survival Predictions

Why Two Survival Predictions Describe a Window

A dynamic prediction often reports the probability that a patient remains event-free at selected future times, conditional on being event-free at a landmark and on the patient’s observed history up to that landmark. A clinical question may instead ask about a particular window: for example, the probability of an event between 12 and 18 months after the landmark. The two endpoint predictions answer that question, provided they use the same conditioning information and event definition.

The key distinction is the population to which the probability refers. One quantity is the probability of experiencing an event in the window among patients who were event-free at the landmark. Another is the probability of experiencing an event in the window conditional on still being event-free at the window’s start. These are different risks and use different calculations.

Definition: Let \(t_0\) be a landmark, \(t_0\leq t_1<t_2\), and let \(\mathcal H_i(t_0)\) denote patient \(i\)’s observed history through the landmark. Write \(G_i(u)=P(T_i>u\mid T_i>t_0,\mathcal H_i(t_0))\) for the conditional survival prediction at future time \(u\), using a fixed fitted model. The window risk relative to the landmark is \(P(t_1<T_i\leq t_2\mid T_i>t_0,\mathcal H_i(t_0))=G_i(t_1)-G_i(t_2)\). The window risk conditional on being event-free at \(t_1\) is \(P(t_1<T_i\leq t_2\mid T_i>t_1,T_i>t_0,\mathcal H_i(t_0))=1-G_i(t_2)/G_i(t_1)\), when \(G_i(t_1)>0\).

Here \(T_i\) is the event time. The convention \((t_1,t_2]\) includes an event at \(t_2\) and excludes one at \(t_1\); for a continuous event-time distribution, the probability of an event at an exact time is ordinarily zero. The prediction \(G_i(u)\) is conditional on survival to \(t_0\), not a population survival probability from the original time origin. This is the setting of conditional survival predictions derived from a joint model.

Deriving the Two Calculations

Among patients event-free at \(t_0\), the event \(T_i>t_1\) includes all those who remain event-free through \(t_2\). The patients who survive past \(t_1\) but not past \(t_2\) are precisely those with an event in \((t_1,t_2]\). Subtracting the probabilities of these nested events gives the landmark-relative window risk:

\begin{aligned} P(t_1<T_i\leq t_2\mid T_i>t_0,\mathcal H_i(t_0)) &=P(T_i>t_1\mid T_i>t_0,\mathcal H_i(t_0))\\ &\quad-P(T_i>t_2\mid T_i>t_0,\mathcal H_i(t_0))\\ &=G_i(t_1)-G_i(t_2). \end{aligned}

For the risk conditional on being event-free at \(t_1\), apply the definition of conditional probability. Given survival to \(t_1\), an event by \(t_2\) is the complement of survival past \(t_2\). The ratio \(G_i(t_2)/G_i(t_1)\) is that conditional probability of surviving past \(t_2\):

P(T_i>t_2\mid T_i>t_1,T_i>t_0,\mathcal H_i(t_0)) =\frac{G_i(t_2)}{G_i(t_1)}.

The condition \(T_i>t_0\) is implied by \(T_i>t_1\), because \(t_1\geq t_0\). Taking the complement therefore gives the conditional-on-window-start risk, \(1-G_i(t_2)/G_i(t_1)\). This ratio should not be confused with the difference: the difference is the probability of the window event in the landmark-survivor group, whereas the ratio-based quantity uses the smaller group that has also survived to \(t_1\).

Worked Example: Risk From 12 to 18 Months After a Landmark

Suppose a fitted joint model gives a hypothetical patient conditional survival prediction of \(G_i(12)=0.86\) and \(G_i(18)=0.71\), with times measured in months after the same time origin. The patient was event-free at landmark \(t_0=6\) months, and both predictions condition on the same history through month 6. Find the event risk during \((12,18]\) among patients event-free at month 6.

The required risk is the difference between the endpoint survival probabilities:

$$ G_i(12)-G_i(18)=0.86-0.71=0.15. $$

Thus the predicted probability of an event from after month 12 through month 18 is 0.15, or 15.0%, among patients event-free at month 6 with the specified history. The calculation is coherent because both survival predictions use the same landmark and history, and survival does not increase between the endpoints.

If instead the question is the risk in that window conditional on remaining event-free at month 12, the calculation is

1-\frac{G_i(18)}{G_i(12)} =1-\frac{0.71}{0.86} =1-0.8256 =0.1744.

That is 17.4% among those who are event-free at month 12. It is larger than 15.0% because it is conditional on surviving to the start of the window, rather than being expressed as a probability in the larger group defined at month 6.

Worked Examples With Different Questions

Worked Example: Interpreting the Risk Set Correctly

For a hypothetical patient, a model gives \(G_i(4)=0.90\) and \(G_i(7)=0.75\), both conditional on being event-free at the same landmark and on the same observed history. A report asks for the chance of an event in \((4,7]\) among patients who were event-free at the landmark. Calculate and interpret it.

Because the requested group is defined at the landmark, use the difference:

P(4<T_i\leq 7\mid T_i>t_0,\mathcal H_i(t_0)) =0.90-0.75=0.15.

The predicted window risk is 15.0% in the landmark-survivor group. The conditional risk among patients additionally known to be event-free at month 4 would instead be

1-\frac{0.75}{0.90} =1-0.8333 =0.1667,

or 16.7%. Neither value is universally preferable: the appropriate one depends on the stated risk set. A report should state that risk set rather than merely label a number “window risk.”

