Research article

Moran random walk with reset and short memory

  • Received: 21 April 2024 Revised: 20 May 2024 Accepted: 29 May 2024 Published: 19 June 2024
  • MSC : 60C05, 60D05, 60G40, 60K15

  • We investigated the statistical properties of the Moran random walk (Yn)n in one dimension, focusing on short memory. Specifically, employing generating function techniques, we determined the cumulative distribution function and the mean of the height Hn. Furthermore, we derived explicit expressions for the distribution, mean, and variance of Yn, along with its asymptotic distribution. Finally, we provided the distribution of the waiting time τh, which represents the number of steps required to reach a specified level h, as the conclusion of our study.

    Citation: Mohamed Abdelkader, Rafik Aguech. Moran random walk with reset and short memory[J]. AIMS Mathematics, 2024, 9(8): 19888-19910. doi: 10.3934/math.2024971

    Related Papers:

    [1] Rafik Aguech . On the central limit theorem for the elephant random walk with gradually increasing memory and random step size. AIMS Mathematics, 2024, 9(7): 17784-17794. doi: 10.3934/math.2024865
    [2] Hui Yang . Large deviations for transient random walks with asymptotically zero drifts. AIMS Mathematics, 2025, 10(4): 8777-8793. doi: 10.3934/math.2025402
    [3] Frithjof Lutscher, Thomas Hillen . Correlated random walks in heterogeneous landscapes: Derivation, homogenization, and invasion fronts. AIMS Mathematics, 2021, 6(8): 8920-8948. doi: 10.3934/math.2021518
    [4] Najmeddine Attia . Some remarks on recursive sequence of fibonacci type. AIMS Mathematics, 2024, 9(9): 25834-25848. doi: 10.3934/math.20241262
    [5] You Lv . Asymptotic behavior of survival probability for a branching random walk with a barrier. AIMS Mathematics, 2023, 8(2): 5049-5059. doi: 10.3934/math.2023253
    [6] He Song, Longmin Wang, Kainan Xiang, Qingpei Zang . The continuity of biased random walk's spectral radius on free product graphs. AIMS Mathematics, 2024, 9(7): 19529-19545. doi: 10.3934/math.2024952
    [7] Luigi Accardi, Amenallah Andolsi, Farrukh Mukhamedov, Mohamed Rhaima, Abdessatar Souissi . Clustering quantum Markov chains on trees associated with open quantum random walks. AIMS Mathematics, 2023, 8(10): 23003-23015. doi: 10.3934/math.20231170
    [8] Andrius Grigutis . Exact expression of ultimate time survival probability in homogeneous discrete-time risk model. AIMS Mathematics, 2023, 8(3): 5181-5199. doi: 10.3934/math.2023260
    [9] Tatpon Siripraparat, Suporn Jongpreechaharn . The exponential non-uniform bound on the half-normal approximation for the number of returns to the origin. AIMS Mathematics, 2024, 9(7): 19031-19048. doi: 10.3934/math.2024926
    [10] Monthira Duangsaphon, Sukit Sokampang, Kannat Na Bangchang . Bayesian estimation for median discrete Weibull regression model. AIMS Mathematics, 2024, 9(1): 270-288. doi: 10.3934/math.2024016
  • We investigated the statistical properties of the Moran random walk (Yn)n in one dimension, focusing on short memory. Specifically, employing generating function techniques, we determined the cumulative distribution function and the mean of the height Hn. Furthermore, we derived explicit expressions for the distribution, mean, and variance of Yn, along with its asymptotic distribution. Finally, we provided the distribution of the waiting time τh, which represents the number of steps required to reach a specified level h, as the conclusion of our study.



    Numerous scientific fields have relied heavily on the study of random processes to gain understanding of a wide range of phenomena, from particle motion to population behavior. The random walk, a mathematical representation of a path made up of a series of random steps, is one of these stochastic processes that has shown to be very significant. The Moran process, so named after the biologist Patrick Moran who first presented it in the context of population genetics, is one prominent variation of the random walk.

    Originating in population genetics, the Moran process provides a fundamental paradigm for comprehending the dynamics of genetic diversity within populations. The Moran process was developed by Patrick Moran in the middle of the 20th century as a straightforward yet effective method for researching evolutionary processes in finite populations.

    Under particular rules governing reproduction, selection, and random genetic drift, individuals interact within a finite population in a discrete-time stochastic process known as the Moran process. In addition to taking into consideration the limiting size of populations, which can have significant long-term effects on genetic variation, Moran's formulation captures the fundamental mechanisms driving evolutionary change, such as genetic drift and selection.

    Under particular rules governing reproduction, selection, and random genetic drift, individuals interact within a finite population in a discrete-time stochastic process known as the Moran process.

    A-part from its use in population genetics, the Moran process has been proven useful in several other disciplines, such as epidemiology, ecology, and sociology. It has been adopted as a model for researching a variety of phenomena that are typified by random interactions and discrete events due to its adaptability and intuitive appeal.

    Since its introduction, several variations of Moran random walk have been introduced from these models: age statistics of the population at a given step [1,2], the height of walks with resets and Moran models [3], the two-dimensional Moran model (number of resets, final altitude, height) [4], and the two-dimensional Moran random walk with time-dependent probability of reset [5]. In this work, we introduce a variation of this model, where we assume a population of size 1 that, at each step, has two possibilities, at random: either it is replaced by a new individual, or it continues to go up with a one-step size.

    Our model is a mixture between a Moran process with a single component and an elephant walk. In this model, the walk at each time only takes steps of size 1 upwards or returns to zero, which is the Moran walk. However, the decision at each step (to go up or return to zero) depends randomly on the previous step: the walk only remembers the previous step, reproducing what is in memory with probability p and the opposite of what is in memory with probability 1p, hence the connection with the elephant walk.

    The examined model in this paper has some similarities with the Markov chain model explored in Brenaud's book [6]. However, there are key differences. In our model, transitions from i to i1 are almost surely (for all i1), and the probability of transitioning from 0 to a state i depends on i1. It is important to note that our model allows for the possibility of moving from i to zero (reset) with a probability, which also depends on i. Additionally, the transition from i, in our model, can lead to either i+1 or 0, in contrast to the Brenaud book where it is limited to i1.

