2
$\begingroup$

Studying homogeneous Poisson process I have the next problems:

  1. Passengers arrive to a bus station according to a homogeneous Poisson process with $\lambda = 4$ per minute. If the bus waits for $20$ passengers to leave, find the probability that each passenger have waited less than $10$ minutes.

  2. Suppose that the buses arrive to the same station according to a homogeneous Poisson porcess with $\lambda = 1$ per minute independent to the passengers' process. Find the probability that when the bus arrives to the station, there already are $n$ passengers waiting at the station.

For the first one I think I've got the answer but I want to verify if my thoughts are correct:

I know that, the interarrival times $T_i \sim Exp(4)$, and that $W_n = \sum_{i=1}^{n} T_i \sim \Gamma(n,4)$. If the bus waits for 20 passengers then the bus waits a total time of $W_{20}\sim \Gamma (20,4)$. If we're looking for the case that each passenger waits less than 10 minutes to leave, does that mean that it's sufficient to look for $\mathbb{P}[W_{20} \leq 10]$?

Now, for the second problem if $\{N_t^{(p)}\}_{t\geq 0}$ denotes the passengers' process and $\{N_t^{(b)}\}_{t\geq 0}$ the buses' process, then because the processes are independent, that means that the interarrival times and jumping times of each process are independent to the other one. If the bus arrives and there are already $n$ passengers waiting for it, does that means $W_{n}^{(p)} < W_1^{(b)}$ ?

If so, I was thinking that the probability the problem is asking for, should be

$$ \mathbb{P}[W_{n}^{(p)} < W_1^{(b)}] = \mathbb{P}[W_{n}^{(p)} < T_1^{(b)}]$$

but then, I'm not sure how to proceed, maybe with law of total probability:

$$\mathbb{P}[W_{n}^{(p)} < T_1^{(b)}] = \int_{0}^{\infty} \mathbb{P}[W_{n}^{(p)} < t \mid T_1^{(b)}= t] \cdot \mathbb{P}[T_1^{(b)}=t] dt$$

but I'm thinking maybe it's simplier than that integral, am I in the right direction o I'm just totally lost?

$\endgroup$

2 Answers 2

2
$\begingroup$

I believe the first question should be $P(W_{20} - W_1 \le 10)$ since you do not care about the time before the first person arrives.


For the second, it may be useful to use results about Poisson superposition and thinning.

Instead of thinking about two independent Poisson processes (passengers and buses), it turns out you can think of a single Poisson process of rate $4+1=5$, and then for each arrival flip a biased coin to decide if it is a passenger arrival (probability $\frac{4}{5}$) or a bus arrival (probability $\frac{1}{5}$).

So, if you want the probability that exactly $n$ people arrive before the first bus arrives, then this is simply $(4/5)^n (1/5)$.

If you want the probability that at least $n$ people arrive before the first bus arrives, then it's $\sum_{k=n}^\infty (4/5)^k (1/5)$ which can be simplified using geometric series to be $(4/5)^n$.

Of course, this approach is contingent on knowing the aforementioned results about Poisson process.


Alternate approach: if you want to follow your approach, the probability that at least $n$ people arrive before the first bus arrives is indeed $P(W_n^{(p)} < W_1^{(b)})$. If you first condition on $W_1^{(b)} = t$, then $$P(W_n^{(p)} < W_1^{(b)} \mid W_1^{(b)} = t) = P(W_n^{(p)} < t) = 1 - \sum_{k=0}^{n-1} e^{-4t} \frac{(4t)^k}{k!}.$$ Then indeed you can integrate as you have set it up. $$P(W_n^{(p)} < W_1^{(b)}) = 1 - \sum_{k=0}^{n-1} \int_0^\infty e^{-4t} \frac{(4t)^k}{k!} \cdot e^{-t} \, dt = 1- \sum_{k=0}^{n-1} (4/5)^k (1/5) \underbrace{\int_0^\infty e^{-5t} \frac{5^{k+1} t^k}{k!}\, dt}_{=1}.$$ This leads to the same answer as above.

$\endgroup$
0
$\begingroup$
  1. Almost but not quite. What you wrote is the probability that all 20 people arrive within 10 minutes of whenever you start the clock (eg when the previous bus leaves). By my reading, the problem wants the probability that no passenger has waited for more than 10 minutes. The passenger who is waiting the longest is the first to arrive, so really what we want is whether the remaining 19 people arrive within 10 minutes of the first one. When the first one arrives doesn't actually matter. So you want $\mathbb{P}[W_{19} \le 10]$.

  2. Yes, that seems right to me, though you don't need the conditional probability in that integral; they're independent so conditioning has no effect. It also might be a bit cleaner to integrate in the other order, but conceptually the same. $$\mathbb{P}[W_n^{(p)} < T_1^{(b)}] = \int_0^\infty \mathbb{P}[W_n^{(p)} < t] \cdot \mathbb{P} [T_1^{(b)} = t]\,dt = \int_0^\infty \mathbb{P}[W_n^{(p)} = t] \cdot \mathbb{P}[T_1^{(b)} \ge t]\,dt$$

$\endgroup$

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.