Abstract

We investigate a variant of the stochastic logistic model that allows individual variation and time-dependent infection and recovery rates. The model is described as a heterogeneous density dependent Markov chain. We show that the process can be approximated by a deterministic process defined by an integral equation as the population size grows.

1. Introduction

The stochastic logistic model, also called the endemic SIS model in the epidemiological context, was first discussed by Weiss and Dishon [1]. This model describes the evolution of an infection in a fixed population of size by a continuous-time Markov chain for the number of infected individuals . The state space is , and the transition rates are given by where and are the infection rate of susceptibles and the recovery rate of infectives, respectively. This simple model has found applications in a variety of fields, including population biology, metapopulation ecology, chemistry, and physics. Properties such as mean epidemic size [2, 3], mean extinction time [1, 4, 5], and quasi-stationary distributions [68] have been extensively studied under different initial conditions.

The stochastic logistic model has an interesting limit property that it can be approximated by deterministic differential equations. In particular, based on Theorem  3.1 of [9], the rescaled process converges in probability uniformly on finite time intervals to the solution of an ordinary differential equation. The issue of differential equation approximations for stochastic processes traces back to the pioneer work of Kurtz [9, 10], where deterministic limit of pure jump density-dependent Markov processes was addressed via Trotter type approximation theorems. Recently, McVinish and Pollett [11] show that a deterministic limit process can be established for a stochastic logistic model with individual variation, where the coefficients of the transition rates of the Markov chain can vary with the nodes. We refer the interested reader to a comprehensive survey [12] for numerous sufficient conditions for this type of convergence.

In this paper, along the above line of study, we investigate a variant of the stochastic logistic model where both individual variation and time-dependent infection and recovery rates are allowed. Specifically, we consider a continuous-time Markov chain on the state space with transition rates given by where , , , and is the -dimensional vector whose th entry is 1, and other entries are 0. Let represent that individual is susceptible and represent that individual is infected. Then is equivalent to the stochastic logistic model by setting , , , and for all , , and .

The above model (2) can be viewed as a generalization of that treated in [11] in twofolds. Firstly, the infection and recovery rates are time varying, incorporating realistic scenarios where the infection and recovery capacities may change over time [13]. As such, the transition rates of the Markov chain are explicitly time dependent, making the previously obtained sufficient conditions no longer applicable.

Secondly, the introduction of functions accommodates needed flexibility for some applications. Indeed, since , the linear form used in [11] implies that no contribution can be made by susceptible individuals. This is only a rough approximation. For one thing, susceptible individuals aware of a disease in their proximity can take measures (such as wearing masks, avoiding public places, and frequent hand washing) to reduce their susceptibility, which in turn affect the epidemic dynamics dramatically [14, 15]. For another, the epidemic progression strongly depends on the contact patterns between susceptible and infected individuals, especially in the network context. Identifying individual role is an interesting and demanding task [16]. In the present framework, each node applies individual contribution (and ), indicating the underlying interaction structure/strength among individuals.

The rest of the paper is organized as follows. We state the main result in Section 2 and provide the proof in Section 3.

2. The Result

In what follows, we assume naturally that for all . We will show that the stochastic logistic model (2) converges weakly to the solution of an integral equation as the population size .

Let be a measurable space with and the Borel -algebra on . Denote by the set of bounded, continuous functions on . Let be the set of -finite measures on . For and , we define the measure-valued nonrandom process and the measure-valued Markov process by For some set , let be the set of right-continuous functions with left-hand limits mapping to . Thus, for , and the sample paths of belong to , since the sample paths of belong to (application of Theorem  16.16 in [17, page 316]).

We assume that the following assumptions hold. (A1), and , where for .(A2) uniformly for , and in as , where and are nonrandom measures, and means weak convergence (i.e., , and converges to in distribution as for all [17, page 316]).(A3) is bounded in .(A4) is Lipschitz continuous.(A5) and as functions of are piecewise constant for .

Theorem 1. If assumptions (A1)–(A5) hold, then the measure-valued Markov process converges weakly to a measure-valued nonrandom process , that is, as , where is the unique solution of with

Let be the space of -integrable functions on . We have the following corollary.

Corollary 2. Suppose that is the unique solution to (5). Then there exists a unique function with such that for all Borel set , and

3. Proofs

In this section, Theorem 1 will be proved through a series of lemmas by tightness and uniqueness arguments [18]. The idea of proofs is similar to that in [11], and we include the complete proofs here, not only for the convenience of the reader but also to convince the reader that the results do hold in our setting.

Lemma 3. The sequence is tight in .

