Why does $ \mathbb{P}(X_n=j) \rightarrow \pi_j$ imply in particular that $\mathbb{P}(X_n=j \mid X_0=i) \rightarrow \pi_j$?

Question

$i,j \in I$ where $I$ is the state space and $(\pi)$ is the stationary/invariant distribution of the Markov Chain $(X_n)$.

Answer 1

If we define $P$ to be our transition matrix and $\delta_i$ to be the starting distribution such that $\delta_i(x)=1$ if $x=i$ and $\delta_i(x)=0$ if $x \neq i$ (i.e. $X_0=i$), then we can say there exists a unique $\pi$ such that $\lim_{n\to\infty} \delta_iP^t = \pi$ if $P$ is irreducible and aperiodic. Obviously then $\delta_iP^t(j)\to\pi_j$ for all $i,j\in I$. So trying to answer your question, I guess I would say that if $\pi$ is the unique stationary distribution for $P$, the chain is irreducible and aperiodic, and consequently the probability of being in state $j$ converges to $\pi_j$ regardless of the starting distribution $\delta_i$. That is, $\lim_{n\to \infty}P(X_n=j)$ exists if the stationary distribution is unique and $\lim_{n\to \infty}P(X_n=j|X_0=i) = \lim_{n\to \infty}P(X_n=j) = \pi_j$. Hopefully that helps.