# What is the limiting probability that there are n people in the facility?

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?

• 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. – Acccumulation Dec 21 '18 at 21:01
• I think, exactly n. Well, what if μ>λ? Is my answer right? – mhk Dec 21 '18 at 21:17
• Possible duplicate of Limiting probability of Markov chain(Terminology) – mhk Dec 22 '18 at 0:07

## 2 Answers

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.

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 $$\pi_n = \rho^n(1-\rho).$$