Fix an aperiodic discrete-time Markov chain $X_0,X_1,\dots$, with $\mathbb P_i$ denoting probabilities according to $X_0=i$. Define the return probability $$p_{ii}^{(n)} = \mathbb P_i[X_n=i].$$

Is it true that $$\sum_{n=0}^\infty |p_{ii}^{(n)}-p_{ii}^{(n+1)}|<\infty\text{?}$$

I am most interested in the null-recurrent case. For transient states there is a positive answer: the return probabilities are themselves summable. I know that for finite state spaces there is exponential convergence to some stationary state, which also implies a positive answer.

  • $\begingroup$ Did you try to compute $p^{(n)}_{ii}$ for the symmetric random walk on $\mathbb Z$ with transitions $i\to i$ and $i\to i\pm1$, for every $i$ in $\mathbb Z$? $\endgroup$
    – Did
    Sep 13, 2017 at 17:37
  • $\begingroup$ @Did: when the self-transition probability is $1/2$ and $i\to i\pm 1$ probabilities each $1/4$, the probabilities are $2^{-2n}\binom{2n}{n}$ which is non-increasing so the differences sum to at most $1$. I don't know about other other self-transition parameters but my guess is they'd behave similarly at least in the long run. $\endgroup$
    – Dap
    Sep 13, 2017 at 17:48
  • $\begingroup$ Right, random walks look like a dead end. Sorry for the noise. $\endgroup$
    – Did
    Sep 13, 2017 at 19:37

1 Answer 1


The answer is no, as shown in the paper "On the problem by Erdős-de Bruijn-Kingman on regularity of reciprocals for exponential series" by Alexander Gomilko, Yuri Tomilov, https://arxiv.org/abs/1701.04357


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