    The structure of this current paper is organized as follows. In Section 2, we present our Moran random walk with short memory. In Section 3, we state our main result concerning the cumulative distribution function of the height statistics associated with our random walk using the moment-generating function. Also, we find the mean of the height statistics. In Section 4, we illustrate numerically our random walk with different lengths: n=10,100,1000,10,000,100,000, and 500,000, for different values of the probabilities p=0.3,0.5,0.85, and r=0.25,0.5,0.8. In Sections 5 and 6, we find the explicit expressions of the mean and the variance of our random walk. In Section 7, we find the closed form of the characteristic function of our random walk. Also, we determine the limit law of our walk using the characteristic function tool. In Section 8, we prove our main results. Finally, in Section 9, we present the conclusions and perspectives of our work.

    Consider the one-dimensional Moran model with short memory Yn. It starts from 0 at time 0 (i.e., P(Y0=0)=1). The Moran model with one dimension is given by the following system: at time 1,

    Y1={1, with probability r,0, with probability 1r,

    and for time n>1, the walk Yn, can only remember the step n1 and decides to repeat the same step with probability p or to do the opposite step with probability 1p, where p(0,1) and r(0,1). We define the event An, at time n, the random walk Yn decides to repeat the movement made at the time n,

    An1:= At time n , the random walk decides to reproduce the same step made at time n-1 .

    We can write the random walk Yn at the time n, by

    Yn=(1+Yn1)1An11{Yn10}+1¯An11{Yn1=0}. (2.1)

    The state space of Yn is {0,,n}.

    Remark 2.1. Some remarks about the probability r, Eq (2.1), and the mean of short memory:

    The probability r is exclusively utilized during the initial step n=1, in which the random walk lacks any recollection, prompting the use of r to initiate the process.

    The event ¯An1 is when, at time n, the random walk decides to reproduce the opposite step made at time n1: more precisely if, for example, at step n1 the random walk decides to go up by a step of size 1 and then, according to event ¯An1, it decides to go to zero at time n.

    By the definition of our model, for all n2,P(An1)=p,

    For all n2, in the event: (An1{Yn1=0})(¯An1{Yn10}), Yn=0.

    By short memory, we mean that the random walk, at any step, remembers only the previous step and not all the previous steps.

    In the paper, E(Yn), Var(Yn), represent respectively, the mean and the variance of Yn.

    This model has found numerous applications in game theory and financial markets. Specifically, in the case of a game where at each step, there is a chance of either winning or losing everything. Each step of the game involves repeating the same action as the previous step with a probability of p, or playing the opposite action with a probability of 1p. In this case the random variable Yn represents the result of the game at step n. Assuming that every time the player wins, their prize pool increases by a dollars, and once they lose, they lose their entire fortune, in this scenario, the player's winnings up to the nth step, amount to aYn. This model can be modeled by a Markov chain. With the aid of [7] and our results, we can obtain findings that appear to be interesting.

    In this section, we want to study the behavior of the height statistics, denoted by Hn, of the random walk Yn. Precisely, we need to study the distribution of Hn. For this objective, we use the moment-generating function, denoted by Ah(z), of the cumulative distribution function, denoted by P(Hnh), of Hn, for all real number z(0,1). We define the height statistics, denoted by Hn=max(Y0,Y1,Y2,,Yn), of the random walk Yn. For all h integer number, we define Ah(.) as the moment-generating function of Hn: for all z(0,1)

    Ah(z)=+n=1P(Hnh)zn.

    The moment-generating function technique has been well developed in [8,9].

    In this reference, the robustness of this technique is highlighted for explaining the distribution of a random variable defined by a recursive equation. We define the following sequences of probabilities: for all integer number h and for all n2,

    an(h)=P(Hnh), (3.1)
    bn(h)=P(Y0h,Y1h,,Yn2h,Yn1=0), (3.2)
    cn(h)=P(Y0h,Y1h,,Yn2h,Yn1=h), (3.3)

    with the initial following conditions,

    ak(h)=1, for all kh,b0(h)=0,b1(h)=1,ck(h)=0, for all kh+1.

    Lemma 3.1. The sequence bn satisfies the following recursion: for all n1,

    bn(h)=(1p)an2(h)+pbn1(h). (3.4)

    Lemma 3.2. The sequence cn satisfies the following recursion: for all nh+3,

    cn(h)=(1p)phP(Hnh2h). (3.5)

    In this section, we present our main result concerning the distribution of the height statistics Hn, of the random walk Yn. We find a closed form of the generating function of Hn and we deduce that the cumulative distribution function, P(Hnh), of Hn.

    Theorem 3.1. The cumulative distribution function of Hn is given by: for all 0hn,

    P(Hnh)=[zn]Ah(z),

    where, for all z(0,1),

    Ah(z)=ph+1(1p)+2z+z31p1pz1ph+1(1p)(z+z2(1p)+z31p1pz)

    is the generating function of the cumulative distribution of Hn.

    In the next corollary, we find the mean of the height Hn using Theorem 3.1.

    Corollary 3.1. The mean of Hn can be obtained as

    E(Hn)=n[zn]n1h=0Ah(z).

    Here, we focus on the waiting time τh for the random walk to reach a given level h. Given a level h.

    Theorem 3.2. The distribution of τh is given by, for all n1,

    P(τh=n)=(1p)ph1cnh1(h).

    Remark 3.1. From Theorem 3.2, we can compute the mean of τh but the expression is very complicated. In fact, from Eq (8.12), we compute the sequence (bn(h))n and finally we use Eq (8.11), to obtain the expression of cn(h).

    In this section, we use the R software to study the influence of the two probabilities p and r on the evolution of the random walk Yn with different lengths. In the first case, we take a small value of the probability r that equals to 0.25 and we change the probability p from 0.3, 0.5 to 0.85 where the lengths of the random walk Yn are equals to 10,100, 1000, 10,000,100,000, and 500,000. For the second case, the probability r is fixed at 0.5, and the probability p is changed from 0.3, 0.5 to 0.85; we treat the random walk Yn with different lengths that are equal to 10,100, 1000, 10,000,100,000, and 500,000. Finally, we finish with a high value of the probability r (0.8) and the probability p changes from 0.3 to 0.85 with the same lengths of Yn.

