Cause-Specific Hazards in Competing Risks

Competing Failure Types

In a competing-risks setting, a unit can experience one of several mutually exclusive failure types, but only the first failure is observed. For example, a component might fail from either corrosion or overheating. Once one type occurs, the other type can no longer be observed as the first failure for that component. The central quantities are therefore not just the survival probability, but also the probability that each cause has occurred by a given time.

A useful mathematical construction assigns a latent failure time to each cause. The observed event time is the minimum of those times, and its cause is whichever latent time attains that minimum. This construction leads to simple formulas when the latent times are independent. It is important, however, to keep the latent failure distributions distinct from the distributions of the observed event and cause.

Definition: For \(K\) failure types, let \(T_j\) be the latent failure time for cause \(j\), and define the observed event time \(T=\min_{1\leq j\leq K}T_j\). If the minimum is attained uniquely, let \(J=j\) when \(T_j=T\). The overall survival function is \(S(t)=P(T>t)\). The cause-\(j\) cumulative incidence function is \(C_j(t)=P(T\leq t,\ J=j)\). If the observed cause-\(j\) subdensity exists, denote it by \(g_j(t)\); it describes the joint occurrence of an event at time \(t\) and cause \(j\).

For the formulas below, suppose the latent times are nonnegative, mutually independent, and have absolutely continuous distributions. Write \(S_j^{\mathrm{lat}}(t)=P(T_j>t)\) for the latent survival function and \(f_j^{\mathrm{lat}}(t)\) for its density. Absolute continuity ensures that ties have probability zero: for independent latent times with densities, the probability that any pair is exactly equal is zero. Thus the cause \(J\) is well defined with probability one.

Observed Survival, Subdensities, and Cumulative Incidence

A unit survives beyond \(t\) exactly when every latent failure time exceeds \(t\). Independence turns that joint event into a product. To fail from cause \(j\) at time \(t\), the latent cause-\(j\) time must be near \(t\), while every other latent time must exceed \(t\). These observations give the basic competing-risks formulas.

Theorem: Under the independent, absolutely continuous latent-time assumptions, the observed survival and cause-\(j\) subdensity are \[ S(t)=\prod_{k=1}^{K}S_k^{\mathrm{lat}}(t), \qquad g_j(t)=f_j^{\mathrm{lat}}(t)\prod_{\substack{k=1\\k\ne j}}^{K}S_k^{\mathrm{lat}}(t). \] Consequently, \[ C_j(t)=\int_0^t g_j(u)\,du. \]

To prove the survival identity, write the event \(T>t\) as the intersection of the events \(T_k>t\), for \(k=1,\ldots,K\). By independence,

P(T>t)=P(T_1>t,\ldots,T_K>t) =\prod_{k=1}^{K}P(T_k>t) =\prod_{k=1}^{K}S_k^{\mathrm{lat}}(t).

For the cause-\(j\) formula, consider a short interval \([t,t+\Delta t)\). The event that \(T\) falls in this interval and \(J=j\) requires \(T_j\in[t,t+\Delta t)\) and \(T_k>t+\Delta t\) for every \(k\ne j\), apart from boundary events whose probabilities are negligible at first order. Independence gives, to first order in \(\Delta t\),

P(t\leq T<t+\Delta t,\ J=j) =f_j^{\mathrm{lat}}(t)\Delta t \prod_{k\ne j}S_k^{\mathrm{lat}}(t)+o(\Delta t).

Dividing by \(\Delta t\) and taking the limit at points where the densities and survivals are defined yields the stated subdensity. Finally, integrating that subdensity from \(0\) to \(t\) gives \(P(T\leq t,J=j)\), the cause-\(j\) cumulative incidence. \(\square\)

The cause-specific hazard is defined using the observed risk set: it is the instantaneous rate of cause-\(j\) failure among units that have not yet failed from any cause. Where \(S(t)>0\), its density form is \(\lambda_j(t)=g_j(t)/S(t)\). The condition \(S(t)>0\) matters: if no units can remain at risk at time \(t\), an observed cause-specific hazard is not defined there.

