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Customers arrive at a store at a Poisson rate of λ and there is a single server with rate μ. The arrival and service times are independent random variables. Customers leave the facility immediately after receiving the service. Find the limiting probability that there are n people in the facility,n=0, 1, 2,... I saw this question in a stochastic book and found a solution but I am not sure. I thought like P0= λ/ μ P1= λ/ μ * P0 … Pn= (λ/ μ)n * P0 Do you think that is right?

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  • $\begingroup$ At least n, or exactly n? If $\mu<\lambda$, then I would think that as $t \rightarrow \infty$, the probability that it will be any particular n would go to 0. $\endgroup$ – Acccumulation Dec 21 '18 at 21:01
  • $\begingroup$ I think, exactly n. Well, what if μ>λ? Is my answer right? $\endgroup$ – mhk Dec 21 '18 at 21:17
  • $\begingroup$ Possible duplicate of Limiting probability of Markov chain(Terminology) $\endgroup$ – mhk Dec 22 '18 at 0:07
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You’re close to the right answer!

If $\lambda > \mu$, then the system does not possess a stationary distribution as the expected number of people in the system grows and grows indefinitely.

If $\lambda < \mu$, you’ll find the long-term probability the queue has $n$ customers, $\pi_n$, is:

$$\pi_n = (1 - \frac{\lambda}{\mu})(\frac{\lambda}{\mu})^n$$

The difference between my answer and yours is only a normalizing factor; both of our answers are invariant measures for the stochastic process, but only the answer above is also a probability distribution.

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This is a $M/M/1$ queue with arrival rate $\lambda$ and service rate $\mu$. Let $X(t)$ be the number of customers in the system at time $t$, then $\{X(t):t\geqslant0\}$ is a continuous-time Markov chain on $\{0,1,\ldots\}$ with generator Q given by $$ q_{ij} = \begin{cases} \lambda,& j=i+1\\ \mu,& j=i-1\\ -(\lambda+\mu),& j=i, i>1\\ -\lambda,& j=i=1\\ 0,& \text{otherwise}. \end{cases} $$ Since this is a birth-death chain ($q_{ij}=0$ when $|i-j|>1$), we have the detailed balance equations $$\lambda \pi_i = \mu\pi_{i+1},\quad i=0,1,2,\ldots $$ This yields the recurrence $\pi_i = \rho^i \pi_0$ where $\rho=\lambda/\mu$. Assuming that $\rho<1$, from the condition $\sum_{i=0}^\infty \pi_i=1$ that a stationary distribution $\pi$ must satisfy, we have $$ \pi_0 = \left(\sum_{i=0}^\infty\rho^i \right)^{-1} = 1-\rho. $$ It follows that the limiting probability $\lim_{t\to\infty}\mathbb P(X(t)=n)$ is given by $$ \pi_n = \rho^n(1-\rho). $$

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