Deriving The Subdistribution Hazard Identity

Why the Identity Matters

A cumulative incidence function describes the probability of experiencing a particular first-event cause by time $t$. A subdistribution hazard expresses the rate of increase of that probability relative to the probability that the cause has not yet occurred. The link between these quantities is useful because it gives the subdistribution hazard a direct mathematical definition from the cumulative incidence function, rather than treating it as an ordinary event-free hazard.

The key denominator is not the probability of remaining free of every event. It is the probability of not having experienced cause $k$. A subject who has already experienced a different first-event cause has not experienced cause $k$, so that subject remains part of the conditioning event in the definition below. The derivation makes this denominator explicit.

Definitions and Assumptions

Definition: Let $T$ be the time to the first event and $J$ its cause, with mutually exclusive causes $1,\ldots,K$. For a specified cause $k$, the cause-specific cumulative incidence function is $F_k(t)=P(T\leq t,J=k)$. When this function has a density, write $f_k(t)=dF_k(t)/dt$. The subdistribution hazard for cause $k$ is $$ \lambda_k^*(t)=\lim_{\Delta t\downarrow 0} \frac{P(t\leq T<t+\Delta t,\ J=k\mid T\geq t\ \text{or}\ (T<t,\ J\ne k))}{\Delta t}, $$ when this limit exists.

The conditioning event says that cause $k$ has not occurred before time $t$. It includes people who are still event-free at $t$, as well as people whose first event before $t$ was a different cause. In the continuous-time setting used for the derivation, whether the endpoint is written $T\geq t$ or $T>t$ does not affect probabilities, because there is no point mass at $t$.

For the identity to hold in its density form, assume that event times have no atoms at the time of interest, $F_k$ is differentiable there, and $P(T\text{ has not yet been cause }k)>0$. The derivation is a population-probability argument; it does not require independence between latent times to different causes. Censoring is not part of the population identity, though it matters when estimating the quantities from censored data.

Deriving the Identity

Let $A_t$ denote the conditioning event: cause $k$ has not occurred before $t$. Its probability is the complement of experiencing cause $k$ before $t$. Under the continuous-time assumption, this is $1-F_k(t)$. The event of cause $k$ during $[t,t+\Delta t)$ is already contained in $A_t$, and its unconditional probability is the increment in the cumulative incidence function over that interval.

Write the denominator.
$P(A_t)=1-F_k(t)$, because $F_k(t)$ is the probability cause $k$ has occurred by time $t$.

Write the interval event probability.
$P(t\leq T<t+\Delta t,\ J=k)=F_k(t+\Delta t)-F_k(t)$. This event implies $A_t$, so adding the condition $A_t$ to this probability does not change it.

Form the conditional rate.
Divide the interval probability by $P(A_t)$, then divide by the interval width $\Delta t$, as in the definition.

Take the limit.
The difference quotient tends to $f_k(t)$, giving the identity.

\begin{aligned} \lambda_k^*(t) &=\lim_{\Delta t\downarrow 0} \frac{P(t\leq T<t+\Delta t,\ J=k\mid A_t)}{\Delta t}\\ &=\lim_{\Delta t\downarrow 0} \frac{F_k(t+\Delta t)-F_k(t)} {\Delta t\{1-F_k(t)\}}\\ &=\frac{f_k(t)}{1-F_k(t)}. \end{aligned}

The second equality uses conditional probability: the joint probability of the interval event and $A_t$ equals the interval event probability, since an event of cause $k$ in that interval cannot follow an earlier event of cause $k$. The final equality follows because $f_k(t)$ is the derivative of $F_k(t)$.

Identity: At times where $F_k$ is differentiable and $F_k(t)<1$, $$ \lambda_k^*(t)=\frac{f_k(t)}{1-F_k(t)}. $$ Equivalently, $f_k(t)=\lambda_k^*(t)\{1-F_k(t)\}$.

Worked Examples

Worked Example: Computing the Identity From a Cumulative Incidence Function

Suppose a hypothetical population has two possible first-event causes. For cause $k$, let $F_k(t)=0.4(1-e^{-0.5t})$, where time is measured in years. Find the subdistribution hazard at $t=2$ years.

Differentiate the cumulative incidence function: $$ f_k(t)=\frac{d}{dt}\{0.4(1-e^{-0.5t})\}=0.2e^{-0.5t}. $$

At $t=2$, $F_k(2)=0.4(1-e^{-1})=0.2528$, and $f_k(2)=0.2e^{-1}=0.07358$ per year. Therefore, $$ \lambda_k^*(2)=\frac{0.07358}{1-0.2528} =\frac{0.07358}{0.7472} =0.09848\ \text{per year}. $$

Thus the subdistribution hazard at 2 years is approximately $0.0985$ per year. This is a rate, not a probability of cause $k$ during a year. The calculation requires a differentiable $F_k$ and a positive denominator, both of which hold here.

Worked Example: Approximating the Identity Over a Short Interval

In a second hypothetical setting, suppose $F_k(t)=0.05t$ for times from 0 to 8 years. At $t=4$, estimate the subdistribution hazard using an interval of width $0.1$ year.

At $t=4$, $F_k(4)=0.20$, so the probability cause $k$ has not yet occurred is $1-0.20=0.80$. The cause-$k$ cumulative incidence increases by $$ F_k(4.1)-F_k(4)=0.05(4.1)-0.05(4)=0.005. $$

The conditional probability of cause $k$ in this interval, given that cause $k$ has not occurred before year 4, is $0.005/0.80=0.00625$. Dividing by the interval width gives $$ \frac{0.00625}{0.1}=0.0625\ \text{per year}. $$

Because this $F_k$ is linear, its density is exactly $0.05$ per year throughout the interval. The identity gives $0.05/(1-0.20)=0.0625$ per year, matching the interval calculation. The values are valid as a cumulative incidence over the stated range: they are nondecreasing and never exceed 1.

