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

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?

• If the two rates are unequal, the system is not at equilibrium. Hence the name "balance". – Did Feb 2 '15 at 10:51
• @Did so in a system where customers are served faster than they arrive, the system will never be in equilibrium? – CodyBugstein Feb 2 '15 at 14:58
• @Did quote from Wikipedia page about M/M/1: The model is considered stable only if λ < μ. Clearly $\lambda$ and $\mu$ do not need to be equal – CodyBugstein Feb 2 '15 at 15:13
• Please read what I wrote, which is also in your post: the two rates must be equal, that is, $\pi_0\lambda=\pi_1\mu$. The rest is of your own invention (but sending me to WP is, how to put it... kinda sweet.). – Did Feb 2 '15 at 17:13
• Let me note that the user who posted an answer basically explained to you the same point. If ever you plan to react to my comment as you did to theirs, just don't. – Did Feb 2 '15 at 17:15

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.

• Can you briefly explain your answer in words? – CodyBugstein Feb 2 '15 at 10:44
• Why did you use a and d? – CodyBugstein Feb 2 '15 at 10:46
• If $\lambda\geq\mu$ then the equation does not hold because there is no limit distribution (The equations ki3i stated still hold for any finite time, but both of the time dependent probabilities will go to zero). Otherwise, it holds and the informal explanation is that what comes in must go out, on average. – QQQ Feb 2 '15 at 11:00
• @Imray, Who said $\lambda$ and $\mu$ must be equal??? The rates being referred to as being equal are $\pi_{0,t}\lambda$ and $\pi_{1,t}\mu$; I prove the equality of these in my answer under an equilibrium assumption. They are rates because they are limits, as $\delta t\to 0$, of a quotient with denominator $\delta t$. – ki3i Feb 2 '15 at 15:42
• @Imray, ...the part you highlighted in yellow does not refer to $\lambda$ and $\mu$. It refers to the rates $\pi_{0,t}\lambda$ and $\pi_{1,t}\mu$ being equal. – ki3i Feb 2 '15 at 16:14

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 $State \ 0$