Why Calculate a Subdistribution Hazard?
A cause-specific cumulative incidence answers an absolute-risk question: by time \(t\), what is the probability that the event of interest has occurred first? A subdistribution hazard gives a different, time-specific summary. It combines the instantaneous density of cause-\(k\) events with the amount of cumulative incidence still remaining before all individuals have experienced cause \(k\). This tutorial focuses on calculating that quantity from the cumulative incidence function and its density at a specified time.
The calculation is useful only when the inputs are understood. The cumulative incidence is a probability, while its density is probability per unit time. Their ratio is a rate, not a probability. The numerical examples below make the distinction explicit and show how to report the result in context.
How to Read the Formula
The numerator \(f_k(t)\) describes how quickly cause-\(k\) cumulative incidence is increasing near time \(t\). Its units are inverse time: if time is measured in years, then \(f_k(t)\) has units of probability per year. The denominator \(1-F_k(t)\) is the proportion of the population that has not yet experienced cause \(k\) by time \(t\). It includes people who remain event-free and, by the subdistribution-risk-set convention, people who have already experienced a competing event.
Thus, the subdistribution hazard is a rate relative to that denominator. It is not the probability that a person experiences cause \(k\) at exactly time \(t\), nor is it the probability of cause \(k\) over an interval. A rate can have a numerical value greater than one when expressed in a chosen time unit; that does not make it a probability.
The calculation assumes a first-event competing-risks setting in which causes are mutually exclusive event types for an individual, and the time origin and units are consistently defined. For the displayed pointwise formula, \(F_k(t)\) must be less than one and \(F_k\) must have a derivative at the time of interest. If the cumulative incidence has a jump or is not differentiable there, an ordinary pointwise density-based hazard is not defined by this formula at that time. In applications, estimated cumulative incidence curves also involve sampling uncertainty, and estimating a derivative can be sensitive to noise and smoothing choices.
Worked Example: A Smooth Cumulative Incidence Function
Suppose a hypothetical model gives the cause-1 cumulative incidence, with time \(t\) measured in years, as \(F_1(t)=0.30(1-e^{-0.20t})\). Calculate and interpret the cause-1 subdistribution hazard at \(t=2\) years.
Step 1: Find the density. Differentiate the cumulative incidence with respect to years: $$ f_1(t)=\frac{d}{dt}\{0.30(1-e^{-0.20t})\}=0.060e^{-0.20t}. $$
Step 2: Evaluate both inputs at two years. Since \(e^{-0.40}\approx0.6703\), $$ F_1(2)=0.30(1-0.6703)=0.09891, \qquad f_1(2)=0.060(0.6703)=0.04022\text{ per year}. $$
Step 3: Divide by one minus the cumulative incidence. $$ \tilde{\lambda}_1(2)=\frac{0.04022}{1-0.09891} =\frac{0.04022}{0.90109} \approx0.04463\text{ per year}. $$
At two years, this hypothetical model therefore has a cause-1 subdistribution hazard of about \(0.04463\) per year. This is a rate based on the cause-1 event density relative to the proportion that has not yet experienced cause 1. It is not a 4.463% probability of cause 1 during the next year.
Calculation and Interpretation in Practice
A reliable calculation follows the same sequence whether the inputs come from a model, a table, or an estimated curve:
Specify which event type is cause \(k\), the time point \(t\), and the time unit.
Use \(F_k(t)\) and \(f_k(t)\) for the same cause, population, and time scale. The density must be a local rate of increase, not the cumulative incidence itself.
Compute \(1-F_k(t)\), then divide \(f_k(t)\) by that value.
State the time unit and explain that the result is tied to the subdistribution risk set, not an interval probability.
Worked Example: A Rational-Form Cumulative Incidence
In another hypothetical setting, time is measured in months and the cause-2 cumulative incidence is \(F_2(t)=0.40t/(t+4)\). Find the cause-2 subdistribution hazard at \(t=4\) months.
Differentiate using the quotient rule: $$ f_2(t)=0.40\frac{(t+4)-t}{(t+4)^2} =\frac{1.6}{(t+4)^2}. $$ At four months, \(F_2(4)=0.40(4/8)=0.20\), and $$ f_2(4)=\frac{1.6}{8^2}=\frac{1.6}{64}=0.0250\text{ per month}. $$ Therefore, $$ \tilde{\lambda}_2(4)=\frac{0.0250}{1-0.20} =\frac{0.0250}{0.80}=0.03125\text{ per month}. $$
The cause-2 subdistribution hazard at four months is \(0.03125\) per month. The density is measured per month, and division by the dimensionless value \(0.80\) leaves the rate in inverse months. The value is not the probability of cause 2 in a one-month interval.
When the Density Is Approximated From Nearby Values
Sometimes a smooth formula for \(F_k(t)\) is unavailable, but the cumulative incidence is known or estimated at nearby times. A central difference can approximate the derivative: $$ f_k(t)\approx \frac{F_k(t+h)-F_k(t-h)}{2h}, $$ where \(h\) is a small time interval. This approximation uses the change across both sides of \(t\); it is not automatically an exact pointwise density. It is most informative when the curve is reasonably smooth over that local interval.
