Conditional Median Survival From a Landmark

What Does a Conditional Median Measure?

At a landmark time $L$, a patient is known to be event-free and has accumulated a biomarker history up to that time. A dynamic prediction may report the patient's probability of remaining event-free over a future horizon. A different summary is the conditional median remaining event-free time: the future duration by which the model predicts that the patient's conditional probability of remaining event-free has fallen to one half.

This is a time measured from the landmark, not from the start of follow-up. It depends on the conditioning information: survival to $L$, the biomarker history observed by $L$, and any other covariates included in the prediction. The conditional event-time density after a landmark describes the full distribution of future event times; the median compresses that distribution to one time point.

Definition: Let $T_i$ be the event time for individual $i$, and let $\mathcal H_i(L)$ denote that individual's observed longitudinal history through landmark time $L$. Define the remaining event-free time as $W_i=T_i-L$, conditional on $T_i>L$ and $\mathcal H_i(L)$. Its conditional survival function is $$ C_i(u\mid L,\mathcal H_i(L)) = P\!\left(T_i>L+u\mid T_i>L,\mathcal H_i(L)\right),\qquad u\geq 0. $$ The conditional median remaining event-free time is the lower conditional median $$ m_i(L)=\inf\left\{u\geq 0:C_i(u\mid L,\mathcal H_i(L))\leq \tfrac12\right\}. $$

The lower-median definition makes the target precise even if the survival curve is flat over some interval. When the conditional distribution is continuous and its survival curve decreases through one half, the median is the solution of $C_i(m_i\mid L,\mathcal H_i(L))=0.5$. A finite solution is not guaranteed: if the curve remains above one half indefinitely, the set in the definition is empty and the median is infinite.

From Conditional Survival to a Median

The required input is the model-based conditional survival curve for the individual at the chosen landmark. It must use the same conditioning information at every future time. In a survival model with fixed covariates $x_i$, the landmark conditioning can be written as a ratio:

C_i(u\mid L,x_i) = \frac{S_i(L+u\mid x_i)}{S_i(L\mid x_i)}.

The ratio expresses the probability of surviving past $L+u$, divided by the probability of surviving past $L$. In a joint model, the prediction is conditional on the longitudinal history as well as survival to the landmark; the conditional curve should be obtained using that joint-model conditioning, rather than by substituting an unrelated marginal survival curve. Once the appropriate curve has been calculated, finding the median is a quantile calculation, not a new model-fitting step.

Theorem: Suppose $C_i(u\mid L,\mathcal H_i(L))$ is continuous and nonincreasing for $u\geq0$, starts at 1, and takes a value at or below 0.5 at some finite time. Then the lower median is finite and is the earliest time at which the curve reaches 0.5. If the curve is strictly decreasing near that crossing, the median is the unique solution to $C_i(u\mid L,\mathcal H_i(L))=0.5$.

The justification follows directly from the definition. Continuity and the starting value of 1 mean the curve cannot get to a value at or below 0.5 without first reaching 0.5. Nonincrease means it cannot cross back above 0.5 afterward. If it is strictly decreasing near the crossing, it takes the value 0.5 only once there. If it never reaches or falls below 0.5, the defining set is empty, so no finite median is available.

Worked Example: Constant Conditional Hazard

Suppose a hypothetical patient's conditional hazard after the landmark is constant at $0.20$ per year. Under this assumption, the conditional survival function for $u$ years beyond the landmark is $C(u)=\exp(-0.20u)$. The hazard is nonnegative, so this curve starts at 1 and decreases continuously; it tends to zero, so a finite median exists.

Set the survival probability to one half and solve:

e^{-0.20m}=0.5,\qquad -0.20m=\log(0.5),\qquad m=\frac{\log(2)}{0.20}=3.466\text{ years}.

