# First Passage Time of a Markov Chain That Converges to Bessel Process

Authors: Moussa Kounta
Journal: Abstract and Applied Analysis
 (2017)

---

## Abstract

We investigate the probability of the first hitting time of some discrete Markov chain that converges weakly to the Bessel process. Both the probability that the chain will hit a given boundary before the other and the average number of transitions are computed explicitly. Furthermore, we show that the quantities that we obtained tend (with the Euclidian metric) to the corresponding ones for the Bessel process.


---

## Body

## 1. Introduction

The study of the probability of the hitting times for stochastic differential equations is an active area of research and they are of great interest in many applications, for example, in finance, the study of path dependent exotic options as barrier options, in percolation theory [1], in optimal control problems [2], and in neuroscience [3]. In this paper we investigate the discrete version of Bessel processes defined by an stochastic differential equation. It is well known [4] that, given a diffusion process defined by a stochastic differential equation, we can produce a discrete Markov chain that converges weakly to the solution of this stochastic differential equation (by making use of a binomial approximation). In this paper we show that the probability of the first passage times and the number of the transitions of this discrete Markov chain tend to the corresponding ones for the continuous times Bessel process.The discrete versions of stochastic processes are interesting in themselves; for instance, in quantum mechanics the motion of a particle should be essentially discontinuous and random. Moreover, in [5] the authors show the application of the discrete version of Cox-Ingersoll-Ross process in hydrology.In this paper, we consider the so-called gambler’s ruin problem for a discrete-time Markov chain that converges to a Bessel process. Phenomena governed by Bessel processes abound in the physical world, as in the case of growth phenomena governed by Stochastic Loewner Evolution (SLE) [1]. Other phenomena include first hitting time of Bessel processes in the study of systems at or near the point of phase transition in statistical physics. In finance, a typical example is the study of stock price. It is well known that in a volatility-stabilized market the stock prices can be represented in terms of Bessel processes. Since the stock price does not vary completely continuously, the discrete formulas that will be derived in the present paper would be interesting.The paper is organized as follows. In Section2, we briefly describe the transition probability derived from the Bessel stochastic differential equation. Our main contribution is in the third section (Section 3.2): We find the explicit formula of the average number of transitions needed to end the game that was impossible to obtain in [6]. We also show the sequence of the probability of the first passage times and the average number of transitions to end the game converges (with Euclidean metric) to the corresponding values in the continuous case.


## 2. Bessel Process Defined by a Stochastic Differential Equation and Simple Binomial Approximation

### 2.1. Preliminaries on Bessel Processes

We consider the Bessel process defined by the following differential equation:(1)dRt=ν+121Rtdt+dBt,whereBtBrownianmotion.Next, let a=ν+1/2 and if there is no ambiguity we remove the index t on R. Assume that R0=r∈1,N, where N∈N (for simplicity), and define (2)τr=inf⁡t>0:Rt=1orN∣R0=r.As is well known (see, e.g., [7], page 220), the probability (3)pr=PRτr=N.satisfies the ordinary differential equation: (4)12p′′r+arp′r=0,withp1=0,pN=1.We easily find that, if a≠1/2(5)pr=r1-2a-N1-2aN1-2a-1for1⩽r⩽N,and when a=1/2, the solutions is (6)pr=ln⁡rln⁡Nfor1⩽r⩽N.Let(7)mu=Eτu.In [7], we see the function mu satisfies the second-order ordinary differential equation: (8)12m′′u+aum′u=-1withm1=mN=0.The general solution of this equation is (9)mu=∫1ur-2a-2∫rNy2ady+c0dr,where(10)c0=2∫1Nr-2a∫rNy2adydr∫1Nr-2adr,a≠12.


