Why Cause-Specific Subdistribution Hazards Do Not Add Across Causes

Why Additivity Depends on the Risk Set

When a first event can occur from several mutually exclusive causes, it is natural to ask whether the hazards for the individual causes add to the overall first-event hazard. For cause-specific hazards, the answer is yes: each cause-specific hazard uses the same event-free risk set. For subdistribution hazards, the answer is generally no. Each cause uses a different denominator, because its subdistribution risk set excludes only events of that particular cause.

That distinction matters when interpreting rates. A sum of cause-specific hazards describes the instantaneous rate of any first event among people still event-free. A subdistribution hazard is built around the occurrence of one specified cause and a denominator that also retains people who have already experienced competing causes. These are different quantities, and adding the latter does not recover the former.

Definitions and the Two Risk Sets

Let $T$ be the time to the first event, and let $J\in\{1,\ldots,K\}$ identify its cause. Assume the causes are mutually exclusive: a person can experience only one first-event cause. Write $F_k(t)=P(T\leq t,J=k)$ for the probability of a first event from cause $k$ by time $t$, and $S(t)=P(T>t)$ for the probability of remaining free of every first event beyond $t$. For the rate formulas below, suppose these functions are differentiable at the time of interest, and the relevant denominators are positive.

Definition: The cause-specific hazard for cause $k$, denoted $\lambda_k(t)$, is the instantaneous rate of a first event from cause $k$ among individuals who are still free of every first event just before $t$. The subdistribution hazard for cause $k$, denoted $\tilde{\lambda}_k(t)$, is the cause-$k$ event density at $t$ divided by the probability of not having experienced cause $k$ by $t$: $$ \lambda_k(t)=\lim_{\Delta t\downarrow 0} \frac{P(t\leq T<t+\Delta t,\ J=k\mid T\geq t)}{\Delta t}, \qquad \tilde{\lambda}_k(t)=\frac{f_k(t)}{1-F_k(t)}, \quad f_k(t)=F_k'(t). $$

The cause-specific risk set contains only people who have not yet had any first event. The denominator $S(t)$ represents that group in the population. By contrast, the subdistribution denominator $1-F_k(t)$ represents everyone who has not yet had cause $k$. It includes people who remain event-free and people who have already experienced a competing cause. It excludes people who have already had cause $k$.

The latter is a cause-specific construction, not a claim that someone who has experienced a competing event remains physically able to experience another first event. In an observed-data implementation, subdistribution risk sets may retain competing-event individuals, with adjustments for censoring. The population definition above makes the denominator difference explicit. Independent censoring is a separate condition for estimation from censored data; it is not what makes the population subdistribution hazards non-additive.

The Additivity Result

The event-free probability is shared across all causes. Since $f_k(t)=S(t)\lambda_k(t)$, the overall first-event hazard is the sum of the cause-specific hazards. The subdistribution hazards instead divide each cause’s event density by its own probability of not yet having that cause.

Result: When the possible first-event causes are mutually exclusive and exhaustive, and the rates exist, $$ \lambda_{\mathrm{all}}(t) =\sum_{k=1}^K\lambda_k(t), \qquad \tilde{\lambda}_k(t) =\frac{S(t)\lambda_k(t)}{1-F_k(t)}. $$ Thus, in general, $$ \sum_{k=1}^K\tilde{\lambda}_k(t) =\sum_{k=1}^K\frac{S(t)\lambda_k(t)}{1-F_k(t)} \ne \lambda_{\mathrm{all}}(t). $$

Here “exhaustive” means every first event belongs to one of the listed causes. The overall hazard is defined among people event-free just before $t$, so the cause-specific hazards have a common denominator. The subdistribution hazards do not: $1-F_k(t)$ varies by cause. There is no general algebraic identity that turns their sum into the overall hazard.

Worked Examples

Worked Example: Two Causes With Equal Subdistribution Denominators

Suppose that at $t=2$ years, a hypothetical population has $S(2)=0.60$, $F_1(2)=0.20$, and $F_2(2)=0.20$. At that instant, suppose the cause-specific hazards are $\lambda_1(2)=0.10$ per year and $\lambda_2(2)=0.05$ per year. Assume these two causes exhaust the possible first events, and that the event-time distributions are smooth at 2 years.

