Buses arrive at 116th and Broadway at the times of a Poisson arrival process with intensity λ arrivals per hour. These may either be M104 buses or M6 buses;
the chance that a bus is an M104 is.6, while the chance that it is an M6 is .4, and the types(M6 or M104) of successive buses are independent.
(a) If I wait for an M104 bus, what is the chance that I will wait longer than x hours?
My approach was to find the probability that at least 1 M104 bus arrived in the first hour by taking the sum of the probabilities of x in the first hour and using the complimentary of the probability that they weren't all M6 busses. Namely
$$P(t=1)=\sum_{k=1}^{\infty} (\frac{\lambda^k\cdot e^{-\lambda}}{k!})\cdot(1-0.4^{k}) = 1 – e^{-\frac{3\lambda}{5}}$$
Then to take that probability to a Geometric distribution of number of trials (h hours) to the first success:
$$P(t=h) = (1-(1-e^{-\frac{3\lambda}{5}}))^{h-1}\cdot(1-e^-\frac{3\lambda}{5})=(e^{\frac{3\lambda}{5}} – 1) e^{-\frac{3h\lambda}{5}} $$
And finally answering as $$P(h>x)=\sum_{h=x+1}^{\infty} (e^{\frac{3\lambda}{5}} – 1) e^{-\frac{3h\lambda}{5}}=e^{-\frac{3\lambda\cdot x}{5}} $$
But this seems tedious and frankly I don't understand how these sums are evaluating like they are. Is there any insight to this? Could I have used some shorter thought processes for the first P(t=1)?
Best Answer
The principle of thinning is well-suited for this type of problem. It roughly says the following. Suppose you have a Poisson process with intensity $\lambda$, and with each arrival you flip a coin independently at random to determine whether the arrival is of type I or type II. It has probability $p$ of being type I, and probability $1 - p$ of being type II. Then this Poisson Process can be decomposed into two independent Poisson Processes describing type I and type II arrivals. The former has intensity $p \lambda$, and the latter $(1 - p) \lambda$.
Applying this to your problem tells us that the arrivals of M104 buses follows a Poisson process of intensity $\frac{3}{5} \lambda$. The first arrival time is exponentially distributed with intensity according to the corresponding Poisson Process intensity, so the probability that you will have to wait more than $x$ hours for the first M104 bus is just $\mathbb{P}(\text{Expon}(3\lambda/5) > x)$.
You may also find useful for similar problems the principle of superposition for Poisson processes, which states the following. Given two independent Poisson processes with intensities $\lambda$ (for type I arrivals) and $\mu$ (for type II arrivals), the combined arrivals process is given by a Poisson process of intensity $\lambda + \mu$. This is effectively the converse of the principle of thinning.