### 2.2. Preliminaries: Binomial Approximation

In this section we recall binomial approximation briefly; for more details please see [4]. We wish to find a sequence of stochastic processes that converges in distribution to process (1) over the time interval 0,T.Take the interval0,T, and chop it into n equal pieces 0,t1,t2,…,tn of length Δt≡T/n,tj=jΔt. Define a sequence Rtj=1+jΔr of binomial approximations from (1), which is constant between nodes, such that at any given node Rtj the process jumps up to Rtj+=Rtj+1 (resp., down to Rtj-=Rtj-1) with probability prtj,rtj+1 (resp., prtj,rtj-1), and stays at the node with probability 1-prtj,rtj+1-prtj,rtj-1. (11)PRtj+1=R+rtj∣Rtj=rtj=prtj,rtj+1,PRtj+1=R-rtj∣Rtj=rtj=prtj,rtj-1,PRtj+1=Rtj=c∣Rtj=rtj=1-prtj,rtj-1-prtj,rtj+1.The local drift of Rtj is given by (12)μΔt=prtj,rtj+1-prtj,rtj-1Δr,and the local second moment of Rtj is given by(13)σΔt=prtj,rtj+1+prtj,rtj-1-prtj,rtj+1-prtj,rtj-12Δr2.On the other hand, (14)limΔt→0⁡ERtj+1-Rtj∣RtjΔt=limΔr,Δt→0⁡μΔtΔt=aRtj,limΔt→0⁡varRtj+1-Rtj∣RtjΔt=limΔr,Δt→0⁡σΔtΔt=1.By solving (12), (13), and (14), we obtain (15)prtj,rtj+1=12A1+artjΔr+Δtartj2,prtj,rtj-1=12A1-artjΔr+Δtartj2.Since 1/rtj is bounded, we can write Δta/rtj2=OΔt; hence we obtain (16)prtj,rtj+1=12A1+aΔrrtj+OΔt,prtj,rtj-1=12A1-aΔrrtj+OΔt.We obtain the following transition probabilities prtj,rtj+1,prtj,rtj-1, and prtj,rtj given by (17)prtj,rtj+1≃12A1+aΔrrtj,prtj,rtj-1≃12A1-aΔrrtj,prtj,rtj≃1-1A.We state the following assumptions, under which Rtj converges weakly to Rt (see [8]).Assumption 1.
With probability 1, a solutionRt of the stochastic integral equation(18)Rt=R0+∫0taRsds+∫0tdBsexists for 0<t<∞, and is distributionally unique.Assumption 2.
For allδ>0 and all T>0,(19)limΔt→0⁡supr⩽δ0⩽t⩽T⁡RΔt+r,t-rt=0,limΔt→0⁡supr⩽δ0⩽t⩽T⁡RΔt-r,t-rt=0.Assumption 3.
For allδ>0 and all T>0,(20)limΔt→0⁡supr⩽δ0⩽t⩽T⁡μΔtr,t-art=0,limΔt→0⁡supr⩽δ0⩽t⩽T⁡σΔtr,t-1=0.Theorem 4.
Under Assumptions1–3, Rtj⇒Rt converge to the solution of (1), where ⇒ denotes weak convergence.Next, we assumeRtj is bounded for any j=0,1,…,n.


## 2.1. Preliminaries on Bessel Processes

We consider the Bessel process defined by the following differential equation:(1)dRt=ν+121Rtdt+dBt,whereBtBrownianmotion.Next, let a=ν+1/2 and if there is no ambiguity we remove the index t on R. Assume that R0=r∈1,N, where N∈N (for simplicity), and define (2)τr=inf⁡t>0:Rt=1orN∣R0=r.As is well known (see, e.g., [7], page 220), the probability (3)pr=PRτr=N.satisfies the ordinary differential equation: (4)12p′′r+arp′r=0,withp1=0,pN=1.We easily find that, if a≠1/2(5)pr=r1-2a-N1-2aN1-2a-1for1⩽r⩽N,and when a=1/2, the solutions is (6)pr=ln⁡rln⁡Nfor1⩽r⩽N.Let(7)mu=Eτu.In [7], we see the function mu satisfies the second-order ordinary differential equation: (8)12m′′u+aum′u=-1withm1=mN=0.The general solution of this equation is (9)mu=∫1ur-2a-2∫rNy2ady+c0dr,where(10)c0=2∫1Nr-2a∫rNy2adydr∫1Nr-2adr,a≠12.


## 2.2. Preliminaries: Binomial Approximation

In this section we recall binomial approximation briefly; for more details please see [4]. We wish to find a sequence of stochastic processes that converges in distribution to process (1) over the time interval 0,T.Take the interval0,T, and chop it into n equal pieces 0,t1,t2,…,tn of length Δt≡T/n,tj=jΔt. Define a sequence Rtj=1+jΔr of binomial approximations from (1), which is constant between nodes, such that at any given node Rtj the process jumps up to Rtj+=Rtj+1 (resp., down to Rtj-=Rtj-1) with probability prtj,rtj+1 (resp., prtj,rtj-1), and stays at the node with probability 1-prtj,rtj+1-prtj,rtj-1. (11)PRtj+1=R+rtj∣Rtj=rtj=prtj,rtj+1,PRtj+1=R-rtj∣Rtj=rtj=prtj,rtj-1,PRtj+1=Rtj=c∣Rtj=rtj=1-prtj,rtj-1-prtj,rtj+1.The local drift of Rtj is given by (12)μΔt=prtj,rtj+1-prtj,rtj-1Δr,and the local second moment of Rtj is given by(13)σΔt=prtj,rtj+1+prtj,rtj-1-prtj,rtj+1-prtj,rtj-12Δr2.On the other hand, (14)limΔt→0⁡ERtj+1-Rtj∣RtjΔt=limΔr,Δt→0⁡μΔtΔt=aRtj,limΔt→0⁡varRtj+1-Rtj∣RtjΔt=limΔr,Δt→0⁡σΔtΔt=1.By solving (12), (13), and (14), we obtain (15)prtj,rtj+1=12A1+artjΔr+Δtartj2,prtj,rtj-1=12A1-artjΔr+Δtartj2.Since 1/rtj is bounded, we can write Δta/rtj2=OΔt; hence we obtain (16)prtj,rtj+1=12A1+aΔrrtj+OΔt,prtj,rtj-1=12A1-aΔrrtj+OΔt.We obtain the following transition probabilities prtj,rtj+1,prtj,rtj-1, and prtj,rtj given by (17)prtj,rtj+1≃12A1+aΔrrtj,prtj,rtj-1≃12A1-aΔrrtj,prtj,rtj≃1-1A.We state the following assumptions, under which Rtj converges weakly to Rt (see [8]).Assumption 1.
With probability 1, a solutionRt of the stochastic integral equation(18)Rt=R0+∫0taRsds+∫0tdBsexists for 0<t<∞, and is distributionally unique.Assumption 2.
For allδ>0 and all T>0,(19)limΔt→0⁡supr⩽δ0⩽t⩽T⁡RΔt+r,t-rt=0,limΔt→0⁡supr⩽δ0⩽t⩽T⁡RΔt-r,t-rt=0.Assumption 3.
For allδ>0 and all T>0,(20)limΔt→0⁡supr⩽δ0⩽t⩽T⁡μΔtr,t-art=0,limΔt→0⁡supr⩽δ0⩽t⩽T⁡σΔtr,t-1=0.Theorem 4.
Under Assumptions1–3, Rtj⇒Rt converge to the solution of (1), where ⇒ denotes weak convergence.Next, we assumeRtj is bounded for any j=0,1,…,n.


