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Journal of Theoretical Probability

, Volume 32, Issue 1, pp 541–543 | Cite as

Correction to: Conservative and Semiconservative Random Walks: Recurrence and Transience

  • Vyacheslav M. AbramovEmail author
Correction
  • 401 Downloads

1 Correction to: J Theor Probab (2018) 31:1900–1922  https://doi.org/10.1007/s10959-017-0747-3

The aim of this note is to correct the errors in the formulation and proof of Lemma 4.1 in [1] and some claims that are based on that lemma. The correct formulation of the aforementioned lemma should be as follows.

Lemma 4.1

Let the birth-and-death rates of a birth-and-death process be \(\lambda _n\) and \(\mu _n\) all belonging to \((0,\infty )\). Then, the birth-and-death process is transient if there exist \(c>1\) and a value \(n_0\) such that for all \(n>n_0\)
$$\begin{aligned} \frac{\lambda _n}{\mu _n}\ge 1+\frac{1}{n}+\frac{c}{n\ln n}, \end{aligned}$$
(1)
and is recurrent if there exists a value \(n_0\) such that for all \(n>n_0\)
$$\begin{aligned} \frac{\lambda _n}{\mu _n}\le 1+\frac{1}{n}+\frac{1}{n\ln n}. \end{aligned}$$
(2)

Proof

Following [2], a birth-and-death process is recurrent if and only if
$$\begin{aligned} \sum _{n=1}^\infty \prod _{k=1}^{n}\frac{\mu _k}{\lambda _k}=\infty . \end{aligned}$$
Write
$$\begin{aligned} \sum _{n=1}^\infty \prod _{k=1}^{n}\frac{\mu _k}{\lambda _k}=\sum _{n=1}^\infty \exp \left( \sum _{k=1}^{n}\ln \left( \frac{\mu _k}{\lambda _k}\right) \right) . \end{aligned}$$
(3)
Now, suppose that (1) holds. Then, for sufficiently large n
$$\begin{aligned} \frac{\mu _n}{\lambda _n}\le 1-\frac{1}{n}-\frac{c}{n\ln n}+O\left( \frac{1}{n^2}\right) , \end{aligned}$$
and since the function \(x\mapsto \ln x\) is increasing on \((0,\infty )\), then
$$\begin{aligned} \begin{aligned} \ln \left( \frac{\mu _n}{\lambda _n}\right)&\le \ln \left( 1-\frac{1}{n}-\frac{c}{n\ln n}+O\left( \frac{1}{n^2}\right) \right) \\&=-\frac{1}{n}-\frac{c}{n\ln n}+O\left( \frac{1}{n^2}\right) . \end{aligned} \end{aligned}$$
Hence, for sufficiently large n
$$\begin{aligned} \sum _{k=1}^n\ln \left( \frac{\mu _k}{\lambda _k}\right) \le -\ln n-c\ln \ln n+O(1), \end{aligned}$$
and thus, by (3), for some constant \(C<\infty \),
$$\begin{aligned} \sum _{n=1}^{\infty }\prod _{k=1}^{n}\frac{\mu _k}{\lambda _k}\le C\sum _{n=1}^{\infty }\frac{1}{n(\ln n)^c}<\infty , \end{aligned}$$
provided that \(c>1\). The transience follows.
On the other hand, suppose that (2) holds. Then, for sufficiently large n
$$\begin{aligned} \frac{\mu _n}{\lambda _n}\ge 1-\frac{1}{n}-\frac{1}{n\ln n}+O\left( \frac{1}{n^2}\right) , \end{aligned}$$
and, consequently,
$$\begin{aligned} \ln \left( \frac{\mu _n}{\lambda _n}\right) \ge \ln \left( 1-\frac{1}{n}-\frac{1}{n\ln n}+O\left( \frac{1}{n^2}\right) \right) . \end{aligned}$$
Similarly to that was provided before, for some constant \(C^\prime \),
$$\begin{aligned} \sum _{n=1}^{\infty }\prod _{k=1}^{n}\frac{\mu _k}{\lambda _k}\ge C^\prime \sum _{n=1}^{\infty }\frac{1}{n\ln n}=\infty . \end{aligned}$$
The recurrence follows. \(\square \)

As \(n\rightarrow \infty \), asymptotic expansion (4.5) obtained in the proof of Lemma 4.2 in [1] guarantees its correctness. However, the corrected version of Lemma 4.1 requires more delicate arguments in the proofs of Lemma 4.2 and Theorem 4.13 in [1]. Specifically, in the proof of Lemma 4.2 instead of limit relation (4.6) we should study the cases \(d=2\) and \(d\ge 3\) separately in terms of the present formulation of Lemma 4.1.

In the formulation of Theorem 4.13 in [1], assumption (4.12) must be replaced by the stronger one:
$$\begin{aligned} \frac{L_n}{M_n}\le 1+\frac{2-d}{n}+\frac{1-\epsilon }{n\ln n}, \end{aligned}$$
for all large n and a small positive \(\epsilon \). In the proof of Theorem 4.13 in [1], we should take into account that for large n
$$\begin{aligned} \frac{\lambda _n(1,d)}{\mu _n(1,d)}=1+\frac{d-1}{n}+O\left( \frac{1}{n^2}\right) \end{aligned}$$
is satisfied (see the proof of Lemma 4.2), and hence,
$$\begin{aligned} \frac{p_n}{1-p_n}\asymp \left[ \frac{\lambda _n(1,d)}{\mu _n(1,d)}\cdot \frac{L_n}{M_n}\right] \le 1+\frac{1}{n}+\frac{1-\epsilon }{n\ln n}+\frac{C}{n^2}, \end{aligned}$$
for a fixed constant C and large n. So, according to Lemma 4.1 the process is recurrent.

Note that the statements of Lemma 4.1 are closely related to those of Theorem 3 in [3] that prove recurrence and transience for the model studied there.

Notes

Acknowledgements

The help of the reviewer is highly appreciated.

References

  1. 1.
    Abramov, V.M.: Conservative and semiconservative random walks: recurrence and transience. J. Theor. Probab. 31(3), 1900–1922 (2018)MathSciNetCrossRefzbMATHGoogle Scholar
  2. 2.
    Karlin, S., McGregor, J.: The classification of the birth-and-death processes. Trans. Am. Math. Soc. 86(2), 366–400 (1957)MathSciNetCrossRefzbMATHGoogle Scholar
  3. 3.
    Menshikov, M.V., Asymont, I.M., Iasnogorodskii, R.: Markov processes with asymptotically zero drifts. Probl. Inf. Transm. 31, 248–261 (1995), translated from Problemy Peredachi Informatsii 31, 60–75 (in Russian)Google Scholar

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Authors and Affiliations

  1. 1.School of Mathematical SciencesMonash UniversityClaytonAustralia
  2. 2.School of ScienceRoyal Melbourne Institute of TechnologyMelbourneAustralia

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