# Limiting distribution and initial distribution of a Markov chain

For a Markov chain (can the following discussion be for either discrete time or continuous time, or just discrete time?),

1. if for an initial distribution i.e. the distribution of $X_0$, there exists a limiting distribution for the distribution of $X_t$ as $t \to \infty$, I wonder if there exists a limiting distribution for the distribution of $X_t$ as $t \to \infty$, regardless of the distribution of $X_0$?
2. When talking about limiting distribution of a Markov chain, is it in the sense that some distributions converge to a distribution? How is the convergence defined?

Thanks!

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1. No, let $X$ be a Markov process having each state being absorbing, i.e. if you start from $x$ then you always stay there. For any initial distribution $\delta_x$, there is a limiting distribution which is also $\delta_x$ - but this distribution is different for all initial conditions.
2. The convergence of distributions of Markov Chains is usually discussed in terms of $$\lim_{t\to\infty}\|\nu P_t - \pi\| = 0$$ where $\nu$ is the initial distribution and $\pi$ is the limiting one, here $\|\cdot\|$ is the total variation norm. AFAIK there is at least a strong theory for the discrete-time case, see e.g. the book by S. Meyn and R. Tweedie "Markov Chains and Stochastic Stability" - the first edition you can easily find online. In fact, there are also extension of this theory by the same authors to the continuous time case - just check out their work to start with.
@Tim Given any initial distribution it admits (if admits) a unique limiting distribution. Thus, if the latter is independent of the former, the latter is unique. In other words, suppose there are two limiting distributions $\pi_1$ and $\pi_2$, then $\|\nu_1 P^n - \pi_1\| \to 0$ and $\|\nu_2 P^n - \pi_2\| \to 0$ for some $\nu_1, \nu_2$ which contradicts with the fact that limit of $\nu_1 P^n$ is the same as of $\nu_2 P^n$ –  S.D. Dec 2 '12 at 9:43