## 3. Discrete Value of the Probability and the Average Number of Transitions of First Passage Time, of the Bessel Process

Let(21)Tj=inf⁡m>0:Rtm=1orN∣R0=1+jΔr,(22)pj=PRTj=N,dj=ERTj=N.In this section, we will compute the quantity pj for j∈1,…,k-1. We will show that pj converges to the function pr for the Bessel process as Δr decreases to zero and k tends to infinity in such a way that 1+kΔr remains equal to N.


### 3.1. Computation of the Probabilitypj

#### 3.1.1. Assuming First ThatΔr=1

Then the state space is1,2,…,N and the transition probabilities become(23)pj,j+1=12A1+aj,pj,j-1=12A1-aj,pj,j=1-1A,for j∈{2,…,N-1}. It is well known that the probability defined in (22) satisfies the following difference equation: (24)pj=pj,j+1pj+1+pj,j-1pj-1+pj,jpj.Equation (24) can be rewritten as(25)wj-j-aj+awj-1=0wherewj=pj-pj-1.We find that the solution of this first-order difference equation that satisfies the boundary condition w1=p2 is given by (26)wj=faΓj+1-aΓj+1+a,wherefa=-1+aΓa1-aΓ-a.We have the following lemma.Lemma 5.
Fora≠1/2, the unique solution of (24) subject to the boundary conditions (4) is given by(27)pj=j+aΓj+1-a/Γj+1+a-1+aΓ2-a/Γ2+aN+aΓN+1-a/ΓN+1+a-1+aΓ2-a/Γ2+aforj=1,2,…,N.Proof.
We have(28)pj+1-pj=fcΓj+1-cΓj+1+c=fcΓ2c∫01tjt-c1-t2c-1dt,pj-p1=∑k=1j-1pk+1-pk=fcΓ2c∫01t-tj1-tt-c1-t2c-1dt.Since p1=0 and Bz,w=∫01tz-11-tw-1dt=∫0∞tz-1/1+tz+wdt=ΓzΓw/Γz+w we obtain,(29)pj=fcΓ2c∫01t1-c1-t2c-2dt-fcΓ2c∫01tj-c1-t2c-2dt=fcΓ2c-1Γ2cc+1Γ2-cΓ2+c-j+cΓj+1-cΓj+c+1.By applying boundary conditions (4), we obtain(30)1=pN=faΓ2a-1Γ2aa+1Γ2-aΓ2+a-N+aΓN+1-aΓN+a+1,faΓ2a-1Γ2a=1a+1Γ2-a/Γ2+a-N+aΓN+1-a/ΓN+a+1,and hence(31)pj=j+aΓj+1-a/Γj+1+a-1+aΓ2-a/Γ2+aN+aΓN+1-a/ΓN+1+a-1+aΓ2-a/Γ2+aforj=1,2,…,N.Now we supposea≠1/2.Lemma 6.
Whena tends to 1/2, the solution becomes (32)pj=Ψj+1/2-2+γ+2ln⁡2ΨN+1/2-2+γ+2ln⁡2,where Ψz=Γ′z/Γz and γ is Euler’s constant.Proof.
Indeed, we have(33)pj=j+aΓj+1-a/Γj+1+a-1+aΓ2-a/Γ2+aN+aΓN+1-a/ΓN+1+a-1+aΓ2-a/Γ2+a=j+aΓj+1-a/j+aΓj+a-1+aΓ2-a/1+aΓ1+aN+aΓN+1-a/N+aΓN+a-1+aΓ2-a/1+aΓ1+aforj=1,2,…,N,lima→1/2⁡pj=lima→1/2⁡d/daΓj+1-a/Γj+a-d/daΓ2-a/Γ1+ad/daΓN+1-a/ΓN+a-d/daΓ2-a/Γ1+a=Ψj+1/2-Ψ3/2ΨN+1/2-Ψ3/2.Since Ψ3/2=2-γ-2ln⁡2 (see [9]), we obtain(34)lima→1/2⁡pj=Ψj+1/2-2+γ+2ln⁡2ΨN+1/2-2+γ+2ln⁡2.