Result: Wherever \(S(t)>0\), the observed cause-specific hazard equals the latent cause-\(j\) hazard: \[ \lambda_j(t)=\frac{g_j(t)}{S(t)} =\frac{f_j^{\mathrm{lat}}(t)}{S_j^{\mathrm{lat}}(t)}. \] This follows by substituting the product formulas and cancelling the positive survival factors. In particular, \(S(t)>0\) ensures that every \(S_k^{\mathrm{lat}}(t)>0\).

The cumulative incidences are related to the observed hazards by

C_j(t)=\int_0^t S(u)\lambda_j(u)\,du,

where the integrand is taken on times at which the observed risk set is nonempty. The factor \(S(u)\) is essential: a cause-specific hazard is a rate among those still event-free, not an unconditional probability. Since the event types are mutually exclusive, \(\sum_{j=1}^K C_j(t)=1-S(t)\).

Worked Calculations

Worked Example: Two Exponential Causes

Suppose two independent latent failure times have exponential distributions with rates \(0.04\) and \(0.06\) per year. Calculate the overall survival at one year, each cause-specific hazard, and each cumulative incidence at one year.

For an exponential lifetime with rate \(r\), the latent survival is \(e^{-rt}\) and the density is \(re^{-rt}\). Thus

\(S_1^{\mathrm{lat}}(t)=e^{-0.04t}\), \(S_2^{\mathrm{lat}}(t)=e^{-0.06t}\), \(f_1^{\mathrm{lat}}(t)=0.04e^{-0.04t}\), and \(f_2^{\mathrm{lat}}(t)=0.06e^{-0.06t}\).

The observed survival is

S(t)=e^{-0.04t}e^{-0.06t}=e^{-0.10t}, \qquad S(1)=e^{-0.10}\approx 0.90484.

The subdensities are \(g_1(t)=0.04e^{-0.10t}\) and \(g_2(t)=0.06e^{-0.10t}\). Dividing by \(S(t)=e^{-0.10t}>0\) gives \(\lambda_1(t)=0.04\) and \(\lambda_2(t)=0.06\). The cumulative incidences at one year are

C_1(1)=\int_0^1 0.04e^{-0.10u}\,du =0.4(1-e^{-0.10})\approx 0.03807,

and

C_2(1)=\int_0^1 0.06e^{-0.10u}\,du =0.6(1-e^{-0.10})\approx 0.05710.

Their sum is approximately \(0.09516\), which equals \(1-S(1)=1-e^{-0.10}\). The cause-specific hazards are constant, but the cause probabilities by one year are smaller than those rates because they account for the possibility of surviving to the event time.

Worked Example: Weibull Causes With a Common Shape

Let two independent latent times have survivals \(S_1^{\mathrm{lat}}(t)=e^{-0.02t^2}\) and \(S_2^{\mathrm{lat}}(t)=e^{-0.03t^2}\) for \(t\geq0\). Find the survival and cumulative incidences at \(t=5\).

Differentiating the latent distribution functions gives \(f_1^{\mathrm{lat}}(t)=0.04t e^{-0.02t^2}\) and \(f_2^{\mathrm{lat}}(t)=0.06t e^{-0.03t^2}\). Therefore,

S(t)=e^{-0.02t^2}e^{-0.03t^2}=e^{-0.05t^2}.

The cause-1 subdensity is \(g_1(t)=0.04t e^{-0.05t^2}\), and the cause-2 subdensity is \(g_2(t)=0.06t e^{-0.05t^2}\). Integrating,

C_1(t)=\int_0^t 0.04u e^{-0.05u^2}\,du =0.4(1-e^{-0.05t^2}),

because the derivative of \(e^{-0.05u^2}\) is \(-0.10u e^{-0.05u^2}\). Similarly,

C_2(t)=\int_0^t 0.06u e^{-0.05u^2}\,du =0.6(1-e^{-0.05t^2}).

