The Subdistribution Hazard Is Not A Probability

Why a Rate Can Be Mistaken for a Probability

A probability is dimensionless and lies between zero and one. A hazard is a rate: it describes how quickly probability accumulates relative to a specified denominator, and it has units of inverse time. The number used to report a hazard therefore depends on the time unit. For example, a rate of 2 per day is numerically greater than one, but it does not mean that an event has probability 2. In competing-risks analysis, this distinction matters for the subdistribution hazard because its denominator is not simply the group currently free of every event.

The cumulative incidence function for cause $k$, written $F_k(t)$, is the probability of experiencing cause $k$ by time $t$ as the first event. If $F_k$ has a density $f_k(t)$, the subdistribution hazard is defined by dividing that density by the probability of not yet having experienced cause $k$. This tutorial focuses on what that definition implies about units, probabilities, and interpretation.

Definition: For a continuously distributed event time with cumulative incidence $F_k(t)$ and density $f_k(t)=dF_k(t)/dt$, the cause-$k$ subdistribution hazard at time $t$ is $$ \widetilde{\lambda}_k(t)=\frac{f_k(t)}{1-F_k(t)}. $$ It is defined where $F_k(t)<1$. Because $f_k(t)$ has units of inverse time and $1-F_k(t)$ is a probability, $\widetilde{\lambda}_k(t)$ has units of inverse time.

The denominator $1-F_k(t)$ is the proportion of the population that has not experienced cause $k$ by time $t$. It includes people who have already experienced a competing event, as well as people who remain event-free. Thus the definition does not describe a probability among only those currently able to experience a first event. The earlier material on the subdistribution risk set develops that denominator’s role; here, the key point is that dividing a density by a probability produces a rate, not a probability.

Deriving the Rate–Probability Distinction

Suppose $F_k$ is differentiable near $t$. Over a short interval of length $\Delta t$, its cumulative incidence changes by approximately $f_k(t)\Delta t$. Dividing that change by the proportion that has not yet experienced cause $k$, and then by the interval length, gives the subdistribution hazard:

\frac{F_k(t+\Delta t)-F_k(t)} {(1-F_k(t))\Delta t} \longrightarrow \frac{f_k(t)}{1-F_k(t)} =\widetilde{\lambda}_k(t) \quad\text{as }\Delta t\downarrow 0.

The numerator in the fraction is a probability increment. After it is divided by $\Delta t$, the result is a probability increment per unit of time. The factor $1-F_k(t)$ is dimensionless, so it does not remove the time unit. A rate can be greater than one for the same reason that a speed can be greater than one kilometre per hour: the number is not itself a probability.

For every interval on which the quantities are defined, the cumulative incidence increment is bounded: $0\leq F_k(t+\Delta t)-F_k(t)\leq 1-F_k(t)$. The corresponding increment divided by $1-F_k(t)$ is therefore between zero and one. But dividing that bounded quantity by a short interval length can produce a value above one. In differential form, the identity is

dF_k(t)=(1-F_k(t))\widetilde{\lambda}_k(t)\,dt.

This equation states how a rate relates to an infinitesimal probability increment. It does not equate the rate with that increment. In particular, $\widetilde{\lambda}_k(t)\,dt$ can serve as a first-order approximation over a sufficiently short interval, but it is not generally the exact probability of an event over a finite interval.

Worked Examples

Worked Example: A Subdistribution Hazard Above One per Day

Hypothetical model. Suppose two mutually exclusive first events have constant cause-specific rates: 2 per day for cause A and 1 per day for cause B. Assume continuous event times, no other event types, and that these rates apply from time zero. Find the cause-A subdistribution hazard at time zero and at $t=0.1$ day, and calculate the probability of cause A in the first 0.1 day.

The overall first-event rate is $2+1=3$ per day, so the probability of remaining free of either event to time $t$ is $e^{-3t}$. Cause A occurs at rate 2 per day among those still event-free. Its density as a first event is therefore

f_A(t)=2e^{-3t}.

Integrating this density gives the cause-A cumulative incidence:

F_A(t)=\int_0^t 2e^{-3u}\,du =\frac{2}{3}(1-e^{-3t}).

At time zero, $f_A(0)=2$ per day and $F_A(0)=0$, so $\widetilde{\lambda}_A(0)=2/(1-0)=2$ per day. This value exceeds one, but it is a rate, not a probability.

At $t=0.1$ day, $e^{-0.3}\approx 0.7408$. Thus $f_A(0.1)=2(0.7408)=1.4816$ per day and $F_A(0.1)=\frac{2}{3}(1-0.7408)=0.1728$. Hence

\widetilde{\lambda}_A(0.1) =\frac{1.4816}{1-0.1728} =\frac{1.4816}{0.8272} \approx 1.791\ \text{per day}.

The probability of cause A in the first 0.1 day is $F_A(0.1)=0.1728$, or about 17.3%. The hazard value $2$ per day at time zero is not a claim of 200% probability. The model assumptions are stated explicitly, and the calculations use a valid competing-risks model with nonnegative rates and mutually exclusive first events.

Worked Example: Converting the Same Rate Between Time Units

Suppose a hypothetical cause-A subdistribution hazard at a particular instant is 0.08 per week. Express this same rate per day and per year, using 7 days per week and 52 weeks per year.

A rate changes numerically when the time unit changes, while the underlying rate does not. There are 7 days in a week, so the rate per day is

0.08\ \frac{1}{\text{week}} \times \frac{1\ \text{week}}{7\ \text{days}} =0.01143\ \frac{1}{\text{day}}.