    From Figures 13, we observe that the maximum level of Yn and the height Hn are dependent on the three parameters: r, p, and n.

    Figure 1.  Evolution of the walk Y with different lengths for r=0.25 and p=0.3.
    Figure 2.  Evolution of the walk Y with different lengths for r=0.25 and p=0.5.
    Figure 3.  Evolution of the walk Y with different lengths for r=0.25 and p=0.85.

    1) When r and n are fixed and p is changed from 0.3, 0.5 to 0.85, we observe that Yn and Hn are always increasing. For example, for fixed n=10 and r=0.25, and changed p from 0.3, 0.5 and 0.85, the maximum level of Yn is increased from 0, 1, to 8 and the height Hn are increased from 2, 3 to 8, respectively. For n=10,000 and r=0.25, and changed p from 0.3, 0.5 and 0.85, the maximum level of Yn is increased from 2, 2, to 10 and the height Hn is increased from 7, 12 to 48, respectively. For n=500,000 and r=0.25, and changed p from 0.3, 0.5 and 0.85, the maximum level of Yn is increased from 7, 14 to 35 and the height Hn is increased from 10, 19 to 69, respectively.

    2) When r and p are fixed and n is changed from 10,100, 1000, 10,000,100,000, to 500,000, we observe that Yn and Hn are always increasing. But the evolution of the walk Yn is very slow. For example, for fixed r=0.25, and p=0.3, and changed n from 10,100, 1000, 10,000,100,000, to 500,000, the maximum level of Yn and the height Hn are increased from 2, 3, 6, 8, 9 to 10, respectively. For example, for fixed r=0.25, and p=0.5, and changed n from 10,100, 1000, 10,000,100,000, to 500,000, the maximum level of Yn and the height Hn are increased from 3, 7, 9, 12, to 18, respectively. When r=0.25, and p=0.85 fixed and changed n from 10,100, 1000, 10,000,100,000, to 500,000, the maximum level of Yn and the height Hn are increased from 8, 18, 34, 48, to 68, respectively.

    Also, Figure 1 shows that the maximum level of the random walk Yn equals 10 when p=0.3, r=0.25, and n=500,000. Figure 2 shows that the maximum level of the random walk Yn equals 18 when p=0.5, r=0.25, and n=500,000. Figure 3 shows that the maximum level of the random walk Yn equals 68 when p=0.85, r=0.25 and n=500,000. That means that if n and p increase, then the maximum level of Yn increases. But this evolution of Yn is slow. Furthermore, the height Hn does not exceed 10, 18, and 68 when r=0.25, p=0.3,0.5,0.85, and n=500,000.

    Figures 46 show that the maximum level of Yn and the height Hn are also dependent on the three parameters: r, p, and n.

    Figure 4.  Evolution of the walk Y with different lengths for r=0.5 and p=0.3.
    Figure 5.  Evolution of the walk Y with different lengths for r=0.5 and p=0.5.
    Figure 6.  Evolution of the walk Y with different lengths for r=0.5 and p=0.85.

    1) When r and n are fixed and p is changed from 0.3, 0.5 to 0.85, we observe that Yn and Hn are always increasing. For example, for fixed n=100, and r=0.5, and changed p from 0.3, 0.5, and 0.85, the maximum level of Yn and the height Hn are increased from 3, 5, to 30, respectively. For n=1000, and r=0.5, and changed p from 0.3, 0.5 and 0.85, the maximum level of Yn and the height Hn are increased from 4, 7 to 40, respectively. For n=500,000 and r=0.5, and changed p from 0.3, 0.5, and 0.85, the maximum level of Yn and the height Hn are increased from 10, 24 to 69, respectively.

    2) When r and p are fixed and n is changed from 10,100, 1000, 10,000,100,000, to 500,000, we observe that Yn and Hn are always increasing. But the evolution of the walk Yn is very slow. For example, for fixed r=0.5, and p=0.3, and changed n from 10,100, 1000, 10,000,100,000, to 500,000, the maximum level of Yn and the height Hn are increased from 1, 3, 4, 7, 9 to 10, respectively. For example, for fixed r=0.5, and p=0.5, and changed n from 10,100, 1000, 10,000,100,000, to 500,000, the maximum level of Yn and the height Hn are increased from 3, 5, 7, 10, 19, to 18, respectively. When r=0.5, and p=0.85 fixed and changed n from 10,100, 1000, 10,000,100,000, to 500,000, the maximum level of Yn and the height Hn are increased from 8, 29, 39, 44, 59, to 68, respectively.

    In addition, Figure 4 shows that the maximum level of the random walk Yn equals 10 when p=0.3, r=0.5, and n=500,000. Figure 5 shows that the maximum level of the random walk Yn equals 24 when p=0.5, r=0.5, and n=500,000. Figure 6 shows that the maximum level of the random walk Yn equals to 68 when p=0.85, r=0.5, and n=500,000. That means that if n and p increase, then the maximum level of Yn increases. But this evolution of Yn is slow. Furthermore, the height Hn does not exceed 10, 24, and 68 when r=0.5, p=0.3,0.5,0.85, and n=500,000.

    Figures 79 show that the maximum level of Yn and the height Hn are also depended on the three parameters: r, p, and n.

    Figure 7.  Evolution of the walk Yn with different lengths for r=0.80 and p=0.3.
    Figure 8.  Evolution of the walk Yn with different lengths for r=0.80 and p=0.5.
    Figure 9.  Evolution of the walk Yn with different lengths for r=0.80 and p=0.85.

    1) When r and n are fixed and p is changed from 0.3, 0.5 to 0.85, we observe that Yn and Hn are always increasing. For example, for fixed n=100, and r=0.8, and changed p from 0.3, 0.5, and 0.85, the maximum level of Yn and the height Hn are increased from 3, 6, to 21, respectively. For n=10,000 and r=0.8, and changed p from 0.3, 0.5, and 0.85, the maximum level of Yn and the height Hn are increased from 7, 11, to 48, respectively. For n=500,000 and r=0.8, and changed p from 0.3, 0.5 and 0.85, the maximum level of Yn and the height Hn are increased from 10, 19, to 68, respectively.