At \(t=5\), the exponent is \(0.05(25)=1.25\). Hence \(S(5)=e^{-1.25}\approx0.28650\), \(C_1(5)=0.4(1-e^{-1.25})\approx0.28540\), and \(C_2(5)=0.6(1-e^{-1.25})\approx0.42810\). The incidences sum to \(0.71350=1-S(5)\), as required.

Worked Example: A Bounded Latent Lifetime and an Empty Risk Set

Let \(T_1\) be exponential with rate \(0.5\), and let \(T_2\) be uniform on \([0,2]\), independently. Determine the observed subdensities and cumulative incidences up to time \(2\), and explain what happens after time \(2\).

For \(0\leq t<2\), \(S_1^{\mathrm{lat}}(t)=e^{-0.5t}\), \(f_1^{\mathrm{lat}}(t)=0.5e^{-0.5t}\), \(S_2^{\mathrm{lat}}(t)=1-t/2\), and \(f_2^{\mathrm{lat}}(t)=0.5\). The observed survival is \(S(t)=(1-t/2)e^{-0.5t}\). The subdensities are

g_1(t)=0.5(1-t/2)e^{-0.5t}, \qquad g_2(t)=0.5e^{-0.5t}.

Integrating the second subdensity gives \(C_2(t)=1-e^{-0.5t}\). For the first, substitute \(v=u/2\), so \(du=2\,dv\):

C_1(t)=0.5\int_0^t(1-u/2)e^{-0.5u}\,du =\int_0^{t/2}(1-v)e^{-v}\,dv =\frac{t}{2}e^{-t/2}.

At \(t=2\), \(S(2)=0\), \(C_1(2)=e^{-1}\), and \(C_2(2)=1-e^{-1}\), so all probability is accounted for. For \(t\geq2\), overall survival remains zero: \(T_2\leq2\) with probability one, so there are no units still at risk after time \(2\). Although the exponential latent survival remains positive, the observed cause-specific hazards are undefined once \(S(t)=0\). This illustrates why positive survival for one latent cause alone is not enough to define an observed hazard.

Interpretation and a Common Pitfall

The cumulative incidence \(C_j(t)\) answers an unconditional probability question: what is the probability that cause \(j\) is the first observed cause by time \(t\)? It is not the same as the latent probability \(P(T_j\leq t)\), because another cause may occur first. Nor is it obtained by treating a cause-specific hazard as a probability and integrating it without accounting for survival.

Incorrect: “The probability of cause \(j\) by time \(t\) is \(\int_0^t\lambda_j(u)\,du\).” This omits the chance that a unit remains event-free until \(u\).
Correction: The cumulative incidence is \(\int_0^t S(u)\lambda_j(u)\,du\). The survival factor weights the cause-specific rate by the probability of still being at risk.
Incorrect: “The observed cause-specific hazard is always the latent cause hazard.” The equality requires the independent latent-time construction and a nonempty observed risk set.
Correction: Under the stated independence assumptions, \(\lambda_j(t)=f_j^{\mathrm{lat}}(t)/S_j^{\mathrm{lat}}(t)\) only where \(S(t)>0\).

The independence assumption is a modeling assumption about latent times; the observed first failure and its cause do not reveal what the other latent failure times would have been. The formulas are useful because they translate a specified latent-time model into observable survival and cause probabilities, while keeping those two levels of description separate.

Check Your Understanding

Use the definitions and formulas to answer the following questions.

In a two-cause model, explain why overall survival is the product of the two latent survivals under independence.
What event does the observed cause-\(j\) subdensity \(g_j(t)\) describe?
Why does the cumulative incidence formula include \(S(u)\) when it is written as an integral of the cause-specific hazard?
For two independent exponential latent times with rates \(a\) and \(b\), what is the overall survival function?
If one latent survival is positive but overall survival is zero, is the observed cause-specific hazard defined? Explain.

Cause-Specific Hazards in Competing Risks

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