= Solution
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> is
$$
h_j(t)=\lim_{\Delta\downarrow0}\frac{\mathbb P(t\leq T<t+\Delta,J=j\mid T\geq t)}{\Delta}.
$$
For absolutely continuous event times the total <hazard function> is $\sum_jh_j(t)$, giving the event-free <survivor function> $S(t)=\exp[-\int_0^t\sum_jh_j(u)\,du]$. Since an event of type $j$ near $t$ has unconditional density $S(t)h_j(t)$, its <cumulative incidence function> satisfies
$$
\boxed{F_j(t)=\mathbb P(T\leq t,J=j)=\int_0^tS(u)h_j(u)\,du.}
$$
This <cause-specific hazard to cumulative incidence formula> does not require <independent> latent times for the competing causes. In general $F_j(t)$ is not $1-e^{-\int_0^th_j}$, since that expression ignores removal by other causes.
Put $a=\theta_A\geq0$, $b=\theta_B\geq0$ and $q=a+b$, with $\tau\geq0$. Here
$$
S(t)=e^{-at-b\min(t,\tau)}.
$$
Integrating $aS(u)$ separately before and after $\tau$ gives, for $q>0$,
$$
\boxed{F_A(t)=\begin{cases}\dfrac{a}{q}(1-e^{-qt}),&0\leq t\leq\tau,\\[4pt]\dfrac{a}{q}(1-e^{-q\tau})+e^{-q\tau}(1-e^{-a(t-\tau)}),&t>\tau.\end{cases}}
$$
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 $F_A=0$.
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