Thus, conditional on being event-free at the landmark and on the information used in this hypothetical prediction, the model's median remaining event-free time is about 3.466 years. This does not say the patient has a 50% chance of an event at exactly 3.466 years. It says the predicted probability of remaining event-free beyond that duration is 0.5 under the stated model.

Solving When the Curve Has No Closed-Form Median

Many conditional survival functions do not yield a convenient algebraic expression for the median. A reliable general method is to find a time below the crossing and a time at or above it, then narrow the interval. For a continuous nonincreasing curve, if $C(a)>0.5$ and $C(b)\leq0.5$, with $a<b$, the median lies in $(a,b]$. Bisection repeatedly evaluates the curve at the interval midpoint and retains the half-interval that still brackets the crossing.

Worked Example: A Time-Varying Hazard After the Landmark

Consider a hypothetical event-time model with cumulative hazard $H(t)=0.04t^2$, where time is measured in years. For a patient event-free at landmark $L=2$ years, assume there are no additional patient-specific terms in this illustrative calculation. The conditional survival $u$ years after the landmark is the survival from time 2 to time $2+u$:

C(u)=\exp\{-[H(2+u)-H(2)]\} =\exp[-0.04((2+u)^2-2^2)] =\exp(-0.16u-0.04u^2).

The exponent's cumulative-hazard increment is nonnegative and increases with $u$, so $C(u)$ is continuous and decreasing. It tends to zero, and therefore crosses one half. Solving the crossing equation gives:

0.16m+0.04m^2=\log(2),\qquad m=\frac{-0.16+\sqrt{0.16^2+4(0.04)\log(2)}}{2(0.04)} \approx 2.618\text{ years}.

A numerical bracket confirms the crossing. At $u=2.500$, the cumulative-hazard increment is $0.16(2.500)+0.04(2.500)^2=0.6500$, so $C(2.500)=e^{-0.6500}\approx0.5220$. At $u=2.750$, it is $0.16(2.750)+0.04(2.750)^2=0.7425$, so $C(2.750)=e^{-0.7425}\approx0.4759$. Hence the root is between 2.500 and 2.750 years. At the unrounded exact root, the increment is $\log(2)\approx0.6931$, and survival is exactly 0.5. The prediction is a median of about 2.618 years after the two-year landmark.

Specify the target.
Fix the landmark, the event-free condition, the biomarker history, the other covariates, and the time units.

Obtain the conditional curve.
Evaluate the model-based probability of remaining event-free for future durations $u$, using the stated conditioning information.

Check whether a finite crossing is supported.
Verify that the curve starts at 1 and decreases to 0.5 or below at a finite future time.

Solve and report.
Use an analytic solution when available; otherwise bracket and numerically locate the first time the curve reaches 0.5. Report that time from the landmark and in context.

When a Finite Median Does Not Exist

A conditional survival curve need not fall below 0.5. For example, suppose a hypothetical model allows a fraction of individuals never to experience the event, and the conditional survival curve is $C(u)=0.4e^{-0.3u}+0.6$. This curve decreases from 1 toward 0.6, so it never reaches one half. The lower median is therefore infinite, not a finite time obtained by extrapolating a numerical solver.

Worked Example: Checking for a Finite Median

For the hypothetical curve above, the hazard among the event-prone part is positive, and the total curve is continuous and decreasing. Its limit is:

\lim_{u\to\infty}C(u)=0.4\lim_{u\to\infty}e^{-0.3u}+0.6=0.6.

Since every finite-time survival probability is greater than 0.6, it is also greater than 0.5. There is no finite $u$ satisfying the lower-median definition. Reporting “the median is 10 years,” for instance, would be wrong even if the curve at 10 years were close to its limit. The appropriate conclusion is that this model's conditional median is not finite.

A related practical issue is a restricted prediction horizon. If a model has only been evaluated through five years and its conditional survival is still above 0.5 at five years, the median has not been shown to occur within the available horizon. One may report that the median is beyond five years or not estimable within that horizon, but should not claim that it is infinite unless the model supports that conclusion over all future times.

