Suppose there are two independent events A and B. The probability that A or B happens is $P_A(t)$ and $P_B(t)$ respectively where $t$ represents the time duration. The probability increases as time duration increases. Both events can happen many times.

Assuming there is a time duration $0\rightarrow T$. What's the probability that event A happens and then B happens in time duration $0\rightarrow T$? In other words, what's the probability that we can find a event series $A,B$ in the duration $T$.

  • $\begingroup$ Is time discrete or continuous? If it is continuous, are $P_A(t)$ and $P_B(t)$ probability densities? $\endgroup$ – Federico Poloni Dec 29 '18 at 9:58
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    $\begingroup$ are these Poisson processes? if not, how are events $A$ at different times correlated? $\endgroup$ – Carlo Beenakker Dec 29 '18 at 10:17
  • $\begingroup$ Time is continuous. $t$ in $P_A(t)$ and $P_B(t)$ means time duration such as 1 minute or 0.5 hour etc. $\lim_{t\rightarrow \infty}P_A(t)=1$. $\lim_{t\rightarrow \infty}P_B(t)=1$ $\endgroup$ – oleotiger Dec 29 '18 at 10:17
  • $\begingroup$ These can be treated as Poisson processes. Memoryless in time domain. $\endgroup$ – oleotiger Dec 29 '18 at 10:21

Let me denote the probability that that there is an event "first $A$ then $B$" by $P_{AB}$ and let me consider $1-P_{AB}$.

One contribution to $1-P_{AB}$ is that the event $A$ does not happen at all in a time $T$, for a Poisson process that probability is $$P_0=\exp\left(-\int_0^T P_A(t)\,dt\right).$$ For the remaining contributions the event $A$ happens at least once, denote by $\tau$ the time at which this first happens. A contribution to $1-P_{AB}$ we require a time interval from 0 to $\tau$ where $A$ does not happen, then $A$ happens, and then from $\tau$ to $T$ the event $B$ does not happen. This gives in total $$1-P_{AB}=P_0+\int_{0}^T \exp\left[-\int_0^\tau P_A(t)\,dt\right]P_A(\tau)\exp\left[-\int_\tau^T P_B(t)\,dt\right]\,d\tau.$$

As a check, let's verify that $P_{AB}=0$ if $P_B(t)$ is identically zero for all times between 0 and $T$. We then have $$P_{AB}(T)=1-\exp\left(-\int_0^T P_A(t)\,dt\right)-\int_0^T \exp\left[-\int_0^\tau P_A(t)\,dt\right]P_A(\tau)\,d\tau.$$ To see that this is indeed zero, first note that $P_{AB}(0)=0$ and then $$\frac{d}{dT}P_{AB}(T)=P_A(T)\exp\left(-\int_0^T P_A(t)\,dt\right)-\exp\left(-\int_0^T P_A(t)\,dt\right)P_A(T)=0,$$ so $P_{AB}(T)=0$ for all $T$ if $P_B\equiv 0$.

  • $\begingroup$ I didn't understand the part $exp[\int_0^{\tau}P_A(t)dt]$. Why is there $exp$? $\endgroup$ – oleotiger Dec 29 '18 at 11:20
  • $\begingroup$ The probability that from $\tau$ to $T$ the event $B$ does not happen is $1-P_B(T-\tau)$? Isn't it? $P_B(T-\tau)$ means that $B$ happens during $\tau$ to $T$. $\endgroup$ – oleotiger Dec 29 '18 at 11:23
  • $\begingroup$ a Poisson process is described by a probability density $P_B(t)$, such that $P_B(t)dt$ is the probability that the event $B$ happens in the infinitesimal time interval from $t$ to $t+dt$. To go from an infinitesimal interval $(t,t+dt)$ to a finite interval $(\tau,T)$ you have to integrate. The appearance of the exponent is explained, for example, in en.wikipedia.org/wiki/… $\endgroup$ – Carlo Beenakker Dec 29 '18 at 12:00
  • $\begingroup$ What if $A$ is not a Poisson process but just a memoryless stochastic process? $\endgroup$ – oleotiger Dec 29 '18 at 12:43
  • $\begingroup$ Let me change the definition of $P_A(t)$. $B$ still is a Poisson process in $0\rightarrow T$. $A$ is no longer Poisson process. $P_A(t)$ means the probability that $A$ happens during $0\rightarrow t$. Then what's the probability that $A,B$ happens during $0\rightarrow T$? $\endgroup$ – oleotiger Dec 29 '18 at 13:16

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