Why Loss to Follow-Up Changes an Event Projection
A survival-based event projection counts participants who are expected to experience the event by a given follow-up time. But an event contributes to an observed event total only if it is ascertained. When a participant is lost to follow-up and no later event information is available, an event occurring after that loss will not be counted. The projection must therefore account for both the event process and the period during which the participant remains observable.
This adjustment depends on what “loss to follow-up” means operationally. If event ascertainment stops at withdrawal, loss can reduce the observed event count. If a registry or other source continues to capture events after withdrawal, the projection should reflect that continued ascertainment instead. The two situations do not have the same event probability.
The formula assumes that dropout prevents subsequent event ascertainment, that the event and dropout times are independent under the projection model, and that the event hazard and dropout distribution used are appropriate for the population being projected. It also assumes the same follow-up horizon \(t\) for each participant in the calculation. If participants have different entry times or follow-up opportunities, the probability must be evaluated for each opportunity and combined appropriately; this tutorial focuses on the loss-to-follow-up adjustment for a specified opportunity.
For continuous event and dropout hazards, write the dropout hazard as \(\mu_C(u)\). Then \(G_C(u)=\exp\{-\int_0^u\mu_C(v)\,dv\}\). If the event hazard is also continuous, \(S_T(u)=\exp\{-\int_0^u\lambda_T(v)\,dv\}\). These expressions make the interpretation explicit: at time \(u\), an observed event requires the participant to have remained event-free and under observation until just before \(u\), followed by an event at \(u\).
Deriving the Adjustment for Constant Hazards
When the event hazard is constant at \(\lambda\) and the dropout hazard is constant at \(\mu\), and both processes are independent, the event and dropout times have exponential survival distributions. At time \(u\), the probability of still being event-free and observed is \(\exp\{-(\lambda+\mu)u\}\). Multiplying by the instantaneous event rate and integrating gives:
By independence, \(S_T(u)G_C(u)=e^{-\lambda u}e^{-\mu u}=e^{-(\lambda+\mu)u}\).
At each time \(u\), multiply the probability of reaching \(u\) event-free and observed by the event hazard \(\lambda\).
For \(t\geq 0\), \(p_{\mathrm{obs}}(t)=\int_0^t \lambda e^{-(\lambda+\mu)u}\,du=\frac{\lambda}{\lambda+\mu}\{1-e^{-(\lambda+\mu)t}\}\).
The result has two useful checks. If \(\mu=0\), there is no loss to follow-up and it reduces to \(1-e^{-\lambda t}\), the ordinary event probability by time \(t\). If \(\mu>0\), then \(p_{\mathrm{obs}}(t)\leq 1-e^{-\lambda t}\): some events that would occur by \(t\) are not observed because follow-up ended first. The dropout hazard is not an event hazard; adding \(\mu\) in the exponential accounts for the chance of leaving observation before an event, not for additional events.
Worked Example: Piecewise Event Hazards With Continuous Dropout
Suppose a hypothetical trial has 300 participants, each with two years of potential follow-up. Assume event and dropout times are independent. The event hazard is \(0.10\) per year during year 1 and \(0.05\) per year during year 2. The dropout hazard is \(0.04\) per year throughout. Loss to follow-up ends event ascertainment. Project the expected number of observed events by two years.
During year 1, the event hazard and dropout hazard are constant. The probability of an observed event during this interval is $$ p_1=\int_0^1 0.10e^{-(0.10+0.04)u}\,du =\frac{0.10}{0.14}(1-e^{-0.14}) \approx 0.09331555. $$
For an event during year 2, a participant must first have remained event-free and observed through year 1. That probability is \(e^{-0.10}e^{-0.04}\). Conditional on reaching year 2, the probability contribution from an event during that year is found by integrating its event hazard against the combined event-free and observation survival over the one-year interval: $$ p_2=e^{-0.10}e^{-0.04}\frac{0.05}{0.05+0.04}(1-e^{-0.09}) \approx 0.04156924. $$
Adding the unrounded interval contributions before rounding gives $$ p_{\mathrm{obs}}(2)\approx 0.09331555+0.04156924=0.13488479, \qquad 300(0.13488479)\approx 40.47\text{ events}. $$
Thus, under these assumptions, the projected expected observed event count is about 40.47 by two years. This is an expectation, not a guarantee that the realized count will be 40 or 41. The calculation also assumes that the stated hazards apply throughout the relevant periods and that dropout does not depend on the participant’s event time.
When Event Ascertainment Continues After Withdrawal
A recorded withdrawal date does not necessarily mark the end of event observation. For example, a participant may stop attending study visits while events continue to be captured through a registry. In that case, treating withdrawal as \(C\) and multiplying by \(G_C(u)\) would undercount events if registry ascertainment actually continues. The relevant observation-ending time is the time at which event information truly becomes unavailable, not necessarily the time of treatment discontinuation or study-visit withdrawal.
If events remain ascertainable for everyone through \(t\), the event probability is \(1-S_T(t)\), even if some participants withdraw from visits. If only a fraction remain observable after withdrawal, the model must describe that ascertainment process; one cannot simply assume either complete loss or complete continued follow-up.
