Starting at $6$ a.m, cars, buses and motorcycles arrive at a highway toll booth according to independent Poisson processes. Cars arrive about once every $5$ minutes, buses about once every $10$ minutes and motorcycles about once every $30$ minutes.
Find the probability that at least one other vehicle arrives between third and fourth car arrival.
If we let $Z_t$ be the number of buses and motorcycles in time interval $t$, we have that $Z_t\sim\text{Poi}(1/10 + 1/30)=\text{Poi}(2/15).$ We also know that the length of the internal time $T$ between cars is $\sim\text{exp(1/5)}.$
According to the book, they say that the sought probability is $P(Z_T>0)$ and they compute it as follows:
\begin{align} P(Z_T>0)=\int_{0}^{\infty}P(Z_T>0 \ | \ T=t)\cdot\frac{e^{t/5}}{5} \ dt=...\tag1 \end{align}
Questions:
- I don't understand why we seek $P(Z_T>0).$ How does this ensure that we end up inbetween car $3$ and car $4$?
- How do they get that integral? Is it that they are actually integrating the joint distribution of $Z$ and $T$ and then conditioning on $T$ and multiplying with its marginal distribution?
$T$ is defined as the time between cars $3$ and $4$ arrivals (actually, the time between any successive cars arrivals).
$Z_{t_1,t_2}$ is the number of buses and motorcycles in time interval $t_1 $ to $t_2$ . But because the process is homogeneous, the probability of this variable only depends on the interval length, so we call it $Z_t$ (with $t=t_2 - t_1$)
Then $Z_T$ is the number of buses and motorcycles that arrive between cars $3$ and $4$ arrivals.
Yes, basically. In general, $P(A)= \int P(A,B) dB=\int P(A |B) P(B) dB$ (marginalizaton and conditional probability). Think of $A$ as the indicator variable of the event of interest ($Z_t >0$) and $B$ as the interval length $T$.