    2) When r and p are fixed and n is changed from 10,100, 1000, 10,000,100,000, to 500,000, we observe that Yn and Hn are always increasing. But the evolution of the walk Yn is very slow. For example, for fixed r=0.8, and p=0.3, and changed n from 10,100, 1000, 10,000,100,000, to 500,000, the maximum level of Yn and the height Hn are increased from 2, 3, 5, 7, 10 to 10, respectively. For example, for fixed r=0.8, and p=0.5, and changed n from 10,100, 1000, 10,000,100,000, to 500,000, the maximum level of Yn and the height Hn are increased from 1, 6, 7, 11, 14, to 19, respectively. When r=0.8 and p=0.85 fixed and changed n from 10,100, 1000, 10,000,100,000, to 500,000, the maximum level of Yn and the height Hn are increased from 8, 21, 34, 44, 70, to 70, respectively.

    In addition, Figure 7 shows that the maximum level of the random walk Yn equals 10 when p=0.3, r=0.8, and n=500,000. Figure 8 shows that the maximum level of the random walk Yn equals 19 when p=0.5, r=0.8, and n=500,000. Figure 9 shows that the maximum level of the random walk Yn equals 70 when p=0.85, r=0.8, and n=500,000. That means that if n and p increase, then the maximum level of Yn increases. But this evolution of Yn is slow. Furthermore, the height Hn does not exceed to 10, 19, and 68 when r=0.8, p=0.3,0.5,0.85, and n=500,000.

    In this section, we study the mean of the random walk Yn. We start by a technical lemma to find a recursive relation between the probability of the random walk Yn that equals 0 at the two successive times n and n1, denoted by pn1(0), and pn(0). Furthermore, we find a closed form of pn(0). In addition, we find a recursive equation between the mean of Yn, the mean of Yn1, and the probability of the random walk Yn at time n1 on 0. Finally, we finish this section by the explicit expression of the mean of Yn. For this, we denote by pn(0)=P(Yn=0) the probability of the random walk Yn, equal to 0 at time n.

    The next lemma is a key lemma to find the explicit form of the first moment (mean) of the random walk Yn. It leads to finding the explicit form of the probability that the random walk Yn equals 0, at time n.

    Lemma 5.1. The sequence of probabilities pn(0), satisfies the following recursive equation: for all n1,

    pn(0)=(1p)+(2p1)pn1(0), (5.1)

    where p0(0)=1. Its explicit expression is given by: for all n1,

    pn(0)=(1r)(2p1)n1+12(1(2p1)n1), (5.2)

    where p0(0)=1.

    Proof. Using the conditional probability: for all n1, we have

    pn(0)=P(Yn=0|Yn1=0)P(Yn1=0)+P(Yn=0|Yn10)P(Yn10)=ppn1(0)+(1p)(1pn1(0))=(1p)+(2p1)pn1(0).

    By, iterating Eq (5.1) n times and using the fact that

    P(Y1=0|Y0=0)=P(Y1=0)=1r,

    we get the desired result.

    Remark 5.1 When r=p in Eq (5.2), we have the special case:

    pn(0)=(1p)(2p1)n1+12(1(2p1)n1)=12(1(2p1)n1).

    The next theorem uses Lemma 5.1 to find the first moment of the walk Yn. We find a recursive equation related E(Yn), E(Yn1) and pn1(0). This relation leads to the determination of the explicit form of Yn.

    Theorem 5.1. The mean of Yn, satisfy the following recursive equation: for all n1,

    E(Yn)=p+pE(Yn1)+(12p)pn1(0). (5.3)

    Its explicit expression is given by:

    E(Yn)=12(1p)(12pn(1r)+(12r)(2p1)n). (5.4)

    Proof. Using Eq (2.1), one has

    E(Yn)=E((1+Yn1)1An11{Yn10}+1¯An11{Yn1=0})=E(1An11{Yn10})+E(Yn11An11{Yn10})+E(1¯An11{Yn1=0})=pP(Yn10)+pE(Yn1)+(1p)P(Yn1=0),

    using Lemma 5.1, then we get

    E(Yn)=p+pE(Yn1)+(12p)pn1(0).

    Iterating Eq (5.3) n times, we obtain

    E(Yn)=p+pE(Yn1)+(12p)pn1(0)=p+(p+pE(Yn2)+(12p)pn2(0))+(12p)pn1(0)=p+p2++pn1n1times+rpn1+(12p)pn1(p1npn1(0)++p1p1(0))=rpn1+ppn1p+(12p)pn1n1k=1pk(0)pk, (5.5)

    where

    pk(0)=(1r)(2p1)k1+12(1(2p1)k1)=12+12(2p1)k1(12r).

    Developing and simplifying

    n1k=1pk(0)pk=n1k=1(12+12(2p1)k1(12r))1pk=12n1k=11pk+12r2(2p1)n1k=1(2p1p)k=12pn1(1p)(1pn1)+(12r)2(1p)pn1(pn1(2p1)n1), (5.6)

    multiplying Eq (5.6) by (12p)pn1, then we get

    (12p)pn1n1k=1pk(0)pk=12p2(1p)(1pn1)+(12r)(12p)2(1p)(pn1(2p1)n1). (5.7)

    Combining Eqs (5.5) and (5.7), then we obtain

    E(Yn)=rpn1+ppn1p+12p2(1p)(1pn1)+(12r)(12p)2(1p)(pn1(2p1)n1). (5.8)

    Replacing the following identities

    ppn1p+12p2(1p)(1pn1)=1pn12(1p),12(1p)pn1(12r)(12p)+rpn1=pn12pn(1r)2(1p),

    in Eq (5.8), we obtain

    E(Yn)=12(1p)(12pn(1r)+(12r)(2p1)n).