Worked Example: Withdrawal With and Without Registry Follow-Up
In a hypothetical cohort of 200 participants, suppose the event hazard is \(0.10\) per year and event times are independent of withdrawal. At one year, 20% withdraw and have no further event ascertainment; the other 80% remain under observation through year 2. Compare the expected number of observed events by year 2 with the projection if a registry continued to capture events for everyone after withdrawal.
For the 80% observed through two years, the probability of an event by year 2 is \(1-e^{-0.10(2)}\). For the 20% whose observation ends at year 1, the observed-event probability is \(1-e^{-0.10(1)}\). Therefore, $$ p_{\mathrm{obs}}(2) =0.80(1-e^{-0.20})+0.20(1-e^{-0.10}) \approx 0.80(0.181269)+0.20(0.0951626) \approx 0.164048. $$ The expected observed count is \(200(0.164048)\approx 32.81\) events.
If a registry instead continued to ascertain events for all participants through year 2, withdrawal would not end event observation. The probability would be \(1-e^{-0.20}\approx0.181269\), giving \(200(0.181269)\approx36.25\) expected events. The difference, about 3.44 events, comes from the ascertainment process, not from changing the event hazard.
Comparing Projections and Checking the Assumptions
A projection that ignores dropout uses the event probability \(1-S_T(t)\). An adjusted projection uses the chance of still being observed at each possible event time. These quantities answer different questions: the first is the probability that the event occurs by \(t\), while the second is the probability that the event occurs by \(t\) and is observed before follow-up ends.
| Projection approach | What it assumes | When it applies |
|---|---|---|
| Event risk without a dropout adjustment | Event ascertainment continues through the full horizon for everyone, or dropout is negligible. | Use when events remain captured after withdrawal or loss is not relevant to the projection. |
| Independent dropout adjustment | Dropout stops event ascertainment, and dropout time is independent of event time under the model. | Use when the dropout process and event survival model are specified and appropriate. |
| Model with dependence or partial ascertainment | The relation between dropout and event risk, or post-withdrawal ascertainment, is explicitly represented. | Needed when the independent-dropout model or complete-loss assumption is not credible. |
Before using the adjusted calculation, check that its inputs describe the trial’s observation process. Useful diagnostics include comparing projected and observed dropout over time, reviewing whether dropout differs by treatment group or measured prognostic factors, and checking whether event ascertainment actually ceases at the recorded dropout date. A mismatch in dropout timing can materially change the projected count because participants lost earlier have less opportunity for events to be observed.
Independence is an assumption about the event and dropout times, conditional on the model and any variables it includes. It may be implausible if worsening health makes both dropout and the event more likely. In such a setting, an unadjusted product of event-free survival and dropout survival may not give the correct joint probability. Stratifying or modeling by relevant predictors can help when appropriate data are available, but it does not automatically remove unmeasured dependence.
Common Mistakes and Limitations
- Counting every dropout as an event. Dropout ends observation; it is not itself the event of interest. The event contribution must be calculated separately.
- Subtracting the dropout percentage from the event probability. The timing of dropout matters. A participant lost late has had more time for an event to be observed than one lost early.
- Applying the adjustment when events are still captured. If a registry continues event ascertainment, loss from visits is not necessarily loss of event follow-up.
- Assuming independence without justification. If dropout is related to event risk, the independent-time formula may misrepresent the joint probability.
- Rounding interval contributions too early. Add values at adequate precision and round the final probability and count; otherwise displayed rounded contributions can give a slightly different sum.
The adjustment produces an expected count under a specified model; it does not by itself describe the uncertainty around that count or the distribution of the date a target is reached. It also does not correct a misspecified event survival curve or dropout process. Those limitations should be kept separate from the arithmetic: precise calculations cannot compensate for inappropriate assumptions.
Check Your Understanding
Try each problem, then compare your reasoning with the solutions below.
- With constant event hazard \(0.08\) per year, dropout hazard \(0.02\) per year, and three years of follow-up, calculate the observed-event probability under independent hazards and the expected count for 500 participants.
- A study loses some participants from clinic visits but continues to ascertain their events through a registry. Should clinic withdrawal automatically be used as the censoring time in the dropout adjustment? Explain.
- State the independence assumption in the event and dropout projection, and give one situation in which it may be doubtful.
- Why is subtracting the overall dropout proportion from the event probability generally not the correct adjustment?
Solutions. (1) The constant-hazard formula gives \(p_{\mathrm{obs}}(3)=\frac{0.08}{0.08+0.02}(1-e^{-(0.08+0.02)3})=0.8(1-e^{-0.30})\approx0.2073\). The expected count is \(500(0.2073)\approx103.7\) events. (2) No. If registry ascertainment continues, withdrawal from visits does not end event observation; the projection should use the registry’s actual follow-up coverage. (3) The model assumes \(T\) and \(C\) are independent, so knowing the dropout time provides no information about the event time under the model. It may be doubtful if worsening health raises both dropout and event risk. (4) Subtraction ignores when participants drop out: earlier loss removes more opportunity to observe an event than later loss. The adjustment must account for remaining observable at each possible event time.