#### 3.1.2. Now, in the General Case WhenΔr>0

We must solve the difference equation:(35)pj=12A1+aΔr1+jΔrpj+1+12A1-aΔr1+jΔrpj-1+1-1Apj,which can be simplified to(36)pj+1-pj=1-aΔr/1+jΔr1+aΔr/1+jΔrpj-pj-1.The boundary conditions become(37)p0=0,pk=1.Next, proceeding as above, we obtain that if a≠1/2, the probability pj is given by (38)pj=1+jΔr+aΔrΓ1+j+1Δr-aΔr/Δr/Γ1+j+1Δr+aΔr/Δr-1+aΔrΓ1+Δr-aΔr/Δr/Γ1+Δr+aΔr/Δr1+kΔr+aΔrΓ1+k+1Δr-aΔr/Δr/Γ1+k+1Δr+aΔr/Δr-1+aΔrΓ1+Δr-aΔr/Δr/Γ1+Δr+aΔr/Δr.Writing(39)n=1+jΔr,N=1+kΔr,in terms of n and N, this expression becomes (40)pn=n+aΔrΓn+Δr-aΔr/Δr/Γn+Δr+aΔr/Δr-1+aΔrΓ1+Δr-aΔr/Δr/Γ1+Δr+aΔr/ΔrN+aΔrΓN+Δr-aΔr/Δr/ΓN+Δr+aΔr/Δr-1+aΔrΓ1+Δr-aΔr/Δr/Γ1+Δr+aΔr/Δr,for n∈{1,1+Δr,…,1+kΔr=N}. The solution reduces to (41)pn=Ψ2n+Δr/2Δr-Ψ2+Δr/2ΔrΨ2N+Δr/2Δr-Ψ2+Δr/2Δrifa=12.We can now state the following proposition.Proposition 7.
Letn=1+jΔr for j∈{0,1,…,k}, with k such that 1+kΔr=N. The probability pn that the discrete-time Markov chain defined in Section 2.1, starting from n, will hit N before 1 is given by (40) if a≠1/2. The value of the probability pn tends to the function in (41) when a tends to 1/2.Next, whena=1/2, by making use of the following approximation Ψz~ln⁡z for large z, we can write that (42)limΔr↓0⁡pn=limΔr↓0⁡ln⁡2n+Δr-ln⁡2+Δrln⁡2N+Δr-ln⁡2+Δr=ln⁡nln⁡Nforn∈1,N.This expression corresponds to the function px given in (6), obtained when a=1/2. Finally, we have(43)Γz+αΓz+β∝zα-β1+O1z,as z tends to infinity (if argz+α<π; see [9]). Hence, in the case a≠0,1/2, we can write that(44)limΔr↓0⁡pn=n1-2a-1N1-2a-1,for1⩽n⩽N.Therefore, we retrieve the formula for pr in (5). In the next section, we will derive the formulas that correspond to the function mr in Section 1.


### 3.2. Computation of the Mean Number of Transitionsdj Needed to End the Game

We now turn to the problem of computing the mean numberdj of transitions that the Markov chain Rtm,m=0,1,…, starting from R0=1+jΔr, takes to reach either 1 or 1+kΔr=N. Unlike the results of [6], we find the explicit formula of the average number of transitions needed to end the game.