The cause-specific hazards add to the overall first-event hazard: $$ \lambda_{\mathrm{all}}(2)=0.10+0.05=0.15\text{ per year}. $$

For cause 1, $f_1(2)=S(2)\lambda_1(2)=0.60(0.10)=0.060$ per year. Its subdistribution denominator is $1-F_1(2)=0.80$, so $$ \tilde{\lambda}_1(2)=\frac{0.060}{0.80}=0.0750\text{ per year}. $$

For cause 2, $f_2(2)=0.60(0.05)=0.030$ per year and $1-F_2(2)=0.80$. Therefore, $$ \tilde{\lambda}_2(2)=\frac{0.030}{0.80}=0.0375\text{ per year}. $$

Their sum is $0.0750+0.0375=0.1125$ per year, not $0.15$ per year. Even though the two subdistribution denominators happen to be equal in this example, each is $0.80$, not the event-free probability $0.60$ used for the overall hazard.

Worked Example: Unequal Cause-Specific Subdistribution Risk Sets

Consider a second hypothetical population at $t=3$ years with $S(3)=0.50$, $F_1(3)=0.35$, and $F_2(3)=0.15$. Suppose the cause-specific hazards are $0.08$ per year for cause 1 and $0.04$ per year for cause 2. Assume the two causes exhaust the possible first events and the relevant rates exist.

The overall first-event hazard is $0.08+0.04=0.12$ per year. The cause-1 event density is $f_1(3)=0.50(0.08)=0.040$ per year, and its denominator is $1-0.35=0.65$. Thus, $$ \tilde{\lambda}_1(3)=\frac{0.040}{0.65}=0.06154\text{ per year}. $$

For cause 2, $f_2(3)=0.50(0.04)=0.020$ per year, while its denominator is $1-0.15=0.85$. Hence, $$ \tilde{\lambda}_2(3)=\frac{0.020}{0.85}=0.02353\text{ per year}. $$

The subdistribution hazards sum to $0.06154+0.02353=0.08507$ per year, which is not the overall hazard of $0.12$ per year. Their denominators differ because the hypothetical population has accumulated different probabilities of the two causes by year 3.

Worked Example: Checking an Additivity Claim

A report gives hypothetical values at one instant: $S(t)=0.70$, $F_1(t)=0.10$, $F_2(t)=0.20$, $\lambda_1(t)=0.04$ per year, and $\lambda_2(t)=0.12$ per year. Assume the two causes exhaust first events and the rates exist. Is the claim “the subdistribution hazards add to the overall first-event hazard” supported by these values?

First, the overall hazard is $0.04+0.12=0.16$ per year. For cause 1, $f_1(t)=0.70(0.04)=0.028$ per year and $1-F_1(t)=0.90$, giving $\tilde{\lambda}_1(t)=0.028/0.90=0.03111$ per year. For cause 2, $f_2(t)=0.70(0.12)=0.084$ per year and $1-F_2(t)=0.80$, giving $\tilde{\lambda}_2(t)=0.084/0.80=0.1050$ per year. Their sum is $0.1361$ per year, not $0.16$ per year. The claimed additivity fails for these values.

This calculation does not say that one hazard type is inherently larger or more clinically important. It shows that the sum depends on the chosen denominators, so the two sums answer different rate questions.

Why the Denominator Difference Produces the Result

Start with the event density.
Among the population, the cause-$k$ event density at time $t$ is $f_k(t)$. Because cause $k$ occurs as a first event only among people event-free just before $t$, $f_k(t)=S(t)\lambda_k(t)$.

Use a shared denominator for cause-specific hazards.
The overall event density is $\sum_k f_k(t)$. Dividing by the event-free fraction gives $\lambda_{\mathrm{all}}(t)=\sum_k f_k(t)/S(t)=\sum_k\lambda_k(t)$.

