Subgeometric Rates of Convergence for Discrete-Time Markov Chains Under Discrete-Time Subordination


In this paper, we are concerned with the subgeometric rate of convergence of a Markov chain with discrete-time parameter to its invariant measure in the f-norm. We clarify how three typical subgeometric rates of convergence are inherited under a discrete-time version of Bochner’s subordination. The crucial point is to establish the corresponding moment estimates for discrete-time subordinators under some reasonable conditions on the underlying Bernstein function.

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The author would like to thank an anonymous referee for careful reading and useful suggestions.

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Correspondence to Chang-Song Deng.

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Financial support through the National Natural Science Foundation of China (11401442, 11831015) is gratefully acknowledged.



If \(\phi :[0,\infty )\rightarrow [0,\infty )\) is a concave function, then it is easy to see that

$$\begin{aligned} \phi (tx)\ge t\phi (x)\quad \text {for all } t\in [0,1]\text { and }x\ge 0. \end{aligned}$$

Lemma 4.1

Let \(\phi :[0,\infty )\rightarrow [0,\infty )\) be a concave function such that \(\phi (1)=1\) and \(\phi \) is differentiable on (0, 1) with \(\phi '|_{(0,1)}\ge 0\). Then

$$\begin{aligned} e^{-\phi (x)}+\phi \left( 1-e^{-x}\right) \ge 1\quad \text {for all }x\ge 0. \end{aligned}$$

In particular, the above inequality holds if \(\phi \) is a Bernstein function (not necessarily with \(\phi (0)=0\)).



$$\begin{aligned} \Phi (x):=e^{-\phi (x)}+\phi \left( 1-e^{-x}\right) ,\quad x\ge 0. \end{aligned}$$

It follows from (4.1) that for \(x\ge 1\),

$$\begin{aligned} 1=\phi (1)=\phi \left( \frac{1}{x}\cdot x\right) \ge \frac{1}{x}\phi (x), \end{aligned}$$


$$\begin{aligned} e^{-\phi (x)}\ge e^{-x},\quad x\ge 1. \end{aligned}$$

Now we obtain from (4.1) and (4.2) that for \(x\ge 1\),

$$\begin{aligned} \Phi (x)&=e^{-\phi (x)}+ \phi \left( \left( 1-e^{-x}\right) \cdot 1\right) \\&\ge e^{-\phi (x)}+ \left( 1-e^{-x}\right) \phi (1)\\&=1+e^{-\phi (x)}-e^{-x}\\&\ge 1. \end{aligned}$$

It remains to consider the case that \(x\in [0,1)\). Since \(\phi \) is concave and differentiable on (0, 1), we know that \(\phi '\) is nonincreasing on (0, 1). For \(x\in (0,1)\), by the elementary inequality that \(1-e^{-x}<x\), we obtain

$$\begin{aligned} \phi '\left( 1-e^{-x}\right) \ge \phi '(x)\ge 0. \end{aligned}$$

Moreover, by (4.1) one has for \(x\in (0,1)\),

$$\begin{aligned} \phi (x)=\phi (x\cdot 1)\ge x\phi (1)=x, \end{aligned}$$

which yields that

$$\begin{aligned} e^{-x}\ge \mathrm {e}^{-\phi (x)},\quad x\in (0,1). \end{aligned}$$

Combining this with (4.3), we find for \(x\in (0,1)\),

$$\begin{aligned} \Phi '(x)=e^{-x}\phi '\left( 1-e^{-x}\right) -e^{-\phi (x)}\phi '(x)\ge 0. \end{aligned}$$

This implies that \(\Phi \) is nondecreasing on (0, 1) and thus for all \(x\in [0,1)\),

$$\begin{aligned} \Phi (x)\ge \Phi (0)= e^{-\phi (0)}+\phi (0)\ge 1, \end{aligned}$$

which completes the proof. \(\square \)

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Deng, C. Subgeometric Rates of Convergence for Discrete-Time Markov Chains Under Discrete-Time Subordination. J Theor Probab 33, 522–532 (2020).

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  • Rate of convergence
  • Subordination
  • Bernstein function
  • Moment estimate
  • Markov chain

Mathematics Subject Classification (2010)

  • Primary 60J05
  • Secondary 60G50