Past exam of the mathematics course of the University of Cambridge / 2017 / iii / Paper 207 / 6

In a competing risks model, the first event has time $T$ and type $J$; observing one type prevents another type from being the first event for that individual. Its cause-specific hazard isFor absolutely continuous event times the total hazard function is ${\sum }_j{h}_j(t)$, giving the event-free survivor function $S(t)=\exp [-{\int }_0^t{\sum }_j{h}_j(u)du]$. Since an event of type $j$ near $t$ has unconditional density $S(t)h_j(t)$, its cumulative incidence function satisfiesThis cause-specific hazard to cumulative incidence formula does not require independent latent times for the competing causes. In general $F_j(t)$ is not $1-e^{-{\int }_0^t{h}_j}$, since that expression ignores removal by other causes.

Put $a={\theta }_A\ge 0$, $b={\theta }_B\ge 0$ and $q=a+b$, with $\tau \ge 0$. HereIntegrating $aS(u)$ separately before and after $\tau$ gives, for $q>0$,After $\tau$, only type $A$ can occur, so the second term is the fraction event-free at $\tau$ multiplied by their subsequent event probability. The expressions agree at $\tau$. If $a=b=0$, both incidence functions are identically zero; if $a=0<b$, this formula gives $F_A=0$.

Only times up to $\tau$ contribute to the cumulative incidence function for type $B$, because its cause-specific hazard vanishes afterwards. With $q={\theta }_A+{\theta }_B>0$,soThis increases before $\tau$ and then remains constant; it includes the protection from type $B$ events produced by earlier type $A$ events. If both intensities vanish, $F_B=0$; if ${\theta }_B=0$, it is also identically zero.

The two event types are disjoint and exhaustive among first events, so their cumulative incidence functions sum to the probability of any event by $t$. Equivalently their derivatives add to $S(t)(h_A(t)+h_B(t))=-S^{\prime }(t)$. ThereforeFor the usual positive disease-death intensity ${\theta }_A>0$, the limit is one: eventually every individual has an event, even though treatment-side-effect events cease after $\tau$. If the permitted boundary ${\theta }_A=0$ is included, however,Then a fraction $e^{-{\theta }_B\tau }$ remains event-free forever, because all hazards vanish after $\tau$. This includes the no-event case ${\theta }_A={\theta }_B=0$. Thus an assertion that the limit always equals one requires ${\theta }_A>0$.

By conditional probability and continuity of event times,When $\tau >0$ and $q={\theta }_A+{\theta }_B>0$, the common factor $1-e^{-q\tau }$ cancels, givingBefore $\tau$, the constant cause-specific hazards divide the total event intensity in these proportions, regardless of how long the interval is. If $\tau =0$ or both intensities are zero, the conditioning event has probability zero and this conditional probability is undefined, not a ratio assigned by convention.