Poisson process has probability with other distribution

A bus platform is now empty of passengers. Passengers arrive according to a Poisson process with rate $\lambda$. Upon arrival the passenger will choose either bus A or bus B and wait in the corresponding line. We denote by $N_1(t)$ the number of passengers that arrive by time t and choose bus A, and by $N_2(t)$ the number of passengers that arrive by time t and choose bus B.

Find $\mathbb P\{N_1(t) = k_1,N_2(t) = k_2\}$ for $k_1, k_2 ≥ 0$ under the following conditions

(i) Let P ∼ Uniform(0, 1), independent of the arrival process of the passengers. Given P = p ∈ (0, 1), each passenger chooses bus A with probability $p$ and choose bus B with probability $1 − p$, independently with the arrival process and the choices of other passengers.

(ii) Let $T_1 > 0$ be the departure time of bus A and $T_2 > 0$ be the departure time of B. If the arrival time of a passenger is $t : t < max(T_1, T_2)$, he will choose bus A with probability $$ p(t):=\frac{\max(\frac{1}{T_1-t},0)}{\max(\frac{1}{T_1 - t},0) + \max(\frac{1}{T_2 - t},0)} \in[0,1]$$ and choose bus B with probability 1 − p(t), independently with the arrival process and the choice of the other passengers.

I have no idea how to handle these type of questions. Is it $Poisson(\lambda \cdot t \cdot A)$, where A is the PDF of the distribution?

Solution 1:

(i) Is easy: The probability that exactly $k$ passengers arrive by time $t$ is $$ {\mathbb P}\{N(t)=k\}=\frac{(\lambda t)^k}{k!}e^{-\lambda t}\,. $$ In case (i), the following conditional probability is from a binomial distribution: $$ {\mathbb P}\{N_1(t)=k_1, N_2(t)=k_2\,|\,N(t)=k\}={k\choose k_1}\,p^{k_1}\,(1-p)^{k_2}\,,\quad\text{ where } k=k_1+k_2\,. $$ Can you calculate now the unconditional probability?

(ii) Seems hard. Where is this from? The arrival time of the $n$-th passenger is $\tau_n=\sigma_1+...+\sigma_n$ where the times $\sigma_i$ are independent and exponentially distributed: $$ {\mathbb P}\{\sigma_i\le t\}=1-e^{-\lambda t}\,. $$ The joint probability density of $\tau_1,...,\tau_k$ can be calculated by differentiating \begin{align} &{\mathbb P}(\tau_1\le t_1,...,\tau_k\le t_k)={\mathbb P}\Big(\sigma_1\le t_1,...,\sigma_1+...+\sigma_k\le t_k\Big)\\[3mm] &\quad={\mathbb P}\Big(\sigma_1\le t_1,\sigma_2\le t_2-\sigma_1,...,\sigma_k\le t_k-\sigma_1-...-\sigma_{k-1}\Big)\\[3mm] &\quad=\int_0^{t_1}\int_0^{t_2-t_1}...\int_0^{t_k-t_1-...-t_{k-1}}\lambda^ke^{-\lambda s_1-...-\lambda s_k}\,ds_k\,...\,ds_2\,ds_1\,. \end{align}

According to the definition (ii) we have \begin{align} &{\mathbb P}\{N_1(t)=k_1, N_2(t)=k_2\,|\tau_1=t_1,...,\tau_k=t_k\}\\ &\quad=\sum_{\{j_1,...,j_{k_1}\}\cup\{i_1,...,i_{k_2}\}=\{1,...,k\}}p(t_{j_1})...p(t_{j_{k_1}})(1-p(t_{i_1}))...(1-p(t_{i_{k_2}}))\,. \end{align} Integrating this w.r.t. the joint probability density of $\tau_1,...,\tau_k$ gives the solution but I don't think this can be done other than numerically.