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The transition probabilities from one node to the other nodes have to sum up to $1$. This is true even if there is only one path out of a certain state. In what follows let $A^n$ the $n^{th}$ state and let $A_{i}^{n+1}$ denote the $(n+1)^{st}$ state of the Markov chain. Also, suppose that $P(A^n)>0$. $(i=1,2, ... m)$ Then $$\sum_{1} ... 1 I see no difference between the two expressions. One says something about Z_0=i_0,Z_1=i_1,\cdots Z_n=i_n for all i_k\in S and the other says (Z_0,...Z_n)=\Sigma for all \Sigma\subset S^{n+1} (assuming you have a typo). You're on a discrete set, so obviously the two imply each other by writing out \Sigma as a union over all tuples in it and using ... 1 It isn't . See Renewal Theory. If you are talking about the time of the next event T_{N+1} given the time of the last event T_N, then the Markov Property is that P(T_{N+1}|T_N,T_{N-1}...)=P(T_{N+1}|T_{N}). Nothing precludes the interarrival times from being any distribution, provided that inter-arriveal times are iid. Note also that the markov ... 1 Let's write out the transition matrix of X when a = 1:$$\mathcal P = \frac{1}{10} \begin{bmatrix} 1 & 3 & 0 & 1 & 5 & 0 \\ 2 & 2 & 1 & 0 & 1 & 4 \\ 6 & 0 & 1 & 1 & 2 & 0 \\ 6 & 0 & 1 & 1 & 0 & 2 \\ 3 & 0 & 4 & 1 & 0 & 2 \\ 3 & 0 & 1 & 4 & ...

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Consider a Markov chain on the integers where it moves right with probability 1. The limit distribution is all zeros, but this doesn't qualify as an invariant distribution because it needs at least one positive entry.

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Note that $P = P_1 \otimes P_2$ is the transition matrix of the process $(X^1,X^2)$ where $X^1$ is a Markov chain of transition matrix $P^1$, $X^2$ is a Markov chain of transition matrix $P^2$, and $X^1$ and $X^2$ are independent. Thus, if $\pi^1$ is stationary for $P^1$ and $\pi^2$ is stationary for $P^2$ then $\pi^1\otimes\pi^2$ is stationary for $P$. ...

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1) We can assume without loss of generality that $X$ starts in $C$, as otherwise we could consider the Markov chain $\{X_\tau, X_{\tau+1},\ldots\}$ where $\tau = \inf\{n\geqslant 0 : X_n\in C\}$. 2) If $X_0\in C$, then $X_n\in C$ for all $n$. Since $C$ is finite, if each state were visited only finitely many times, this would contradict the fact that \$X_0, ...

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