    Remark 5.2. When r=p, we have the special case:

    E(Yn)=12(1p)(12pn(1p)+(2p1)n+1).

    Corollary 5.1. Asymptotically, the mean of the random walk Yn converges to 12(1p) for all r(0,1) and p(0,1).

    Proof. The proof follows directly from Eq (5.4).

    In this section, we find the explicit form of the variance of the random walk Yn. For this objective, we start with a key lemma to find a recursive equation between the three following quantities: E(Y2n), E(Yn), and n1k=1pkE(Yk). In addition, we need to calculate n1k=1pkE(Yk), combining with the explicit form of the first moment of Yn defined in Theorem 5.1 to obtain an explicit form of the second moment of Yn, denoted by E(Y2n). Finally, we state the closed form of the variance of Yn.

    We start by a technical lemma to find the recursive equation between: E(Y2n), E(Yn), and n1k=1pkE(Yk).

    Lemma 6.1. The second moment of Yn, satisfies the following recursive equation: n1,

    E(Y2n)=E(Yn)+2pnn1k=1pkE(Yk). (6.1)

    Proof. Using Eq (2.1), one has

    Y2n=((1+Yn1)1An11{Yn10}+1¯An11{Yn1=0})2=(1+2Yn1+Y2n1)1An11{Yn10}+1¯An11{Yn1=0},

    hence

    E(Y2n)=E((1+2Yn1+Y2n1)1An11{Yn10}+1¯An11{Yn1=0})=pP(Yn10)+2pE(Yn1)+pE(Y2n1)+(1p)P(Yn1=0)=p+2pE(Yn1)+pE(Y2n1)+(12p)P(Yn1=0),

    using Lemma 5.1, then we get

    E(Y2n)=p+2pE(Yn1)+pE(Y2n1)+(12p)pn1(0).

    Iterating Eq (6.1) n times, we obtain

    E(Y2n)=p+2pE(Yn1)+pE(Y2n1)+(12p)pn1(0)=p+2pE(Yn1)+(12p)pn1(0)+p(p+2pE(Yn2)+pE(Y2n2)+(12p)pn2(0))=rpn1+n1k=1pk+(12p)pn1n1k=1pk(0)pk+2pnn1k=1pkE(Yk), (6.2)

    and using Eq (5.5), we deduce

    E(Y2n)=E(Yn)+2pnn1k=1pkE(Yk). (6.3)

    In the next lemma, we need to calculate n1k=1pkE(Yk) using the explicit form of the mean of the walk Yn, defined in (5.4). This lemma is a technical lemma used to find the explicit expression of the second moment of Yn.

    Lemma 6.2. One has: for all n2,

    2pnn1k=1pkE(Yk)=p(1p)2{12rpn2n(1r)(1p)pn1(12r)(2p1)n}. (6.4)

    Proof. Applying Eq (5.4), we have

    n1k=1pkE(Yk)=12(1p)n1k=1pk{(12pk(1r)+(12r)(2p1)k)}=12(1p){n1k=1pkn1k=12pkpk(1r)+(12r)n1k=1(2p1p)k},

    developing and after some calculations, we have

    n1k=1pk=1pn1pn1(1p),n1k=1(2p1p)k=2p1(1p)pn1(pn1(2p1)n1),

    and get

    n1k=1pkE(Yk)=12(1p){2(n1)(1r)+1pn1pn1(1p)+(12r)(2p1)pn1(1p)pn1(12r)(2p1)n(1p)pn1}=12(1p)2pn1{2(n1)(1r)(1p)pn1+1pn1+(12r)(2p1)pn1(12r)(2p1)n}. (6.5)

    Define the sequence In, for all n1, by

    In=2(n1)(1r)(1p)pn1+1pn1+(12r)(2p1)pn1,

    developing and simplifying,

    In=2(n1)(1r)(1p)pn1+1pn1+(12r)(2p1)pn1=12rpn2n(1r)(1p)pn1, (6.6)

    replacing Eq (6.6) in Eq (6.5), we obtain

    n1k=1pkE(Yk)=12pn1(1p)2{12rpn2n(1r)(1p)pn1(12r)(2p1)n}, (6.7)

    multiplying Eq (6.7) by 2pn, then we get

    2pnn1k=1pkE(Yk)=p(1p)2{12rpn2n(1r)(1p)pn1(12r)(2p1)n}.

    Now, we present the explicit form of the second moment of the random walk Yn. It is used to find the explicit form of the variance of Yn.

    Proposition 6.1. The explicit form of the second moment of Yn, is given by: n1,

    E(Y2n)=12(1p)2{(1+p)2pn(1rp+3pr)+(13p)(12r)(2p1)n4n(1r)(pnpn+1)}. (6.8)

    Proof. Replacing Eqs (5.4) and (6.4) in Eq (6.1), then we get

    E(Y2n)=12(1p)(12pn(1r)+(12r)(2p1)n)+p(1p)2(12rpn2n(1r)(1p)pn1(12r)(2p1)n)=12(1p)2(An+Bn), (6.9)

    where the two sequences An and Bn are defined by

    An=(1p)(12pn(1r)+(12r)(2p1)n),Bn=2p(12rpn2n(1r)(1p)pn1(12r)(2p1)n),

    adding An and Bn, and regrouping the terms

    An+Bn=((1p)(12r)2p(12r))(2p1)n4np(1r)(pn1pn)+(1p)2pn(1r)(1p)+2p(12rpn)=(13p)(12r)(2p1)n4np(1r)(pn1pn)+(1+p)2pn(1rp+3pr), (6.10)

    combining Eqs (6.9) and (6.10), then we obtain

    E(Y2n)=12(1p)2{(1+p)2pn(1rp+3pr)+(13p)(12r)(2p1)n4np(1r)(pn1pn)}.

    In the next corollary, we present the explicit expression of the variance of Yn using Theorem 5.1 and Lemma 6.1.

