A basic question on irreducible periodic markov chain For an irreducible periodic (period $2$) Markov Chain I know that both of the following two quantities are same and equal to $\pi(i)$:
$$ \lim_{n\to \infty} \frac{1}{2}(p_n(j,i) + p_{n+1}(j,i))$$
$$ \lim_{n \to \infty} \frac{1}{2}(\phi P^{n+1} + \phi P^{n+2})_i$$
where $\phi$ is the initial distribution.
But, I could not understand the meaning of the above two quantities.
 A: The first expression is a special case of the second.  Both represent the fraction of the time the system spends in state $i$ in the long run, the second when there is an arbitrary initial distribution, the first when the system starts in a particular state, $j.$
If a finite chain is irreducible and aperiodic, then the probability it is found in state $i$ after $n$ transitions approaches $\pi(i)$ as $n$ gets large, for any choice of initial distribution.
If the finite chain is irreducible and periodic, then the large $n$ limit of $p_n(j,i)$ does not exist.  Nevertheless, you can still say that, in the long term, the fraction of the time the system spends in state $i$ is $\pi(i)$.
If the chain has period $2,$ the set of states can be partitioned into two parts, $A$ and $B,$ such that the probability of a single-step transition from a state in $A$ to a state in $A$ or from a state in $B$ to a state in $B$ is $0.$  The transition matrix can therefore be written
$$
P=\begin{bmatrix}0 & R\\ S & 0\end{bmatrix},
$$
where $R$ contains the probabilities of a transition from a state in $A$ to a state in $B,$ and $S$ contains the probabilities of a transition from a state in $B$ to a state in $A.$  Then $p_{2n}(j,i)=0$ if $j$ and $i$ are in different parts and $p_{2n+1}(j,i)=0$ if $j$ and $i$ are in the same part.  Since
$$
P^2=\begin{bmatrix}RS & 0\\ 0 & SR\end{bmatrix},
$$
the chain $P^2$ is reducible.  The chains $(RS)$ and $(SR)$ do have steady-state probabilities; if $j$ and $i$ are in the same part, then $p_{2n}(j,i)$ tends toward the steady-state value for state $i;$ if $j$ and $i$ are in different parts, then $p_{2n+1}(j,i)$ tends toward the same steady-state value.
So in the expression
$$\frac{1}{2}\left(p_n(j,i)+p_{n+1}(j,i)\right),
$$
one of the terms will be $0$ and the other will tend toward a steady-state value.  Hence the expression tends toward half the steady-state value, reflecting the fact that half the time, it is impossible for the system to be in state $i,$ and half the time, the probability the system will be in state $i$ is approaching the steady-state value.
The situation is similar when there's an initial distribution $\phi.$  Let $a$ be the sum of the elements of $\phi$ corresponding to states in $A,$ and let $b$ be defined similarly for states in $B.$  So $a+b=1;$ $a$ and $b$ represent the probabilities the system is initially in a state of $A$ or a state of $B.$  Then the $i^\text{th}$ component of $\phi P_{2n}$ will approach $a$ times the steady-state value for state $i$ if $i\in A,$ and $b$ times the steady-state value for state $i$ if $i\in B.$  Likewise, the $i^\text{th}$ component of $\phi P_{2n+1}$ will approach $b$ times the steady-state value for state $i$ if $i\in A,$ and $a$ times the steady-state value for state $i$ if $i\in B.$  Your second expression contains the average of these probabilities, which makes sense, since each applies half the time.  Again, the limiting value is half the steady state value for state $i.$