Interpretation, Alternatives, and Diagnostics

The median remaining event-free time is useful when a single time summary is needed and the survival curve has a finite half-survival crossing. It is less sensitive to the far tail of the distribution than the mean remaining event-free time, but it discards information about the curve before and after the crossing. A fixed-horizon risk answers a different question: for a chosen duration $h$, the conditional event risk is $1-C_i(h\mid L,\mathcal H_i(L))$. The conditional event-time density after a landmark gives a distributional description rather than just a quantile. These summaries are alternatives, not interchangeable estimates of one quantity.

The median is also not the absolute event time. If a predicted median is $m_i(L)$ and the landmark is $L$, the corresponding median event time measured from the original time origin is $L+m_i(L)$. Keep the units consistent: if $L$ is recorded in months, both the median duration and the absolute time must be expressed in months before adding them.

Before reporting a median, inspect the conditional survival curve. It should equal 1 at $u=0$, remain between 0 and 1, and be nonincreasing. A value above 1 or a rise in survival with future time usually signals a coding, conditioning, or numerical problem. Check that the crossing is bracketed by evaluations on opposite sides of 0.5, and refine the bracket until the time precision is adequate. If the curve is available only at a coarse time grid, state that the median is approximate to the grid resolution or interpolate using a method justified by the model.

These checks do not establish that the joint model is correct. A poorly specified longitudinal trajectory, event-time process, or association between them can produce a smooth and numerically precise but biased conditional curve. The prediction also depends on the chosen landmark and history. Finally, a point estimate of the median does not by itself quantify uncertainty from estimated model parameters or latent quantities. Report uncertainty separately when available, and do not mistake numerical precision for predictive certainty.

Common Mistake

Wrong reasoning: “The median survival is 2.6 years, so the patient has an event at 2.6 years with probability one half.”
Correction: The conditional survival of 0.5 at 2.6 years means the model predicts a one-half probability of remaining event-free beyond that duration, conditional on the landmark information. It does not assign probability one half to an event occurring at that exact time.
Wrong reasoning: “The median from the landmark is 2.6 years, so the event-time median is 2.6 years from baseline.”
Correction: The duration is measured from the landmark. If the landmark is at time 2 years, the corresponding absolute time is 4.6 years from baseline.
Wrong reasoning: “The survival curve is close to one half at the end of the plotted window, so the median must be there.”
Correction: A finite median requires a supported crossing of 0.5. A curve that remains above 0.5 at the end of the window may have a later crossing, or may never cross at all.

Key takeaway: The conditional median remaining event-free time is the earliest future duration at which the properly conditioned survival curve reaches or falls below one half. Verify the curve, confirm that a finite crossing exists, and report the duration from the landmark.

Check Your Understanding

Try each question, then check the solutions below.

A hypothetical conditional survival curve is $C(u)=e^{-0.35u}$, with $u$ measured in months. Find the conditional median remaining time.
A decreasing conditional survival curve has $C(3)=0.62$ and $C(8)=0.44$. What can you conclude about its median without further evaluations?
A patient reaches a landmark at 6 months and has a predicted median remaining event-free time of 2.4 years. What is the corresponding median time from baseline, in years?
For a model with conditional survival $C(u)=e^{-0.1u}$, where $u$ is in years, determine the median and explain whether it is finite.

Solutions. (1) Set $e^{-0.35m}=0.5$, giving $m=\log(2)/0.35=1.980$ months. (2) The curve is above 0.5 at 3 and below 0.5 at 8, so continuity and monotonicity place the crossing in $(3,8]$; the exact median needs further evaluation. (3) Six months is 0.5 years, so the corresponding time from baseline is $0.5+2.4=2.9$ years. (4) The curve decreases continuously to zero, so the median is finite: $m=\log(2)/0.1=6.931$ years.