#### 3.2.1. The CaseΔr=1

IfΔr=1, the state of the Markov chain is the set 1,2,…,N. Conditioning on the result of the first transition of the chain and taking this first transition into account, we obtain that the quantity dj=ERTj=N satisfies the second-order linear, nonhomogeneous difference equation (45)dj=pj,j+1dj+1+pj,j-1dj-1+pj,jdj+1forj=2,…,N-1,where the probabilities pj,j+1, pj,j-1, and pj,j are defined, respectively, in (17), with the boundary conditions:(46)d1=dN=0.We have(47)qj-j-aj+aqj-1=-2Ajj+awhereqj=dj+1-dj.Let r1j=j-a/j+a, r2j=-2Aj/j+a.Now, by making use of the formula of Batchelder [10] (5, p.62), we obtain that a formal particular rational solution Rj∗ of (47) is given by (48)Rj∗=-∑k=0∞r2j+k∏i=0kr1j+i=2AΓj-aΓj+a∑k=0∞j+kΓj+a+kj+k-aΓj-a+k=2AΓj-aΓj+a∑k=0∞Γj+a+k+1Γj-a+k+1-a∑k=0∞Γj+a+kΓj-a+k+1.Then (49)Rj∗=2AΓj-aΓj+aΓj+a+1Γj-a+1Γ1Fj+a+1,1;j-a+1,1-aΓj+aΓ1Γj-aFj+a,1;j-1,1,where Fa,b;c;z=Γc/ΓaΓb∑n=0∞Γa+nΓb+n/Γc+nzn/n!, hypergeometric function.Proposition 8.
In the caseΔr=1, the mean number dj of transitions needed by the Markov chain, starting from j, to reach either 1 or N can be expressed as follows: (50)dj=∑k=1j-1α0Γk+1-aΓk+1+a+Rk∗,for j=1,…,N, where the constant α0 is given by (51)α0=-∑k=1N-1Rk∗∑k=1N-1Γk+1-a/Γk+1+a.


#### 3.2.2. The CaseΔr>0

To obtain the solution to our problem in the case whenΔr>0, we must find a particular solution of (52)dj=12A1+aΔr1+jΔrdj+1+12A1-aΔr1+jΔrdj-1(53)+1+1Adj+1,for j=1,…,k-1, with d0=dk=0; proceeding as above by letting dj+1-dj=qj, we can rewrite this second-order linear, nonhomogeneous difference equation as follows: (54)qj+1-1+jΔr-aΔr1+jΔr+aΔrqj=-2A1+jΔr1+jΔr+aΔr.We deduce from the preceding subsection that (55)Rj∗=-2AΓ1/Δr+j-aΓ1/Δr+j+a∑n=0∞1Δr+n+kΓ1/Δr+k+n+aΓ1/Δr+k+n-a+1.To complete the work, we now show that dj×Δt converges to the function mr when we choose Δt=Δr2/A. To do so, we express the general solution of (53) in terms of r=1+jΔr and write dj as dr: (56)d′r≃α0ΔrΓr/Δr-a+1Γ1/Δr+a+1Γ1/Δr+a+1Γr/Δr-a+1-2AΓr/Δr-aΓr/Δr+a∑n=0∞r+nΔr/ΔrΓr+nΔr/Δr+aΓr+nΔr/Δr-a+1,since for z→∞ and argz+a<π (see [9]) (57)Γz+aΓz+b≃za-b1+O1z.Then for Δr small enough, we may write that (58)d′r≃α0Δrr-2a1+OΔrr1Δr-2a×1+OΔr-2ArΔr-2a1+OΔrr×∑n=0∞1Δrr+nΔrr+nΔrΔr2a-11+OΔrr.Next, let(59)Sr=∑n=0∞1Δrr+nΔrr+nΔrΔr2a-11+OΔrr=1+OΔrr1Δr2a+1∑n=0∞r+nΔr2aΔr.Now, let l=N-r/Δr, where [x] is the least integer greater than or equal to x. Then, for small Δr, we may write that (60)Sr=liml→∞⁡1+OΔrr1Δr2a+1∑n=0lr+nΔr2aΔr=1+OΔrr1Δr2a+1∫rNy2ady.Thus, from what precedes, we deduce that (61)limΔr→0⁡d′rΔr22A=α0ΔrΔr2Ar-2a-2r-2a∫rNy2ady,and we get (62)limΔr→0⁡d′rΔt=r-2aα0ΔrΔrΔt-2r-2a∫rNy2ady.Hence, if we let Du=limΔr→0⁡duΔt, we can write (by making use of the boundary condition D1=0) that (63)Du=∫1ur-2alimΔr→0⁡α0ΔrΔrΔt-2r-2a∫rNy2adydr.Finally, to satisfy the boundary condition DN=0, we find that the constant α0Δr must be chosen so that Du=mu. That is, we retrieve formula (9) for the function mu.


