buses arrive at a bus stop according to a non-homogeneous Poisson process with the rate
function $$\mu(t) = 1 + t$$
I have some idea on how to do this but I don't know if I'm correct.
Q1: Assuming the first bus arrives at bus stop at time t = 2, what is the probability that one has to wait more than 3 time units for the second bus to arrive?
So if the first bus arrive at t = 2, then is the probability of one has to wait more than 3 time units from the second bus to arrive just $$P(N([2,5]) = 0)?$$
Q2: Assuming that 2 buses has arrived at the bus stop at t = 6, find the probability that the first bus arrived at the bus stop at t = 2?
So $$N([0,6]) = 2$$, and $$N(]0,2]) = 1,$$
$$P(N([0,2]) = 1|N([0,6]) = 2) = \frac{P(N([0,2]) = 1, P(N([2,6]) = 1)}{P(N([0,6]) = 2)}$$
Is this correct?
Best Answer
Let $t_0,t_1,\ldots ,t_n\in[2,5]$ such that $2=t_0<t_1<\ldots<t_{n-1}<t_n=5$. For each $1\leq i \leq n$ take $\Delta t_i=t_i-t_{i-1}$ and let $X_i$ denote the number of arrivals that occur in the time interval $[t_{i-1},t_i)$. Notice how $X_i$ is approximately $\text{Poisson}\Big(\mu(t_i)\Delta t_i \Big)$ and that $X_1, \ldots ,X_n$ are independent. Hence $$P\Big(\text{No arrivals in} [2,5]\Big) \approx \prod_{i=1}^nP(X_i=0)=\exp{\bigg(-\sum_{i=1}^n\mu(t_i)\Delta t_i\bigg)}$$ Taking $n \longrightarrow \infty $ gives us $$P\Big(\text{No arrivals in} [2,5]\Big)=e^{-\int_2^5\mu(t)dt}=\frac{1}{e^{13}\sqrt{e}}$$ In fact, if $N\Big([a,b]\Big)$ denotes the number of arrivals in $[a,b]$, then it turns out that $$N\Big([a,b]\Big) \sim \text{Poisson}\bigg(\int_a^b\mu(t)dt\bigg)$$