    Corollary 6.1. The variance of Yn, is given by

    Var(Yn)=14(1p)2(14p(12r)(2p1)n+4(1r)(12r)(2p2p)n4n(1r)(pnpn+1)+4(12r)pn+14(1r)2p2n(12r)2(2p1)2n). (6.11)

    Proof. Applying Eqs (5.4) and (6.8), we have

    E(Y2n)+E(Yn)=12(1p)2{12pn(1rp+2pr)+(12p)(12r)(2p1)n2n(1r)(pnpn+1)}, (6.12)

    developing and simplifying

    (E(Yn))2=14(1p)2(12pn(1r)+(12r)(2p1)n)2=14(1p)2(14pn(1r)+4p2n(1r)2+(12r)2(2p1)2n+2(12r)(2p1)n4(1r)(12r)pn(2p1)n), (6.13)

    combining Eqs (6.12) and (6.13), then we have

    Var(Yn)=14(1p)2(14p(12r)(2p1)n+4(1r)(12r)(2p2p)n4n(1r)(pnpn+1)+4(12r)pn+14(1r)2p2n(12r)2(2p1)2n).

    Remark 6.1. When r=p, we have the special case:

    Var(Yn)=14(1p)2(1+4p2(32p)(2p1)n+14n(1p)2pn+4(12p)pn+14(1p)2p2n(2p1)2(n+1)).

    In this section, we define the characteristic function of the random walk Yn, denoted by Φn(t), for all tR. Furthermore, we find a recursive equation between Φn(t), Φn1(t), pn(0), and pn1(0). By iteration, we determine the explicit expression of Φn(t). In the next step, we study the convergence of Φn(t) asymptotically. Finally, we finish this section by the limit distribution of the random walk Yn, denoted by Y. We start with the definition of the characteristic function, denoted by Φn(t),

    Φn(t)=E(eitYn)=nk=0eitkP(Yn=k),tR. (7.1)

    In the next theorem, we present an important result concerning the closed form of the characteristic function of the random walk Yn. It is used to determine the distribution of Yn.

    Theorem 7.1. The characteristic function Φn(t), of Yn satisfies the recursive equation:

    Φn(t)=peitΦn1(t)+(12p)pn1(0)eit+pn(0),n1.

    Its explicit expression is given by: for all n1,

    Φn(t)=rpn1eint+(1r)(2p1)n1+12(1(2p1)n1)+(1p)eit2(1peit)(1pn1ei(n1)t)+(1p)(12r)eit2(12p+peit)(pn1ei(n1)t(2p1)n1), (7.2)

    where Φn(0)=1.

    Proof. Using Eqs (2.1) and (7.1), then we have: for all tR

    Φn(t)=E(e(1+Yn1)itIAn1I{Yn10})+E(eitI¯An1I{Yn1=0})+P(Yn=0)=E(e(1+Yn1)itIAn11{Yn10})+eitP(¯An1{Yn1=0})+P(Yn=0)=peitE(eYn1itI{Yn10})+eitP(¯An1{Yn1=0})+P(Yn=0). (7.3)

    Using the following identities

    E(eYn1itI{Yn10})=Φn1(t)pn1(0),P(¯An1{Yn1=0})=(1p)pn1(0),

    Eq (7.3) can be written as

    Φn(t)=peitΦn1(t)+(12p)pn1(0)eit+pn(0). (7.4)

    Iterating Eq (7.1) n times, we get

    Φn(t)=rpn1eint+(1r)pn1ei(n1)t+(12p)pn1eintn1k=1eiktpk(0)pk+pneintnk=2eiktpk(0)pk. (7.5)

    Rewriting Eq (7.5), we obtain

    Φn(t)=pn(0)+rpn1eint+(1p)pn1eintn1k=1eiktpk(0)pk, (7.6)

    applying Eq (5.6), we have

    n1k=1eiktpk(0)pk=12n1k=11(peit)k+12r2(2p1)n1k=1(2p1peit)k=12pn1ei(n1)t(1peit)(1pn1ei(n1)t)+(12r)(12p+peit)pn1ei(n1)t(pn1ei(n1)t(2p1)n1), (7.7)

    multiplying Eq (7.7) by (1p)pn1eint, then we get

    (1p)pn1eintn1k=1eiktpk(0)pk=(1p)eit2(1peit)(1pn1ei(n1)t)+(1p)(12r)eit2(12p+peit)(pn1ei(n1)t(2p1)n1). (7.8)

    Combining Eqs (7.6) and (7.8), we obtain

    Φn(t)=rpn1eint+(1r)(2p1)n1+12(1(2p1)n1)+(1p)eit2(1peit)(1pn1ei(n1)t)+(1p)(12r)eit2(12p+peit)(pn1ei(n1)t(2p1)n1).

    Now, we can deduce the distribution of the random walk Yn from Theorem 7.1.

    Corollary 7.1. The distribution of the random walk Yn, is given by: for all n1,

    P(Yn=k)={12(1(2p1)n1),k=0,12(pk+1pk+2)(1(2p1)k1),1kn1,rpn1,k=n.

    Proof. From Eq (7.6), we have

    Φn(t)=nk=0eitkP(Yn=k)=pn(0)+rpn1eint+(1p)pn1eintn1k=1ei(nk)ktpk(0)pk=pn(0)+rpn1eint+(1p)pn1n1l=1eiltpl(0)pl,

    we conclude the proof by identification of the coefficients of eitk,k=0,,n.

    We prove in the next corollary that the characteristic function of the random walk Yn converges to ΦY(t), representing the characteristic function of the limit law of the random walk Yn. Also, we find the explicit expression of ΦY(t).

    Corollary 7.2. The distribution of Yn converges asymptotically to the random variable Y, with the characteristic function, denoted by ΦY(t), given by

    ΦY(t)=limnΦn(t)=12(1peit)(1+eit2peit). (7.9)

    Proof. From Eq (7.2), we have

    Φn(t)=rpn1eint+(1r)(2p1)n1+12(1(2p1)n1)+(1p)eit2(1peit)(1pn1ei(n1)t)+(1p)(12r)eit2(12p+peit)(pn1ei(n1)t(2p1)n1),

    passing to the limit, we get

    ΦY(t)=limnΦn(t)=12(1peit)(1+eit2peit).

    The result presented in Corollary 7.2 is used to find the distribution of the random variable Y=limnYn.