## 3.1. Computation of the Probabilitypj

### 3.1.1. Assuming First ThatΔr=1

Then the state space is1,2,…,N and the transition probabilities become(23)pj,j+1=12A1+aj,pj,j-1=12A1-aj,pj,j=1-1A,for j∈{2,…,N-1}. It is well known that the probability defined in (22) satisfies the following difference equation: (24)pj=pj,j+1pj+1+pj,j-1pj-1+pj,jpj.Equation (24) can be rewritten as(25)wj-j-aj+awj-1=0wherewj=pj-pj-1.We find that the solution of this first-order difference equation that satisfies the boundary condition w1=p2 is given by (26)wj=faΓj+1-aΓj+1+a,wherefa=-1+aΓa1-aΓ-a.We have the following lemma.Lemma 5.
Fora≠1/2, the unique solution of (24) subject to the boundary conditions (4) is given by(27)pj=j+aΓj+1-a/Γj+1+a-1+aΓ2-a/Γ2+aN+aΓN+1-a/ΓN+1+a-1+aΓ2-a/Γ2+aforj=1,2,…,N.Proof.
We have(28)pj+1-pj=fcΓj+1-cΓj+1+c=fcΓ2c∫01tjt-c1-t2c-1dt,pj-p1=∑k=1j-1pk+1-pk=fcΓ2c∫01t-tj1-tt-c1-t2c-1dt.Since p1=0 and Bz,w=∫01tz-11-tw-1dt=∫0∞tz-1/1+tz+wdt=ΓzΓw/Γz+w we obtain,(29)pj=fcΓ2c∫01t1-c1-t2c-2dt-fcΓ2c∫01tj-c1-t2c-2dt=fcΓ2c-1Γ2cc+1Γ2-cΓ2+c-j+cΓj+1-cΓj+c+1.By applying boundary conditions (4), we obtain(30)1=pN=faΓ2a-1Γ2aa+1Γ2-aΓ2+a-N+aΓN+1-aΓN+a+1,faΓ2a-1Γ2a=1a+1Γ2-a/Γ2+a-N+aΓN+1-a/ΓN+a+1,and hence(31)pj=j+aΓj+1-a/Γj+1+a-1+aΓ2-a/Γ2+aN+aΓN+1-a/ΓN+1+a-1+aΓ2-a/Γ2+aforj=1,2,…,N.Now we supposea≠1/2.Lemma 6.
Whena tends to 1/2, the solution becomes (32)pj=Ψj+1/2-2+γ+2ln⁡2ΨN+1/2-2+γ+2ln⁡2,where Ψz=Γ′z/Γz and γ is Euler’s constant.Proof.
Indeed, we have(33)pj=j+aΓj+1-a/Γj+1+a-1+aΓ2-a/Γ2+aN+aΓN+1-a/ΓN+1+a-1+aΓ2-a/Γ2+a=j+aΓj+1-a/j+aΓj+a-1+aΓ2-a/1+aΓ1+aN+aΓN+1-a/N+aΓN+a-1+aΓ2-a/1+aΓ1+aforj=1,2,…,N,lima→1/2⁡pj=lima→1/2⁡d/daΓj+1-a/Γj+a-d/daΓ2-a/Γ1+ad/daΓN+1-a/ΓN+a-d/daΓ2-a/Γ1+a=Ψj+1/2-Ψ3/2ΨN+1/2-Ψ3/2.Since Ψ3/2=2-γ-2ln⁡2 (see [9]), we obtain(34)lima→1/2⁡pj=Ψj+1/2-2+γ+2ln⁡2ΨN+1/2-2+γ+2ln⁡2.


### 3.1.2. Now, in the General Case WhenΔr>0

We must solve the difference equation:(35)pj=12A1+aΔr1+jΔrpj+1+12A1-aΔr1+jΔrpj-1+1-1Apj,which can be simplified to(36)pj+1-pj=1-aΔr/1+jΔr1+aΔr/1+jΔrpj-pj-1.The boundary conditions become(37)p0=0,pk=1.Next, proceeding as above, we obtain that if a≠1/2, the probability pj is given by (38)pj=1+jΔr+aΔrΓ1+j+1Δr-aΔr/Δr/Γ1+j+1Δr+aΔr/Δr-1+aΔrΓ1+Δr-aΔr/Δr/Γ1+Δr+aΔr/Δr1+kΔr+aΔrΓ1+k+1Δr-aΔr/Δr/Γ1+k+1Δr+aΔr/Δr-1+aΔrΓ1+Δr-aΔr/Δr/Γ1+Δr+aΔr/Δr.Writing(39)n=1+jΔr,N=1+kΔr,in terms of n and N, this expression becomes (40)pn=n+aΔrΓn+Δr-aΔr/Δr/Γn+Δr+aΔr/Δr-1+aΔrΓ1+Δr-aΔr/Δr/Γ1+Δr+aΔr/ΔrN+aΔrΓN+Δr-aΔr/Δr/ΓN+Δr+aΔr/Δr-1+aΔrΓ1+Δr-aΔr/Δr/Γ1+Δr+aΔr/Δr,for n∈{1,1+Δr,…,1+kΔr=N}. The solution reduces to (41)pn=Ψ2n+Δr/2Δr-Ψ2+Δr/2ΔrΨ2N+Δr/2Δr-Ψ2+Δr/2Δrifa=12.We can now state the following proposition.Proposition 7.
Letn=1+jΔr for j∈{0,1,…,k}, with k such that 1+kΔr=N. The probability pn that the discrete-time Markov chain defined in Section 2.1, starting from n, will hit N before 1 is given by (40) if a≠1/2. The value of the probability pn tends to the function in (41) when a tends to 1/2.Next, whena=1/2, by making use of the following approximation Ψz~ln⁡z for large z, we can write that (42)limΔr↓0⁡pn=limΔr↓0⁡ln⁡2n+Δr-ln⁡2+Δrln⁡2N+Δr-ln⁡2+Δr=ln⁡nln⁡Nforn∈1,N.This expression corresponds to the function px given in (6), obtained when a=1/2. Finally, we have(43)Γz+αΓz+β∝zα-β1+O1z,as z tends to infinity (if argz+α<π; see [9]). Hence, in the case a≠0,1/2, we can write that(44)limΔr↓0⁡pn=n1-2a-1N1-2a-1,for1⩽n⩽N.Therefore, we retrieve the formula for pr in (5). In the next section, we will derive the formulas that correspond to the function mr in Section 1.