Worked Example: Recovering Cumulative Incidence From a Constant Subdistribution Hazard

Suppose, hypothetically, that a cause-$k$ subdistribution hazard is constant at $0.08$ per year over the first 5 years, and $F_k(0)=0$. Find $F_k(5)$.

From the identity, $dF_k(t)/dt=0.08\{1-F_k(t)\}$. Rearranging and integrating from 0 to 5 gives $$ \int_0^{F_k(5)}\frac{du}{1-u}=\int_0^5 0.08\,dt, \qquad -\log\{1-F_k(5)\}=0.40. $$

Therefore $1-F_k(5)=e^{-0.40}=0.6703$, so $$ F_k(5)=1-e^{-0.40}=0.3297. $$

The implied cumulative incidence by 5 years is $0.3297$, or about 33.0%. This calculation is justified while $F_k(t)<1$, so the denominator in the identity remains positive. A constant subdistribution hazard is a mathematical assumption for this example, not a claim that such a rate is appropriate for every population.

What the Identity Does—and Does Not—Say

The denominator $1-F_k(t)$ is specific to cause $k$. It is not the event-free probability $P(T\geq t)$, which excludes anyone who has already experienced any first-event cause. For comparison, the cause-specific hazard uses the event-free population as its conditioning group. The subdistribution hazard instead links the instantaneous increase in $F_k$ to the proportion that has not yet experienced cause $k$.

Quantity	Numerator or rate linked to	Relevant probability in the denominator
Cause-specific hazard for cause $k$	Cause-$k$ event rate among those still event-free	Probability of having experienced no first event by $t$
Subdistribution hazard for cause $k$	Density $f_k(t)$ of the cause-$k$ cumulative incidence	$1-F_k(t)$, probability cause $k$ has not occurred by $t$

These are different denominators and different rates. The identity does not say the subdistribution hazard is a probability, nor does it imply that a change in that hazard is a causal effect. It is an algebraic relation for the cumulative incidence and its derivative under the stated conditions.

Diagnostics and Failure Modes

The identity also provides checks on calculations. A valid cumulative incidence function must be nondecreasing, so wherever its derivative exists, $f_k(t)\geq 0$. Provided $F_k(t)<1$, the resulting subdistribution hazard must therefore also be nonnegative. If a computed hazard is negative, check the differentiation, time ordering, units, and whether a noisy numerical estimate of $F_k$ has become nonmonotone.

Small denominator: If $F_k(t)$ is close to 1, $1-F_k(t)$ is small. Even modest errors in estimating $f_k(t)$ or $F_k(t)$ can then produce a highly unstable ratio.
Atoms or tied event times: If the event-time distribution has point masses, a density and derivative may not describe the full change at those times. The continuous-time identity as written should not be applied uncritically at a jump.
Finite intervals treated as instantaneous: The conditional probability over a nonzero interval is an average over that interval. It approximates the instantaneous rate only when the interval is sufficiently short and the relevant quantities do not change sharply within it.
Confusing event-free survival with $1-F_k$: Replacing $1-F_k(t)$ with the probability of no event of any cause changes the denominator and no longer gives the subdistribution hazard.

In applications, $F_k$ and its derivative are usually estimated rather than known. Smoothing or differencing a cumulative incidence estimate can amplify sampling variability, especially near the boundary where $1-F_k(t)$ is small. The population identity remains exact under its assumptions, but a noisy plug-in calculation need not be a reliable estimate.

Common Mistake

Common mistake: “A person who has experienced a competing event is no longer eligible for cause $k$, so the denominator must be the probability of remaining event-free.” The correction is that this changes the conditioning event. For the subdistribution hazard, the denominator is $1-F_k(t)$: it counts the probability of not yet having experienced cause $k$, including those whose earlier first event was another cause. The event-free denominator belongs to the cause-specific hazard.

Check Your Understanding

Use the identity and show the key calculation in each answer.

If $F_k(t)=0.3(1-e^{-0.2t})$, find $f_k(3)$, $F_k(3)$, and $\lambda_k^*(3)$. Time is in years.
At a time $t$, suppose $F_k(t)=0.40$ and $f_k(t)=0.09$ per month. Calculate the subdistribution hazard and state its units.
Why is $1-F_k(t)$, rather than the event-free probability, the denominator in the identity?
Suppose a proposed estimate has $f_k(t)<0$ while $F_k(t)<1$. What should be checked?

Solutions. (1) $f_k(t)=0.06e^{-0.2t}$, so $f_k(3)=0.06e^{-0.6}=0.03293$ per year. Also $F_k(3)=0.3(1-e^{-0.6})=0.1354$. Thus $\lambda_k^*(3)=0.03293/(1-0.1354)=0.03808$ per year. (2) $\lambda_k^*(t)=0.09/(1-0.40)=0.15$ per month. (3) The conditioning event is that cause $k$ has not occurred; it includes people who previously experienced a different first-event cause. Its probability is $1-F_k(t)$. (4) Since a valid cumulative incidence is nondecreasing, its density cannot be negative where it exists. Check the derivative, data ordering, units, and whether noise has made the estimated cumulative incidence nonmonotone.

Quantity	Numerator or rate linked to	Relevant probability in the denominator
Cause-specific hazard for cause \(k\)	Cause-\(k\) event rate among those still event-free	Probability of having experienced no first event by \(t\)
Subdistribution hazard for cause \(k\)	Density \(f_k(t)\) of the cause-\(k\) cumulative incidence	\(1-F_k(t)\), probability cause \(k\) has not occurred by \(t\)