Proof. Recall that [17, Theorem  16.27, page 324] is tight in if and only if is tight in for every . According to Aldous's tightness criterion [18, Theorem  16.10, page 178], the tightness of holds, if the following two conditions are satisfied. (A) For each , , and , there exist a and an such that if is a discrete -stopping time satisfying , then (B) For each ,
To show (A), we express the generator of by for all real-valued functions with . Hence, using (3), we obtain Applying Dynkin's formula to (see, e.g., [19, Proposition  1.7, page 162]), we have that is a martingale with respect to the filtration , and . Moreover, is square integrable. The condition (A) holds if for each , , and , there exist a and an such that for any discrete -stopping time satisfying ,
To see (15), we note that by using (3). It follows from Markov's inequality and (17) that From assumptions (A1) and (A4), we obtain . Therefore, we can find and such that (15) is satisfied by using the assumptions (A1), (A3), (A4), and the fact that .
To prove (16), we need to introduce the quadratic variation process for . Since is a square integrable martingale, it follows from Proposition  6.1 in [19, page 79] that is also a martingale. By using Markov's inequality and martingale properties, we obtain
Applying Dynkin's formula to similarly as in the derivation of (14), we have that is a martingale, where with
By using (14) and (20), we obtain for any , It follows from (3) and the assumption (A5) that has piecewise constant sample paths, and is a continuous finite variation process. Therefore, (14), (23), and the fact imply that From (19), (24), and martingale properties, we obtain The assumption (A1) implies that , and hence . Therefore, as in the derivation of (15), we can choose some and such that (16) is satisfied.
Now, the only thing remaining to verify is the condition (B). Since and the assumption (A3) holds, there exists a constant such that for all and . From Markov's inequality and the assumption (A1), we have as . The proof of Lemma 3 is complete.

Lemma 4. For any , in as .

Proof. By the corollary in [18, page 28], it is sufficient to show that as , where means convergence in probability. Here, by definition, the space is the so-called Skorohod space , and is the metric on it, which defines the Skorohod topology (see, e.g., [18, page 168]). Theorem  16.7 in [18, page 174] further implies that it is sufficient to show that for each . The metric generates the topology of Skorohod space (see [18, pages 166 and 125] for definitions).
Since , by Doob's martingale inequality, we have for any , where the first equality uses the fact that is a martingale with expectation 0, and the second equality follows from (24). It then follows from the derivation in (25) that which concludes the proof of Lemma 4.

From Lemma 3, the sequence is tight, and then, there exists a weakly convergent subsequence in . We still denote it by in the next lemma for convenience.

Lemma 5. If as , then for any , in as , where

Proof. For each , define a random process by From Theorem  3.1 in [18, page 28] and the arguments in the proof of Lemma 4, it is sufficient to show the following two conditions: (A) as , for each ,(B) as .
To show (A), note that From the argument following (17), there is some constant satisfying Hence, by the dominated convergence theorem, as , since and the assumption (A2) holds.
To show (B), it is sufficient to show that [18, Theorem  2.1, page 16] as , for any bounded, uniformly continuous function . By the definition of and the assumption , we only need to show that is a continuous function of mapping to , invoking the dominated convergence theorem. Furthermore, it is sufficient to show that is a continuous function of mapping to . In the following, we take one term in as an example to show the continuity with respect to . Other terms can be shown analogously.
Define Suppose that in holds. We need to show in as . Indeed, since and is continuous for any , we have in and in , as . Hence, from the assumptions (A3) and (A4), we obtain in . It follows from [18, Equation (12.14), page 124] that for all but countably many . The dominated convergence theorem then yields in supremum norm on finite time intervals and hence in .

Lemma 6. The solution to (5) is unique in .

Proof. Suppose that and are two solutions to (5), and they are the limits of two weakly convergent subsequences of , respectively. By the assumption (A2), . We need to show that for all .
From (3), we have for any . Therefore, by the assumption (A2), hold. Define . In view of (5), we obtain The assumption (A4) indicates that for some constant . Therefore, using (38) we can derive where is some constant. To see this, note that by the assumptions (A1), (A2), and (38), we have for any . The assumption (A4) then implies
Again employing the assumption (A2), we have for all . Hence, by using the assumption (A3), the inequality (42) becomes for some constant . A simple application of Gronwall's lemma yields for all , which concludes the proof.

Proof of Theorem 1. By Lemma 4 and Lemma 5, the limit of any weakly convergent subsequence of must satisfy (5). By Lemma 6, we find that the sequence must itself converge weakly to that unique solution (see the corollary in [18, page 59]). The proof of Theorem 1 is thus completed.

Proof of Corollary 2. For any open set , we obtain from (38) that where is taken as a continuous function upwardly converging to the indicator function of . By using the dominated convergence theorem, we know that . A regularity property (see, e.g., [17, page 18, Lemma  1.34]) implies that for all . This means that is absolutely continuous with respect to for any . An application of the Radon-Nykodym theorem yields the existence of such that (8) holds with . Now the result follows straightforwardly from Theorem 1.

Acknowledgments

The author expresses his sincere gratitude to the anonymous referee and the editor for careful reading of the original paper and useful comments that helped to improve presentation of results.