    Theorem 7.2. The distribution of the random variable Y, with ΦY(t), its characteristic function is given by:

    P(Y=k)={12,k=0,12(pk+(12p)k1),k>0.

    Remark 7.1. Our model is different from the model in [4]. In [4], the distribution of Y was proved by a shifted geometric with parameter p, which is different from the distribution given in Theorem 7.2.

    Proof of Theorem 7.2. Starting from Eq (7.9), we have

    E(eitY)=limnΦn(t)=12(1peit)(1+eit2peit)=12k=0(peit)k+12(12p)k=0pkei(k+1)t=12+12k=1(pk+(12p)pk1)eitk.

    Since

    ΦY(t)=E(eitY)=P(Y=0)+k=1P(Y=k)eitY. (7.10)

    The proof is finished by identifying Eqs (7.2) and (7.10).

    Proof of Lemma 3.1. Applying Eq (3.2),

    bn(h)=P(Y0h,Y1h,,Yn2h,Yn1=0)=hk=0P(Y0h,Y1h,,Yn3h,Yn2=k,Yn1=0)=hk=1P(Y0h,Y1h,,h,Yn3h,Yn2=k,Yn1=0)+P(Y0h,Y1h,,Yn3h,Yn2=0,Yn1=0). (8.1)

    Define two sequence of probabilities dn(h) and gn(h), for all n1, by

    dn(h)=P(Y0h,Y1h,,Yn3h,Yn2=k,Yn1=0),gn(h)=P(Y0h,Y1h,,Yn3h,Yn2=0,Yn1=0),

    using the conditional probability, we have

    dn(h)=P(Y0h,Yn1=0|{Yn2=k})P(Y1h,,Yn3h,Yn2=k)=(1p)P(Y0h,Y1h,,Yn3h,Yn2=k), (8.2)

    and

    gn(h)=P(Yn1=0|{Yn2=0})P(Y0h,Y1h,,Yn3h,Yn2=0)=pP(Y0h,Y1h,,Yn3h,Yn2=0). (8.3)

    Replacing Eqs (8.2) and (8.3) in Eq (8.1),

    bn(h)=(1p)hk=1P(Y1h,,Yn3h,Yn2=k)+pP(Y1h,,Yn3h,Yn2=0),

    using Eqs (3.1) and (3.2), then we get, for all n2,

    bn(h)=(1p)an2(h)+pbn1(h).

    Proof of Lemma 3.2. Define the following events

    Un1,h={Y0h,,Ynh2h,Ynh1=0,Ynh=1,,Yn1=h}, (8.4)
    Hnh2={Y0h,Y1h,,Ynh2h}, (8.5)

    observe that

    Un1,hHnh2.

    From Eqs (3.2), (8.4) and (8.5), we have: for all nh+3,

    cn(h)=P(Y0h,Y1h,,Yn2h,Yn1=h)=P(Y0h,Y1h,,Ynh2h,Ynh1=0,Ynh=1,,Yn1=h)=P(Un1,h|Hnh2)P(Hnh2)=(1p)phP(Hnh2h).

    Then, Eq (3.5) is verified.

    Now, we are able to proof our main result concerning the distribution of the height statistics Hn.

    Proof of Theorem 3.1. Applying Eq (3.1), we have

    an(h)=P(Hnh)=P(Y0h,Y1h,Y2h,,Ynh)=hk=0P(Y0h,Y1h,Y2h,,,Yn1=k,Ynh)=hk=1P(Y0h,Y1h,Y2h,,Yn1=k,Ynh)+P(Y0h,Y1h,Y2h,,Yn1=0,Ynh),=P(Rn,0,h)+hk=1P(Rn,k,h), (8.6)

    where Rn,k,h, Sn,k,h and Tn,h, are defined by: for all n1, 0kh and 0hn

    Rn,k,h={Y0h,Y1h,,Yn1=k,Ynh},Sn,k,h={Y0h,Y1h,,Yn1=k},Tn,h={Ynh}.

    Using the conditional probability

    P(Rn,k,h)=P(Tn,h|Sn,k,h)P(Sn,k,h)=[1P(¯Tn,h|Sn,k,h)]P(Sn,k,h),where¯Tn,h={Ynh}, (8.7)

    then, we have

    hk=1P(Rn,k,h)=hk=1P(Sn,k,h)hk=1P(¯Tn,h|Sn,k,h)P(Sn,k,h)=hk=1P(Hn1h)hk=1P(¯Tn,h|Sn,h,h)P(Sn,h,h). (8.8)

    Replacing Eq (3.3) in (8.8), we get

    hk=1P(Rn,k,h)=an1(h)pcn(h). (8.9)

    If, in Eq (8.7) we take k=0, we have

    P(Rn,0,h)=P(Tn,h|Sn,0,h)P(Sn,0,h)=bn(h), (8.10)

    where

    P(Sn,0,h)=P(Y0h,Y1h,,Yn2h,Yn1=0)=bn(h),P(Tn,h|Sn,0,h)=P({Ynh}|{Y0h,Y1h,,Yn2h,Yn1=0})=1.

    Combining Eqs (8.6), (8.9), and (8.10), we deduce that, for all 0hn,

    an(h)=P(Hnh)=an1(h)pcn(h)+bn(h),

    Using Lemmas 3.1 and 3.2, we obtain the following recursion

    an(h)=an1(h)+bn(h)pcn(h). (8.11)

    In order to obtain the explicit form of an(h), we use the moment-generating functions, let

    Ah(z)=+n=1an(h)zn,Bh(z)=+n=1bn(h)zn,Ch(z)=+n=1cn(h)zn.

    By Eq (3.4), we have

    Bh(z)=z2(1p)(1+Ah(z))1pz, (8.12)

    and from Eq (8.11), we have

    Ah(z)=z(1+Ah(z))+z2(1p)Ah(z)+z+z+n=2bn(h)znph+1(1p)(1+Ah(z))=z(1+Ah(z))+z2(1p)Ah(z)+z+zBh(z)ph+1(1p)(1+Ah(z)).

