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Corrigendum: “Extended generator and associated martingales for M/G/1 retrial queue with classical retrial policy and general retrial times”

Published online by Cambridge University Press:  02 February 2024

S. Meziani
Affiliation:
Department of Probability and Statistics, Faculty of Mathematics, University of Sciences and Technology USTHB, Algeria
T. Kernane*
Affiliation:
Department of Probability and Statistics, Faculty of Mathematics, University of Sciences and Technology USTHB, Algeria Laboratory of Research in Intelligent Informatics and Applied Mathematics (RIIMA), University of Sciences and Technology USTHB, Algeria
*
Corresponding author: Tewfik Kernane; Email: [email protected]
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Abstract

Type
Erratum
Creative Commons
Creative Common License - CCCreative Common License - BY
This is an Open Access article, distributed under the terms of the Creative Commons Attribution licence (http://creativecommons.org/licenses/by/4.0), which permits unrestricted re-use, distribution and reproduction, provided the original article is properly cited.
Copyright
© The Author(s), 2024. Published by Cambridge University Press.

In this Corrigendum, we correct an error in Theorem 3 and its proof of [Meziani, S. & Kernane, T. (2023). Extended generator and associated martingales for M/G/1 retrial queue with classical retrial policy and general retrial times. Probability in the Engineering and Informational Sciences 37(1):206–213.].

We have confused the expectation of the residual service time $E[Y(t)]$ with the expectation of the total service time. Let S a generic random variable representing the duration of the service time of a customer in the system considered. The residual service time Y(t) at time t is given by:

\begin{equation*} Y(t)=S-t\mid S \gt t. \end{equation*}

Hence, from Guess et al. [Reference Guess and Proschan2]

\begin{align*} \mu_{Y}(t)&=E[Y(t)]\\ &=\frac{1}{(1-F(t))}\int_{t}^{\infty}(1-F(u))du, \end{align*}

where F is the distribution function of the service time.

Replace the statement of Theorem 3 with the following:

Theorem 0.1. The conditional expectation of the number of blocked customers N(t) given $ N(0)=n_{0}$ and $ Y(0)=y_{0}$ $($when $ Y(t)\in \mathbb{E}_{1}\cup \partial^{*}\mathbb{E}_{1})$ is given by:

\begin{equation*} E[N(t)|N(0)=n_{0}]= \begin{cases} n_{0}+ \lambda t & \text{for} \;\;t \in [0,\tau_{0}];\\ n_{0}+\lambda t+\dfrac{1}{\mu_{1}} (y_{0}-t-\mu_{Y}(t)) & \text{for}\;\; t \in [0,\tau_{1}], \end{cases} \end{equation*}

where $ \mu_{1}=\int\limits_{0}^{\infty}ydF(y)$ and $\mu_{Y}(t)$ is the mean residual service time at time t.

Proof. The equation (5.1) in the proof is replaced by:

(5.1)\begin{equation} E[N(t)\mid N(0)=n_{0}]=n_{0}+\lambda t-\frac{1}{\mu_{1}}E[Y(t)]. \end{equation}

Replace the mean residual service time in the proof by the following:

\begin{equation*}E\lbrack Y(t)\rbrack= \begin{cases}\begin{array}{ll}0,&\text{for }t\in\lbrack0,\tau_0\rbrack;\\\mu_Y(t),&\text{for }t\in\lbrack0,\tau_1\rbrack.\end{array}\end{cases} \end{equation*}

Remark: Integrability conditions (4.7) and (4.8) in the paper can be stated without taking the expectation as in Dassios and Zhao [Reference Dassios and Zhao1], p. 817, with any consequence on the results of the paper.

References

Dassios, A. & Zhao, H. (2011). A dynamic contagion process. Advances in Applied Probability 43(3): 814846.10.1239/aap/1316792671CrossRefGoogle Scholar
Guess, F. & Proschan, F. (1988). 12 mean residual life: theory and applications. Handbook of statistics 7 215224.10.1016/S0169-7161(88)07014-2CrossRefGoogle Scholar