I have a Markov chain $\mathbf{A} = (A_0, A_1, \ldots)$ with state space $\{0, \ldots, n\}$ which converges towards a stationary distribution $\pi$. There are a lot of well-known results on upper-bounding the convergence rate. However, I'm interested in getting a lower bound.

In detail, the problem looks like this: The transition probability is given as

$p_{ij} = {n \choose i}\left(1-({1\over 2})^j\right)^i\left(({1\over 2})^j\right)^{n-i}$ for $j\neq 0$,

$p_{ij} = {n \choose i}\left(1-({1\over 2})^n\right)^i\left(({1\over 2})^n\right)^{n-i}$ for $j = 0$,

and the initial distribution $A_0$ is $(1, 0, \ldots, 0)^T$, i.e., the chain starts in state $0$ with probability $1$.

Given this information, is it possible to derive a lower bound on the convergence rate? Since I'm particularly interested in state 0, I would like to come up with something like this:

$\lvert \mathbb{P}(A_k = 0) - \pi_0 \rvert \geq \ldots$ some function of $k$ and $n$.

Any hints on how to approach this would be appreciated. Please also speak up if you think that it is unlikely that such a closed-form expression exists and I should stop wasting my time on this problem. I'm a computer scientist, not a mathematician, so it's quite possible that I've overlooked something obvious.


1 Answer 1


See Chapter 7 of Markov Chains and Mixing Times by Levin, Peres, and Wilmer.

  • $\begingroup$ You beat me to it! $\endgroup$ Aug 31, 2010 at 20:49
  • $\begingroup$ Thanks! I cannot use the techniques mentioned there yet since I don't have a closed-form bound on $\pi_0$, but your answer has been helpful. $\endgroup$
    – Heinzi
    Sep 14, 2010 at 12:35

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