Limiting distribution of a Markov Chain

Def A Markov chain $(X_n{)}_{n\ge 0}$ is said to admit a limiting distribution if the following conditions are satisfied:

1. the limits

$${\pi }_j:=\underset{n\to \infty }{\lim }P(X_n=j|X_0=i)$$

exists for all $i,j\in I$, and 2. they form a probability distribution on $I$, i.e.

$$\sum _{j\in I}\underset{n\to \infty }{\lim }{P}_i(X_n=j)=1,\forall i\in I$$

Remark If condition $1$ is satisfied and the state space $I$ is finite, then condition $2$ is always satisfied (in this case we can exchange the limit with the summation.)

Example We can compute the limiting distribution of the two state MC with transition matrix

$$P=\left(\begin{array}{cc}1-a& a\\ b& 1-b\end{array}\right)$$

we need to compute ${\lim }_{n\to \infty }{P}^n$, to do so we can diagonalize it and get

$$P^n=\frac{1}{a+b}\left(\begin{array}{cc}b+a(1-a-b{)}^n& a(1-(1-a-b{)}^n)\\ b(1-(1-a-b{)}^n)& a+b(1-a-b{)}^n\end{array}\right)$$

if $a=b=0$ then $P=I$, we have a trivial MC. With the hypotesis $(1-a-b)<1$ the limit exists

$$\underset{n\to \infty }{\lim }{P}^n=\frac{1}{a+b}\left(\begin{array}{cc}b& a\\ b& a\end{array}\right)=\left(\begin{array}{cc}\frac{b}{a+b}& \frac{a}{a+b}\\ \frac{b}{a+b}& \frac{a}{a+b}\end{array}\right)$$

which also show that the limiting probability distribution doesn’t care about the starting state $X_0$.