## 3.1.1. Assuming First ThatΔr=1

Then the state space is1,2,…,N and the transition probabilities become(23)pj,j+1=12A1+aj,pj,j-1=12A1-aj,pj,j=1-1A,for j∈{2,…,N-1}. It is well known that the probability defined in (22) satisfies the following difference equation: (24)pj=pj,j+1pj+1+pj,j-1pj-1+pj,jpj.Equation (24) can be rewritten as(25)wj-j-aj+awj-1=0wherewj=pj-pj-1.We find that the solution of this first-order difference equation that satisfies the boundary condition w1=p2 is given by (26)wj=faΓj+1-aΓj+1+a,wherefa=-1+aΓa1-aΓ-a.We have the following lemma.Lemma 5.
Fora≠1/2, the unique solution of (24) subject to the boundary conditions (4) is given by(27)pj=j+aΓj+1-a/Γj+1+a-1+aΓ2-a/Γ2+aN+aΓN+1-a/ΓN+1+a-1+aΓ2-a/Γ2+aforj=1,2,…,N.Proof.
We have(28)pj+1-pj=fcΓj+1-cΓj+1+c=fcΓ2c∫01tjt-c1-t2c-1dt,pj-p1=∑k=1j-1pk+1-pk=fcΓ2c∫01t-tj1-tt-c1-t2c-1dt.Since p1=0 and Bz,w=∫01tz-11-tw-1dt=∫0∞tz-1/1+tz+wdt=ΓzΓw/Γz+w we obtain,(29)pj=fcΓ2c∫01t1-c1-t2c-2dt-fcΓ2c∫01tj-c1-t2c-2dt=fcΓ2c-1Γ2cc+1Γ2-cΓ2+c-j+cΓj+1-cΓj+c+1.By applying boundary conditions (4), we obtain(30)1=pN=faΓ2a-1Γ2aa+1Γ2-aΓ2+a-N+aΓN+1-aΓN+a+1,faΓ2a-1Γ2a=1a+1Γ2-a/Γ2+a-N+aΓN+1-a/ΓN+a+1,and hence(31)pj=j+aΓj+1-a/Γj+1+a-1+aΓ2-a/Γ2+aN+aΓN+1-a/ΓN+1+a-1+aΓ2-a/Γ2+aforj=1,2,…,N.Now we supposea≠1/2.Lemma 6.
Whena tends to 1/2, the solution becomes (32)pj=Ψj+1/2-2+γ+2ln⁡2ΨN+1/2-2+γ+2ln⁡2,where Ψz=Γ′z/Γz and γ is Euler’s constant.Proof.
Indeed, we have(33)pj=j+aΓj+1-a/Γj+1+a-1+aΓ2-a/Γ2+aN+aΓN+1-a/ΓN+1+a-1+aΓ2-a/Γ2+a=j+aΓj+1-a/j+aΓj+a-1+aΓ2-a/1+aΓ1+aN+aΓN+1-a/N+aΓN+a-1+aΓ2-a/1+aΓ1+aforj=1,2,…,N,lima→1/2⁡pj=lima→1/2⁡d/daΓj+1-a/Γj+a-d/daΓ2-a/Γ1+ad/daΓN+1-a/ΓN+a-d/daΓ2-a/Γ1+a=Ψj+1/2-Ψ3/2ΨN+1/2-Ψ3/2.Since Ψ3/2=2-γ-2ln⁡2 (see [9]), we obtain(34)lima→1/2⁡pj=Ψj+1/2-2+γ+2ln⁡2ΨN+1/2-2+γ+2ln⁡2.


