Probability distribution over time of an event dependent on a prior event

Question

Suppose I have an event $A$ that can only occur once in a experiment. A large ensemble of experiments reveals that $A$ occurs at a rate $r\, \mathrm{d}t$. For simplicity take $r$ as a constant. The cdf is

$$P_A(t)=1-e^{-rt}$$

Now take a second event $B$, which can also only occur once. $B$ is impossible until $A$ occurs, but repeated measurements on a large ensemble reveal that, in cases where $A$ has occurred, $B$ occurs at a rate $x r\, \mathrm{d}t$. If $A$ takes place at $t_1$. $$P_B(t|t_1)=1-e^{-xr(t-t_1)}$$ (sorry if this notation is bad - hopefully my meaning is clear).

What is the unconditional cdf of $B$ over time? My thought was that it should be

$$P_B(t)=\int_0^t \mathrm{d}t_1 \frac{\mathrm{d}P_A}{\mathrm{d}t_1}(t_1)P_b(t|t_1)$$ but this doesn't seem to be correct. If $x=1$, this reduces to $P_A(t)$ which can't be right.

Answer 1

You were on the right track. But if you carefully do the integral you present, the answer for $x=1$ is $$ P_B(t) = 1 - e^{-rt} - rte^{-rt} $$ which is not the same as $P_A(t)$.

Your likely mistake was to discard a constant $re^{-rt}$ when integrating $dt_1$.

Answer 2

In other words... one assumes that $A$ is exponential with parameter $r$ and that $B-A$ is independent of $A$ and exponential with parameter $xr$, then, for every nonnegative $t$, $$f_B(t)=\int_0^tf_A(s)f_{B-A}(t-s)\mathrm dt=\int_0^tr\mathrm e^{-rs}xr\mathrm e^{-xr(t-s)}\mathrm ds=xr^2\mathrm e^{-xrt}\int_0^t\mathrm e^{-r(1-x)s}\mathrm ds.$$ If $x\ne1$, $$f_B(t)=\frac{xr}{(1-x)}\mathrm e^{-xrt}\int_0^tr(1-x)\mathrm e^{-r(1-x)s}\mathrm ds=\frac{xr}{(1-x)}\mathrm e^{-xrt}(1-\mathrm e^{-r(1-x)t}),$$ that is, $$f_B(t)=xr\frac{\mathrm e^{-xrt}-\mathrm e^{-rt}}{1-x}.$$ The case $x=1$ yields $$f_B(t)=r^2t\mathrm e^{-rt}.$$