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Consider a airport exit with people leaving with a poisson process rate of 1 per minute. From the exit they can catch a cab, which arrives with a poisson process of 2 per minute. A person will wait no matter how many other customers are present but if a cab does not find a customer upon their arrival they have leave due not being allowed to wait.

a) If there are no people waiting at time 0, find the expected time for two customers waiting for the first time.
b) Find the average number of waiting customers
c) Calculate the limiting probability of each state 𝑖, 𝜋𝑖
d) How does the overall system look in a generator matrix?

I found the expected time to be 2 minutes due to 1+1 minutes of arrival in a) and the average number of waiting customers to be 1 in b). If those two are not incorrect how do I solve questions c) and d)?

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Here is (c). For $j\geq 1$ take $X_j$ to be the number of people waiting immediately following the $j^{\text{th}}$ arrival of either a cab or a passenger to the airport terminal. First notice $$P(X_{j+1}=0|X_j=0)=2/3$$ $$P(X_{j+1}=1|X_{j}=0)=1/3$$ On the other hand, we see for $i\geq 1$ that $$P(X_{j+1}=i-1|X_j=i)=2/3$$ $$P(X_{j+1}=i+1|X_j=i)=1/3$$ From the total law of probability, $$\pi_0=\frac{2 \pi_0}{3}+\frac{2 \pi_1}{3}$$ Similarly for $i\geq 1$ $$\pi_i=\frac{\pi_{i-1}}{3}+\frac{2\pi_{i+1}}{3}$$ Solving this system of equations subject to the condition that $\sum_{i=0}^{\infty}\pi_i=1$ yields $\pi_i=\frac{1}{2^{i+1}}$.

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