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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.

user27118
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  • It would be easier to check your computation of the integral if you showed how you did it. – David K Oct 23 '14 at 21:44
  • Do you how I came up with the integral expression or the result I got when x=1? – user27118 Oct 23 '14 at 21:47
  • I mean how you computed that $P_B(t) = P_A(t)$ when $x = 1,$ although if there is something to say about why that integral should give $P_B(t),$ it wouldn't hurt to say that too. – David K Oct 23 '14 at 21:53
  • But you have answers that have deduced that information already, so maybe it no longer matters. – David K Oct 23 '14 at 21:54

2 Answers2

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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 $dt_1$.

Mark Fischler
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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 $$f_B(t)=r^2t\mathrm e^{-rt}.$$

Did
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  • This provides the pdf - getting the cdf from this is not trivial (to me at least with my feeble integrating skills). But it is useful to confirm the results from the integral in the question by differentiating. – user27118 Oct 23 '14 at 22:18
  • Surely you can integrate this PDF to get the CDF... – Did Oct 23 '14 at 22:20
  • Yes for constant r - with te^-rt = d/dr e^-rt - but someone trying to google the answer might not. Also not clear in the answer if this is equivalent to the formulation in the question, although with some checking it seems to be. The question version is important because it is much easier in the general case r(t). – user27118 Oct 23 '14 at 22:29
  • Not sure that I am following what you are saying, but this is not important at all. – Did Oct 23 '14 at 22:40