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Consider a DTMC with state space $\{1, 2,\ldots {}{} \}$. Now we want to calculate the probability that state 1 is followed by state 2 in the long run i.e, $P(X_n=1, X_{n+1}=2)$ as $n$ tends to infinity. Now if we evaluate this expression then it comes as $\pi_{1}p_{12}$. I have basic conceptual doubt here. If a DTMC reaches state 1 in the long-run then it will stay there for ever. Then the required probability should be zero. I know that I am making a mistake, but not able to find where.

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What makes you think it will stay there forever? It doesn't (necessarily) stay there forever if it starts there, after all. –  Jonathan Christensen Jan 11 '13 at 1:24
You should probably try to state rigorously what is meant by the phrase "in the long run". –  Nate Eldredge Jan 11 '13 at 1:54
Indeed, if $P(X_n=1)\to\pi(1)$ then $P(X_n=1,X_{n+1}=2)\to\pi(1)p(1,2)$. –  Did Jan 11 '13 at 11:13
0% accept rate? –  Did Jan 11 '13 at 11:13

1 Answer 1

If a DTMC reaches state 1 in the long-run then it will stay there for ever.

This is vague and, most likely, incorrect. The word "steady" in "steady-state analysis" refers not to the value of $X_n$ being steady, but to the probabilities $P(X_n=k)$, for each fixed $k$. So, the probability of $P(X_n=1)$ may converge to $\pi(1)$ but this does not mean that $X_n$ will stay at $1$. As Did remarked, in this scenario $P(X_n=1,X_{n+1}=2)\to\pi(1)p(1,2)$ as $n\to\infty$.

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