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In time domain $0\rightarrow T$, there are two independent events $A$ and $B$. $B$ follows Poisson Process with density $\lambda$. It's easy to get $P_B(t)$ which denotes $P_B(N(\tau+t)-N(\tau)\geq 1)$ in a certain time duration $t$.

$P_A(t)$ denotes $P_A(N(t)-N(0)\geq 1)=P_A(N(t)-0\geq 1)$. $P_A(t)$ means the probability that $A$ happens after a certain time duration $t$ from time $0$. $P_A(0)=0, P_A(+\infty)=1, P_A(t_1)\leq P_A(t_2)\text{ under } t_1< t_2$ obviously.

If we don't know what $P_A(t)$ is exactly, we just know there is a $P_A(t)$.

Can we answer the following question with only $P_A(t)$ and Poisson Process $B$? If not, is more information about $A$ needed?

What is the probability $P(A\rightarrow B)$ that in time duration $0\rightarrow T$ event $A$ happens and then event $B$ happens(both events happens and $B$ happens after $A$). How to represent $P(A\rightarrow B)$ with $P_A(t)$ and any information about Poisson Process $B$?

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  • $\begingroup$ Is more infomation about $A$ needed? Any problem with this question? $\endgroup$
    – oleotiger
    Commented Jan 14, 2019 at 9:43

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I follow the reasoning from your previous question.

One contribution to $1-P_{A\rightarrow B}$ is that the event $A$ does not happen at all in a time $T$, that probability is $1-P_A(T)$. For the remaining contributions to $1-P_{A\rightarrow B}$ the event $A$ happens at least once, denote by $\tau$ the time at which this first happens, with probability density $dP_A/d\tau$. Then from $\tau$ to $T$ the event $B$ does not happen, with probability $e^{-(T-\tau)\lambda}$. This gives in total $$1-P_{A\rightarrow B}=1-P_A(T)+\int_{0}^T \frac{dP_A(\tau)}{d\tau}e^{-(T-\tau)\lambda}\,d\tau$$ $$\Rightarrow P_{A\rightarrow B}=\lambda\int_0^T e^{-(T-\tau)\lambda}P_A(\tau) \,d\tau.$$

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