Buses leave a bus station according to a Poisson process
$\{B_t\}$ of rate $\lambda$, while travellers reach the same station according to (an independent) Poisson process $\{T_t\}$ of rate $\mu$. Travellers get on the bus and leave the station as soon as a bus departures.$(1)$ Let $N$ be the random variable, expressing the number of passengers on a departuring bus. Find $P(N=k),~k=0,1,2,\ldots$
$(2)$ Let $\Lambda$ be the random variable, expressing the number of departuring buses till the time the $1000$th traveller has reached the station. Find $P(\Lambda=k),~k=0,1,2,\ldots$
Attempt.
(1) In order for $N$ to be $k$ there must be $k$ travellers reaching the bus station and no bus leaving in between. The time needed for k travellers to reach the station is $W\sim Gamma(k,\mu)$, so $$P(N=k)=P(B_W=0)=\int_{0}^{+\infty}P(B_w=0)f_W(w)dw=\int_{0}^{+\infty}e^{-\lambda w} \frac{\mu^k}{(k-1)!}w^{k-1}e^{-\mu w}dw.$$
(2) The time needed for the $1000$th traveller to reach the station is $W\sim Gamma(1000,\mu)$, so $$P(\Lambda=k)=P(B_W=k)=\int_{0}^{+\infty}P(B_w=k)f_W(w)dw=\int_{0}^{+\infty}e^{-\lambda w}\frac{(\lambda w)^k}{k!} \frac{\mu^{1000}}{999!}w^{999}e^{-\mu w}dw.$$
Am I on the right path? Thank you!
Best Answer
For the first part: At time $W\sim \Gamma(k,\mu)$, $k$ travellers have arrived to the bus station. Thus we know that if $B_W = 0$, then at least $k$ passengers will leave with the first bus*. So you are really evaluating $P(N\geq k)$. Your final integral in (1) should be easily computed by recognising what looks like the density of another gamma distribution. To generalise the thinking from the 'first bus' to 'a bus', it is of course important that a Poisson process $L$ satisfies, that $L(t)-L(s)$ is equal in distribution to $L(t-s)$ and that it is independent of $L(l)_{l\leq s}$. Just have this in mind: Showing it for the first bus is enough.
For the second part: Seems all fine.
*: Maybe some additional considerations should be given to this, after all the Poisson processes only have càdlàg sample paths.