There are 52 weeks in a year, so the rate per year is

0.08\ \frac{1}{\text{week}} \times \frac{52\ \text{weeks}}{1\ \text{year}} =4.16\ \frac{1}{\text{year}}.

The rate is greater than one when expressed per year, but that does not imply a probability above one. It is the same instantaneous rate expressed using a longer time unit. The calculation assumes only a consistent conversion of units; it makes no assumption that the rate stays constant over a week or a year. A probability for a specified interval would require information about the rate throughout that interval and the cumulative incidence at its start.

Worked Example: A Finite-Interval Probability Is Not the Rate Times the Interval

Suppose that, over a hypothetical interval, a cause-$k$ subdistribution hazard is constant at 2 per day, and at the start of the interval $F_k=0.30$. Calculate the cumulative incidence after one day and compare the one-day increment with the hazard.

From the definition, $dF_k/dt=(1-F_k)\widetilde{\lambda}_k$. With a constant rate of 2 per day, this becomes $dF_k/dt=2(1-F_k)$. Writing $G(t)=1-F_k(t)$ gives $dG/dt=-2G$, whose solution over an interval of length $\Delta t$ is $G(\Delta t)=G(0)e^{-2\Delta t}$. Therefore,

1-F_k(t+\Delta t) =(1-F_k(t))e^{-2\Delta t}.

For $\Delta t=1$ day, the starting value is $1-F_k(t)=0.70$. Thus

F_k(t+1) =1-0.70e^{-2} \approx 1-0.70(0.1353) =0.9053.

The one-day increase in cumulative incidence is $0.9053-0.30=0.6053$, which is a probability increment between zero and one. It is not $2$, and the rate times one day, $2\times1=2$, is not a valid finite-interval probability. Here the one-day increment divided by the initial $1-F_k(t)$ is $0.6053/0.70=0.8647$, also between zero and one. The calculation assumes the subdistribution hazard remains constant over this interval; it illustrates why even a constant rate must be translated into a probability using the rate’s relationship to $F_k$.

Interpretation, Alternatives, and Diagnostics

A cause-specific hazard and a subdistribution hazard are both rates, but their denominators differ. The cause-specific hazard concerns the event-free risk set at time $t$. The subdistribution hazard is defined using $1-F_k(t)$, which includes people who have already had a competing event. Neither rate is a probability. The cumulative incidence $F_k(t)$, by contrast, is the probability of cause $k$ having occurred by time $t$; it is often the more direct quantity when the scientific question asks about absolute event risk by a specified time.

A practical interpretation should name the cause, the time point, and the time unit. For example, “the cause-A subdistribution hazard at this time is 1.79 per day” is a rate statement. “The probability of cause A by 0.1 day is 0.173” is a probability statement. They describe different quantities and cannot be substituted for each other. When a particular finite-interval risk is the target, calculate or report the cumulative incidence at the interval endpoint rather than treating the instantaneous hazard as that risk.

Several checks help detect errors:

Check the units. A subdistribution hazard should carry units such as per day or per year. If it is reported as a probability without an interval, the interpretation needs correction.
Check the cumulative incidence. A valid $F_k(t)$ is nondecreasing and lies between zero and one. A negative density or a decreasing cumulative incidence signals an invalid calculation or model.
Check the interval. A rate times a long interval is not generally an exact event probability. It may even exceed one, while the actual probability cannot.
Check the denominator. Replacing $1-F_k(t)$ with the proportion currently free of all events changes the hazard definition.
Check the time scale. Converting per day to per year changes the numerical value of the rate, not the underlying probability of an event over a specified physical interval.

A common mistake is to see a subdistribution hazard of 1.4 per day and conclude that “the probability of the event is 140%.” This is wrong because the hazard is a derivative-scaled rate, with inverse-time units. The correction is to report it as 1.4 per day and, if an event probability is needed, calculate the cumulative incidence over the time interval of interest. A related mistake is to assume that a rate must lie below one because a probability does. That bound applies to probabilities, not rates.

The definition presumes a continuous-time setting in which the cumulative incidence has a density at the time being considered. If event times are recorded only in discrete intervals or the cumulative incidence is not differentiable at a point, an instantaneous subdistribution hazard may not be defined there in this form. In that situation, report interval probabilities or use a method defined for the data’s time scale rather than forcing an instantaneous-rate interpretation. These limitations reinforce the central distinction: the subdistribution hazard is a useful rate summary, but it is not itself an absolute-risk probability.

Key takeaway: $\widetilde{\lambda}_k(t)=f_k(t)/(1-F_k(t))$ has units of inverse time. It can exceed one, and its numerical value depends on the time unit. Event probabilities remain between zero and one and must be obtained from cumulative incidence over a specified time horizon.

Check Your Understanding

Answer each question, then compare with the solutions below.

If a subdistribution hazard is 1.5 per day at a given instant, is the event probability at that instant 1.5? Explain.
Convert 0.20 per week to a rate per day, using 7 days per week.
If $F_k(t)=0.40$ and $f_k(t)=0.12$ per month, calculate the subdistribution hazard at $t$, including its units.
Why is the product of a rate and a long interval not generally an exact event probability?

Solutions. (1) No. The value is a rate of 1.5 per day, not a probability; a probability requires a specified time interval and cumulative-incidence calculation. (2) $0.20/7=0.02857$ per day. (3) $\widetilde{\lambda}_k(t)=0.12/(1-0.40)=0.12/0.60=0.200$ per month. (4) A rate describes the instantaneous change in cumulative incidence relative to $1-F_k(t)$; over a finite interval both the rate and denominator may matter. The resulting probability increment must remain between zero and one, whereas a rate multiplied by a long interval can exceed one.

The Subdistribution Hazard Is Not A Probability

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