Worked Example: Checking Risks Across Adjacent Windows

For a hypothetical landmark prediction, suppose \(G_i(3)=0.88\), \(G_i(6)=0.70\), and \(G_i(9)=0.56\). Calculate landmark-relative event risks in \((3,6]\), \((6,9]\), and \((3,9]\), then check whether the adjacent-window risks add to the full-window risk.

For the first window, the risk is \(0.88-0.70=0.18\), or 18.0%. For the second it is \(0.70-0.56=0.14\), or 14.0%. For the full window it is \(0.88-0.56=0.32\), or 32.0%. The check is

$$ (0.88-0.70)+(0.70-0.56)=0.18+0.14=0.32=0.88-0.56. $$

The landmark-relative risks add because they refer to disjoint event windows within the same group defined at the landmark. By contrast, risks conditional on being event-free at each window’s start do not add directly. For the second window, that conditional risk is

1-\frac{0.56}{0.70}=0.20.

It is 20.0% among those event-free at month 6, not 14.0% among the original landmark group. Multiplying by the probability of surviving to month 6 recovers the landmark-relative risk: \(0.70(0.20)=0.14\).

Assumptions, Checks, and Alternatives

The arithmetic is simple, but the predictions must be comparable. Before subtracting or taking a ratio, check that both endpoints use the same patient, landmark, observed history, fitted model, time origin, and event definition. If a new biomarker measurement or other information was incorporated for one prediction but not the other, the values do not describe a single survival curve under one conditioning set. Recalculate them consistently before forming a window risk.

A valid event-free survival curve is non-increasing in time and lies between zero and one. Thus, for \(t_1<t_2\), it should satisfy \(1\geq G_i(t_1)\geq G_i(t_2)\geq0\). The difference must lie between zero and one. If \(G_i(t_1)=0\), a conditional-on-\(t_1\) ratio is undefined: the model assigns zero probability to reaching that start time, so the conditional risk is not available from these predictions. Small violations of monotonicity can arise from numerical approximation or a prediction implementation error; do not silently truncate a negative risk without investigating the cause.

The endpoint calculation is an alternative to directly integrating a predicted event-time distribution over the window. For an event-free survival function, both approaches represent the same probability when the predictions are coherent: the decrease in survival across the window is the probability of an event there. The endpoint approach is convenient when the model reports survival predictions at specific horizons. The ratio approach answers a different, also useful question: risk among individuals still event-free when the window begins.

These formulas concern the event represented by \(T_i\) and the model’s event-free survival. If the question concerns one particular cause in the presence of competing events, a survival difference for all events does not by itself give that cause-specific cumulative incidence. The competing-event structure must be handled by an appropriate cumulative-incidence prediction. Also, endpoint point predictions yield a point estimate of window risk, not an uncertainty interval. Uncertainty in model estimates and patient-specific predictions should be propagated when uncertainty intervals are required; subtracting endpoint confidence limits separately is not generally a valid interval calculation.

Common Mistakes

Using a ratio for the wrong risk set. \(1-G_i(t_2)/G_i(t_1)\) is conditional on being event-free at \(t_1\). For risk among patients event-free at the landmark, use \(G_i(t_1)-G_i(t_2)\).
Subtracting predictions with different conditioning information. Endpoint values must belong to the same conditional survival curve. Predictions after different histories cannot be combined as if they were one curve.
Calling the survival probability itself the window risk. \(G_i(t_2)\) is the probability of surviving past \(t_2\) under the conditioning set; it is not the event probability between \(t_1\) and \(t_2\).
Adding conditional interval risks. Risks conditional on surviving to different window starts use different risk sets. Landmark-relative interval risks add across adjacent windows; conditional risks need to be weighted by survival to their respective starts.

Key takeaway: Given conditional survival predictions from one coherent curve, the window event probability among landmark survivors is \(G_i(t_1)-G_i(t_2)\). If the target group is instead those event-free at \(t_1\), use \(1-G_i(t_2)/G_i(t_1)\) and state that conditioning explicitly.

Check Your Understanding

For each question, identify the relevant risk set before calculating.

A model gives \(G_i(5)=0.82\) and \(G_i(8)=0.67\). What is the risk in \((5,8]\) among landmark survivors?
Using the same predictions, what is the risk in \((5,8]\) conditional on being event-free at month 5?
Why must the two endpoint predictions use the same landmark and observed history?
If \(G_i(2)=0.91\), \(G_i(5)=0.78\), and \(G_i(8)=0.62\), what are the landmark-relative risks in \((2,5]\) and \((5,8]\), and what is the risk in \((2,8]\)?
What should be investigated if a prediction gives \(G_i(5)=0.70\) and \(G_i(8)=0.74\)?

Solutions. (1) The landmark-relative risk is \(0.82-0.67=0.15\), or 15.0%. (2) Conditional on being event-free at month 5, it is \(1-0.67/0.82=0.1829\), or 18.3%. (3) Different conditioning histories generally define different survival curves, so their endpoint difference is not the probability for one consistent risk set. (4) The adjacent risks are \(0.91-0.78=0.13\) and \(0.78-0.62=0.16\); the full-window risk is \(0.91-0.62=0.29\), equal to \(0.13+0.16\). (5) Survival has increased over time, contrary to the required non-increasing property of a coherent event-free survival curve; check the prediction inputs, time ordering, and numerical implementation.