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I'm supposed to determine the stationary distribution, when it exists, for a birth and death process having constant parameters $\lambda_n=\lambda$ for $n=0,1,2,...$ and $\mu_n=\mu$ for $n=1,2,...$

My attempt:

This looks like a steady-state distribution which would give us the following system of equations:

$\lambda_{n-1} p_{n-1} + \mu_{n+1} p_{n+1} - (\lambda_n \mu_n)p_n=0$ for $n \ge 1$

$\mu_1 p_1 - \lambda_0 p_0=0$

Solving the system to get:

$p_n=\frac{\lambda_0 \lambda_1 \dots \lambda_{n-1}}{\mu_1 \mu_2 \dots \mu_n}p_0$

So $p_0$=$$(1+ \sum _{n \ge 1}^{ } \frac{\lambda_0 \lambda_1 \dots \lambda_{n-1}}{\mu_1 \mu_2 \dots \mu_n})^{-1}$$

I feel like I haven't explained this correctly though or that I don't have the right answer. Any suggestions?

user193429
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1 Answers1

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I think you're on the right track though I'll provide a full answer just for clarity.

The problem you describe is rather classical, as it corresponds to finding the stationary distribution of the $M/M/1$ queue. It can be shown that this Markov chain is reversible with respect to the stationary distribution, $\pi$, which gives us the so-called stationary balance equations, $$ \lambda \pi_{n-1} = \pi_{n} \mu.$$

I'm using the fact that $\lambda_n = \lambda$ and $\mu_n = \mu$ (i.e. as you describe, the birth and death rates are independent of state). Applying reversibility over and over gives us that $$\pi_n = \rho^n \pi_0,$$ where $\rho=\lambda/\mu$. Finally, imposing the normalization condition $$\sum_{k=0}^\infty \pi_k = 1$$ gives you that $\pi_0 = (1-\rho)$ and hence $$\pi_n = (1-\rho)\rho^n.$$ Of course, a stationary distribution exists iff $\rho < 1$, since this is necessary and sufficient for the convergence of the geometric series.

megaads92
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