Worked Example: Approximating a Density From Nearby Times
A hypothetical estimated cause-1 cumulative incidence is \(0.1182\) at 4.9 years and \(0.1218\) at 5.1 years. Use these values to approximate the cause-1 subdistribution hazard at 5 years, given \(F_1(5)=0.1200\).
The time gap is \(5.1-4.9=0.2\) years, so the central-difference estimate of the density is $$ \widehat f_1(5)\approx\frac{0.1218-0.1182}{0.2} =\frac{0.0036}{0.2} =0.0180\text{ per year}. $$ The denominator is \(1-F_1(5)=1-0.1200=0.8800\). The resulting approximate subdistribution hazard is $$ \widehat{\tilde{\lambda}}_1(5)\approx\frac{0.0180}{0.8800} =0.02045\text{ per year}. $$
The approximate rate at five years is \(0.02045\) per year. The word “approximate” matters: the calculation treats the average slope from 4.9 to 5.1 years as a local estimate of the derivative at 5 years. With noisy estimates, a wider or narrower interval can change this slope, so the result should not be reported with more precision than the curve supports.
Why the Identity Holds
The formula can be justified directly from the subdistribution survival function, defined here as \(\tilde S_k(t)=1-F_k(t)\). This quantity is the complement of the cause-\(k\) cumulative incidence. Where \(\tilde S_k(t)>0\) and the derivative exists, the corresponding hazard rate is the negative derivative of its log: $$ -\frac{d}{dt}\log\{\tilde S_k(t)\} =-\frac{d}{dt}\log\{1-F_k(t)\}. $$
By the chain rule, $$ -\frac{d}{dt}\log\{1-F_k(t)\} =-\frac{-f_k(t)}{1-F_k(t)} =\frac{f_k(t)}{1-F_k(t)}. $$
This gives the stated identity. The derivation also explains the conditions: the logarithm requires \(1-F_k(t)>0\), and the derivative calculation requires \(F_k\) to be differentiable at \(t\). If \(F_k(t)=1\), the denominator is zero and the formula does not give a finite hazard.
Diagnostics, Alternatives, and Common Pitfalls
Before interpreting a computed value, check that both inputs refer to the same cause and time. Confirm that \(F_k(t)\) lies between zero and one, and that it does not decrease as time increases. A negative density obtained from a purported cumulative incidence curve signals a data, calculation, or smoothing problem. Also check units: if \(f_k(t)\) is per year, the result must be per year, not per month.
A frequent error is to divide the density by \(F_k(t)\), or to treat \(f_k(t)\) itself as the subdistribution hazard. The denominator is \(1-F_k(t)\), not \(F_k(t)\), and the numerator is a density rather than a cumulative probability. Another error is to call the result the probability of cause \(k\) in the next time interval. The hazard is a rate at a time; an interval probability requires additional information about how the rate behaves across that interval.
The main alternative summaries answer different questions. The cumulative incidence \(F_k(t)\) directly reports the probability of experiencing cause \(k\) by time \(t\). A cause-specific hazard describes the instantaneous rate among individuals who have not yet experienced any event. The subdistribution hazard instead uses the denominator \(1-F_k(t)\). These summaries are related, but they are not interchangeable. In particular, a single subdistribution hazard value does not by itself provide the absolute probability by a future time.
The pointwise formula also has practical limits. A density is a derivative, so small irregularities in an estimated cumulative incidence curve can produce unstable local slopes. Finite differences depend on the chosen time window, while smoothing introduces its own choices. Near a point where \(F_k(t)\) is close to one, the denominator is small, so modest errors in either input may have a relatively large effect on the ratio. A numerical value should therefore be accompanied by a clear time unit, an explanation of whether the density is exact or estimated, and appropriate uncertainty assessment when the inputs are estimated from data.
Check Your Understanding
Work through each question, then compare your reasoning with the solutions.
- At a specified time, \(F_k(t)=0.25\) and \(f_k(t)=0.030\) per year. Calculate the subdistribution hazard and state its units.
- A cause-\(j\) cumulative incidence is \(0.10\) at time \(t\), and its density is \(0.018\) per month. What is the subdistribution hazard at that time?
- Why is \(f_k(t)\) divided by \(1-F_k(t)\), rather than by \(F_k(t)\), in the subdistribution hazard formula?
- A cumulative incidence curve is estimated at times immediately before and after \(t\), but it is visibly irregular. What should be checked before interpreting a finite-difference hazard estimate?
Solutions. (1) \(\tilde{\lambda}_k(t)=0.030/(1-0.25)=0.030/0.75=0.0400\) per year. This is a rate, not a 4% interval probability. (2) \(\tilde{\lambda}_j(t)=0.018/(1-0.10)=0.018/0.90=0.0200\) per month. (3) The denominator is the complement of the cause-\(k\) cumulative incidence: the subdistribution risk set is represented by \(1-F_k(t)\), not by the proportion that has already experienced cause \(k\). (4) Check the time spacing, the curve values, and whether the local slope is stable to reasonable choices of interval or smoothing. An irregular estimated curve can make the derivative—and therefore the hazard estimate—unstable.