-1
$\begingroup$

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 $T$.

$\endgroup$
  • $\begingroup$ 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
  • 1
    $\begingroup$ 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
0
$\begingroup$

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 $P_B\equiv 0$.

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking "Post Your Answer", you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.