Assume bursts arrive as a Poisson process and the number of instances from a bursts follow a certain distribution $G$ (Assume it's a geometric if it really matters). Now, I need to calculate the probability of having three instances in a time span $[0,T]$. There are multiple scenarios where how this may happen:
- One arrival and 3 instances bursts from that single arrival.
- Two arrivals, where two instances burst from a one arrival and only one instance bursts from the other.
- Three arrivals, where a one instance bursts from each arrival.
I have no problem in calculating the probability of the first one, however I am struggling to calculate the second and the third scenarios. The reason is that the probability of having exactly two arrivals is $(\lambda t)^2 e^{\lambda T} \over 2!$ and I can't handle each event separately. For instance:
P[one arrival] * P[two bursts] * P[one arrival] * P[one bursts]
How to handle the second and third cases?
Say that $a$ is the number of arrivals (the average number of arrivals in the given interval is $\lambda$), $b$ is the number of bursts for one arrival, $G(b)$ is the probability of having $b$ bursts during one arrival: for each arrival you have to extract a number $b$ from $G$.
You want to calculate the probability that the total number of bursts observed in the given interval sum up to a number $b_{tot}=3$.
I assume that arrivals with no bursts are not possible. So, to have $b_{tot} =3$ you have three possibilities: $(a=1, b_1=3)$ or $(a=2, b_1+b_2=3)$ or $(a=3, b_1+b_2+b_3=3)$. Then,
$$ P(a=3,b_1=1,b_2=1,b_3=1) = e^{-\lambda} \frac{\lambda^3}{3!} G(1)^3$$ $$ P(a=2,b_1=1,b_2=2) = e^{-\lambda} \frac{\lambda^2}{2!} G(2)G(1)$$ $$ P(a=2,b_1=2,b_2=1) = e^{-\lambda} \frac{\lambda^2}{2!} G(2)G(1)$$ $$ P(a=1,b_1=3) = e^{-\lambda} \frac{\lambda^3}{3!} G(3)$$
Therefore,
$$ P(b_{tot}=3) = P(a=1,b_1=3) + 2 P(a=2,b_1=2,b_2=1) + \\ +P(a=3,b_1=1,b_2=1,b_3=1) $$