In M/M/1 Markov process, why must entering and leaving the zero state be equal?

Question

According to the image below, which I snipped from this article, the rate of leaving State 0 and the rate of arriving into State 0 are equal. The article simple asserts this but does not explain why.

Conceptually, this makes no sense to me, as there are many cases where $\lambda$ is greater or less than $\mu$.

Is there something I am missing? Why must they be equal?

Answer 1

Define:

(i) $P_{{\bf a},t}{}:={}P(1\,{\bf a}\mbox{rrival in }(t,t+\delta t]\,\vert \, \mbox{system in state 0 at time }t )$;

(ii) $P_{{\bf c},t}{}:={}P(1\,{\bf c}\mbox{ompletion in }(t,t+\delta t]\,\vert \, \mbox{system in state 1 at time }t )\,;$

(iii) $\pi_{i,t}{}:={}P(\mbox{ system in state }i\mbox{ at time }t)\,.$

At equillibrium, for each system state, the probability of the system being in the given state does not change with time. Therefore, $$ \lim\limits_{\delta t\to 0}\dfrac{\pi_{0,t+\delta t}{}-{}\pi_{0,t}}{\delta t}{}={}0\,. $$

This means that,

$$ \begin{eqnarray*} &\overbrace{\pi_{0,t+\delta t}}^{\begin{array}{c}probability\,system\\is\,in\,state\,0\,at\,time\\t+\delta t\end{array}}{}={}\overbrace{\pi_{0,t}\left(1{}-{}P_{{\bf a},t}\right)}^{\begin{array}{c}probability\,system\\undergoes\,no\,change\\from\,state\,0\,in\,(t,t+\delta t]\end{array}}{}+{}\overbrace{\pi_{1,t}P_{{\bf c},t}}^{\begin{array}{c}probability\,system\\undergoes\,change\\from\,state\,1\,to\,0\,in\,(t,t+\delta t]\end{array}}&\newline &&\newline {}\implies{}&\lim\limits_{\delta t\to 0}\dfrac{\pi_{0,t+\delta t}{}-{}\pi_{0,t}}{\delta t}{}={}-\pi_{0,t}\lim\limits_{\delta t\to 0}\dfrac{P_{{\bf a},t}}{\delta t}{}+{}\pi_{1,t}\lim\limits_{\delta t\to 0}\dfrac{P_{{\bf c},t}}{\delta t}\newline &&\newline {}\implies{}&0{}={}-\pi_{0,t}\lambda{}+{}\pi_{1,t}\mu\,,& \end{eqnarray*} $$

from which the assertion follows.

Answer 2

I figured out the answer to my confusion.

To recap, I was wondering why it must be that $$P_0\lambda = P_1\mu$$

It seemed strange to me, since I knew that $\lambda$ and $\mu$ could be different from one another and $P_0$, $P_1$ could be anything else. But I asked, why should the outflow of State $0$ be equal to the inflow to $0$ via State $0$.

Solution

What I realized is that $State \ 1$ is directly dependent on the outflow from $State \ 0$. Therefore the probability of $State \ 1$ even ever existing is built in to the expression $P_0\lambda$. So it makes sense to say that the flow back from $1$ - given that we are in State 1 - to $0$ is be equal to the flow from $0$ to $1$

To put it into words:

The probability of being in $State \ 0$ and moving to $State \ 1$ is equal to the probability of being in $State \ 1$ and moving to $State \ 0$