# How to calculate the probability of 2 events happening in time series under only cdf information?

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 $$0$$. $$P_A(0)=0, P_A(+\infty)=1, P_A(t_1)\leq P_A(t_2)\ under\ t_1< t_2$$ 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 $$B$$?

• Is more infomation about $A$ needed? Any problem with this question? Jan 14, 2019 at 9:43

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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