    Finally, we conclude that

    Ah(z)=+n=1P(Hnh)zn=ph+1(1p)+2z+z31p1pz1ph+1(1p)(z+z2(1p)+z31p1pz),

    and that, for all 0hn,

    P(Hnh)=[zn]Ah(z).

    Finally, we proof our second main result about the waiting time τh.

    Proof of Theorem 3.2. For all h and for all n1, we have

    {τh=n}={Y0=0,Y1h1,,Yn1h1,Yn=h}={Y1h1,,Ynh2h1,Ynh1h1,Ynh=0,Ynh+1=1,,Yn=h}.

    Then,

    P(τh=n)=P(Y1h1,,Yn1h1,Ynh1h1,Ynh=0,Ynh+1=1,,Yn=h)=P(Yn=h,Yn1=h1,,Ynh+1=1|Ynh=0)×P(Y1h1,,Yn1h1,Ynh1h1,Ynh=0)=(1p)ph1P(Y1h1,,Yn1h1,Ynh1h1,Ynh=0)=(1p)ph1cnh1(h).

    In this paper, we explore the statistical properties of the one-dimensional Moran random walk with short memory. Specifically, we investigate the cumulative distribution function and the mean of the height statistics Hn using generating function techniques. Our analysis leads to the explicit form of the generating function Ah(z) of Hn. Additionally, we establish that the cumulative distribution function P(Hnh) of Hn is equivalent to the coefficient of zn in the power series Ah(z). Furthermore, we derive explicit expressions for the mean and variance of the random walk Yn. Utilizing characteristic function tools, we determine the distribution of Yn and deduce its limit distribution.

    For the remainder of our study, we plan to investigate a variant of the Moran random walk model with long memory. Specifically, at each step n, the walk randomly and uniformly selects a past step k, and decides to either repeat the same movement as step k (move up by +1 or return to 0) with probability p, or to perform the opposite of step k with probability 1p. In our opinion, this model is highly intriguing as it combines elements of both the elephant random walk [10,11,12] and the Moran walk [3,4,5].

    Mohamed Abdelkader: Conceptualization, Methodology, Software, Formal analysis, Writing-original draft; Rafik Aguech: Conceptualization, Software, Formal analysis, Writing-original draft. All authors have read and approved the final version of the manuscript for publication.

    The authors declare they have not used Artificial Intelligence (AI) tools in the creation of this article.

    The first author extends appreciation to King Saud University in Riyadh, Saudi Arabia for funding this research work through Researchers Supporting Project number (RSPD2024R1068).

    All authors declare no conflicts of interest in this paper.



    [1] Y. Itoh, H. M. Mahmoud, Age statistics in the Moran population model, Stat. Probab. Lett., 74 (2005), 21–30. https://doi.org/10.1016/j.spl.2005.04.028. doi: 10.1016/j.spl.2005.04.028
    [2] Y. Itoh, H. M. Mahmoud, D. Takahashi, A stochastic model for solitons, Random Struct. Algor., 24 (2004), 51–64. https://doi.org/10.1002/rsa.10106 doi: 10.1002/rsa.10106
    [3] R. Aguech, A. Althagafi, C. Banderier, Height of walks with resets, the Moran model, and the discrete Gumbel distribution, Seminaire Lotharingien de Combinatoire, 87B (2024), 12.
    [4] M. Abdelkader, On the height of one-dimensional random walk, Mathematics, 11 (2023), 4513. https://doi.org/10.3390/math11214513 doi: 10.3390/math11214513
    [5] R. Aguech, M. Abdelkader, Two-dimensional Moran model: Final altitude and number of resets, Mathematics, 11 (2023), 3774. https://doi.org/10.3390/math11173774 doi: 10.3390/math11173774
    [6] P. Bremaud, Markov chains: Gibbs fields, Monte Carlo simulation and queues, Springer Cham, 2020. https://doi.org/10.1007/978-3-030-45982-6
    [7] J. Filar, K. Vrieze, Competitive Markov decision processes, New York: Springer, 1997. https://doi.org/10.1007/978-1-4612-4054-9
    [8] E. A. Bender, Z. Gao, Part sizes of smooth supercritical compositional structures, Combin. Probab. Comput., 23 (2014), 686–716. https://doi.org/10.1017/S0963548314000315 doi: 10.1017/S0963548314000315
    [9] X. Gourdon, Largest component in random combinatorial structures, Discrete Math., 180 (1998), 185–209. https://doi.org/10.1016/S0012-365X(97)00115-5 doi: 10.1016/S0012-365X(97)00115-5
    [10] B. Bercu, A martingale approach for the elephant random walk, J. Phys. A: Math. Theor., 51 (2018), 015201. https://doi.org/10.1088/1751-8121/aa95a6 doi: 10.1088/1751-8121/aa95a6
    [11] R. Aguech, On the central limit theorem for the elephant random walk with gradually increasing memory and random step size, AIMS Mathematics, 9 (2024), 17784–17794. https://doi.org/10.3934/math.2024865 doi: 10.3934/math.2024865
    [12] R. Aguech, M. El Machkouri, Gaussian fluctuations of the elephant random walk with gradually increasing memory, J. Phys. A: Math. Theor., 57 (2024), 065203. http://dx.doi.org/10.1088/1751-8121/ad1c0d doi: 10.1088/1751-8121/ad1c0d
  • This article has been cited by:

    1. Mohamed Abdelkader, Three-Dimensional Moran Walk with Resets, 2024, 16, 2073-8994, 1222, 10.3390/sym16091222
  • Reader Comments
  • © 2024 the Author(s), licensee AIMS Press. This is an open access article distributed under the terms of the Creative Commons Attribution License (http://creativecommons.org/licenses/by/4.0)
通讯作者: 陈斌, bchen63@163.com
  • 1. 

    沈阳化工大学材料科学与工程学院 沈阳 110142

  1. 本站搜索
  2. 百度学术搜索
  3. 万方数据库搜索
  4. CNKI搜索

Metrics

Article views(1267) PDF downloads(222) Cited by(1)

Figures and Tables

Figures(9)

Other Articles By Authors

/

DownLoad:  Full-Size Img  PowerPoint
Return
Return

Catalog