Use a cause-specific denominator for each subdistribution hazard.
For cause $k$, the denominator is $1-F_k(t)$. Therefore the sum is $\sum_k f_k(t)/(1-F_k(t))$, with a different denominator for each term. It does not reduce to $\sum_k f_k(t)/S(t)$ in general.

This justification uses only the definitions and the factorization $f_k(t)=S(t)\lambda_k(t)$. It does not rely on a model such as proportional hazards. The earlier material on subdistribution rates emphasizes their interpretation; here the central check is whether the risk sets match the rate being summed.

Interpretation, Alternatives, and Diagnostics

A useful diagnostic is to write the denominator next to every hazard before combining rates. For cause-specific hazards, the denominator is the same event-free group for every cause. For cause-$k$ subdistribution hazards, it is the group that has not yet had cause $k$, including people with prior competing events. In a finite data set, the cause-specific risk sets are shared across causes, whereas the subdistribution risk sets retain competing-event individuals and therefore differ by cause.

A common mistake is to reason, “The causes are mutually exclusive, so their subdistribution hazards must add to the overall hazard.” Mutual exclusivity does not make the denominators equal. The correction is to check the defining ratio: subdistribution hazard $k$ divides $f_k(t)$ by $1-F_k(t)$, while the overall first-event hazard divides the sum of the cause densities by $S(t)$. These denominators generally differ.

If the scientific question concerns the instantaneous rate of any first event among people still event-free, cause-specific hazards are the natural rates to combine. If the question concerns one cause’s rate under the subdistribution definition, report that cause’s subdistribution hazard and its interpretation; do not add it to the other causes’ subdistribution hazards as though they formed a partition of a common event-free risk set. The distinction between instantaneous cause shares among events and hazard additivity is also important: a share among events is a ratio of cause-specific hazards, not a claim that subdistribution hazards add.

The subdistribution hazard is a rate scale, not itself a probability. Its value depends on the cause-specific event history through $F_k(t)$, as well as on the instantaneous event density. It may be useful for a cause-specific comparison, but it does not describe a physical transition rate among only those currently event-free. Nor does a lack of additivity imply an error in estimation: the sum is simply not generally an overall first-event hazard.

Finally, the formulas require well-defined first-event causes and positive denominators. At a time when $1-F_k(t)=0$, the cause-$k$ subdistribution hazard ratio is undefined; when $S(t)=0$, the usual event-free cause-specific hazard is likewise not defined by that ratio. Tied or simultaneous event types require a stated convention for defining causes and risk sets. Censoring must also be handled appropriately when estimating these quantities, but censoring is not a competing event and does not change the population denominator identities.

Check Your Understanding

Answer each question, then compare your reasoning with the solutions.

At a time $t$, suppose $S(t)=0.80$, $F_1(t)=0.10$, $F_2(t)=0.10$, $\lambda_1(t)=0.05$ per year, and $\lambda_2(t)=0.03$ per year. Find the overall hazard and both subdistribution hazards.
Why does a person who has already experienced cause 1 belong to the cause-2 subdistribution denominator, but not to the cause-specific risk set?
True or false: if all first events belong to one of two mutually exclusive causes, then the two subdistribution hazards add to the overall first-event hazard. Explain.
Which hazard type should be summed to obtain the overall first-event hazard, and what common feature makes that sum work?

Solutions. (1) The overall hazard is $0.05+0.03=0.080$ per year. The cause-1 density is $0.80(0.05)=0.040$ per year, so $\tilde{\lambda}_1=0.040/(1-0.10)=0.04444$ per year. The cause-2 density is $0.80(0.03)=0.024$ per year, so $\tilde{\lambda}_2=0.024/0.90=0.02667$ per year. Their sum is $0.07111$ per year, not $0.080$. (2) The cause-2 denominator includes everyone who has not yet experienced cause 2, including people with a prior competing cause 1. The cause-specific risk set includes only people still free of every first event. (3) False. Each subdistribution hazard uses its own denominator $1-F_k(t)$, whereas the overall hazard uses $S(t)$. (4) Sum the cause-specific hazards. They share the same event-free risk set and therefore the same denominator $S(t)$.