## 3.1.2. Now, in the General Case WhenΔr>0

We must solve the difference equation:(35)pj=12A1+aΔr1+jΔrpj+1+12A1-aΔr1+jΔrpj-1+1-1Apj,which can be simplified to(36)pj+1-pj=1-aΔr/1+jΔr1+aΔr/1+jΔrpj-pj-1.The boundary conditions become(37)p0=0,pk=1.Next, proceeding as above, we obtain that if a≠1/2, the probability pj is given by (38)pj=1+jΔr+aΔrΓ1+j+1Δr-aΔr/Δr/Γ1+j+1Δr+aΔr/Δr-1+aΔrΓ1+Δr-aΔr/Δr/Γ1+Δr+aΔr/Δr1+kΔr+aΔrΓ1+k+1Δr-aΔr/Δr/Γ1+k+1Δr+aΔr/Δr-1+aΔrΓ1+Δr-aΔr/Δr/Γ1+Δr+aΔr/Δr.Writing(39)n=1+jΔr,N=1+kΔr,in terms of n and N, this expression becomes (40)pn=n+aΔrΓn+Δr-aΔr/Δr/Γn+Δr+aΔr/Δr-1+aΔrΓ1+Δr-aΔr/Δr/Γ1+Δr+aΔr/ΔrN+aΔrΓN+Δr-aΔr/Δr/ΓN+Δr+aΔr/Δr-1+aΔrΓ1+Δr-aΔr/Δr/Γ1+Δr+aΔr/Δr,for n∈{1,1+Δr,…,1+kΔr=N}. The solution reduces to (41)pn=Ψ2n+Δr/2Δr-Ψ2+Δr/2ΔrΨ2N+Δr/2Δr-Ψ2+Δr/2Δrifa=12.We can now state the following proposition.Proposition 7.
Letn=1+jΔr for j∈{0,1,…,k}, with k such that 1+kΔr=N. The probability pn that the discrete-time Markov chain defined in Section 2.1, starting from n, will hit N before 1 is given by (40) if a≠1/2. The value of the probability pn tends to the function in (41) when a tends to 1/2.Next, whena=1/2, by making use of the following approximation Ψz~ln⁡z for large z, we can write that (42)limΔr↓0⁡pn=limΔr↓0⁡ln⁡2n+Δr-ln⁡2+Δrln⁡2N+Δr-ln⁡2+Δr=ln⁡nln⁡Nforn∈1,N.This expression corresponds to the function px given in (6), obtained when a=1/2. Finally, we have(43)Γz+αΓz+β∝zα-β1+O1z,as z tends to infinity (if argz+α<π; see [9]). Hence, in the case a≠0,1/2, we can write that(44)limΔr↓0⁡pn=n1-2a-1N1-2a-1,for1⩽n⩽N.Therefore, we retrieve the formula for pr in (5). In the next section, we will derive the formulas that correspond to the function mr in Section 1.


## 3.2. Computation of the Mean Number of Transitionsdj Needed to End the Game

We now turn to the problem of computing the mean numberdj of transitions that the Markov chain Rtm,m=0,1,…, starting from R0=1+jΔr, takes to reach either 1 or 1+kΔr=N. Unlike the results of [6], we find the explicit formula of the average number of transitions needed to end the game.


### 3.2.1. The CaseΔr=1

IfΔr=1, the state of the Markov chain is the set 1,2,…,N. Conditioning on the result of the first transition of the chain and taking this first transition into account, we obtain that the quantity dj=ERTj=N satisfies the second-order linear, nonhomogeneous difference equation (45)dj=pj,j+1dj+1+pj,j-1dj-1+pj,jdj+1forj=2,…,N-1,where the probabilities pj,j+1, pj,j-1, and pj,j are defined, respectively, in (17), with the boundary conditions:(46)d1=dN=0.We have(47)qj-j-aj+aqj-1=-2Ajj+awhereqj=dj+1-dj.Let r1j=j-a/j+a, r2j=-2Aj/j+a.Now, by making use of the formula of Batchelder [10] (5, p.62), we obtain that a formal particular rational solution Rj∗ of (47) is given by (48)Rj∗=-∑k=0∞r2j+k∏i=0kr1j+i=2AΓj-aΓj+a∑k=0∞j+kΓj+a+kj+k-aΓj-a+k=2AΓj-aΓj+a∑k=0∞Γj+a+k+1Γj-a+k+1-a∑k=0∞Γj+a+kΓj-a+k+1.Then (49)Rj∗=2AΓj-aΓj+aΓj+a+1Γj-a+1Γ1Fj+a+1,1;j-a+1,1-aΓj+aΓ1Γj-aFj+a,1;j-1,1,where Fa,b;c;z=Γc/ΓaΓb∑n=0∞Γa+nΓb+n/Γc+nzn/n!, hypergeometric function.Proposition 8.
In the caseΔr=1, the mean number dj of transitions needed by the Markov chain, starting from j, to reach either 1 or N can be expressed as follows: (50)dj=∑k=1j-1α0Γk+1-aΓk+1+a+Rk∗,for j=1,…,N, where the constant α0 is given by (51)α0=-∑k=1N-1Rk∗∑k=1N-1Γk+1-a/Γk+1+a.


### 3.2.2. The CaseΔr>0

To obtain the solution to our problem in the case whenΔr>0, we must find a particular solution of (52)dj=12A1+aΔr1+jΔrdj+1+12A1-aΔr1+jΔrdj-1(53)+1+1Adj+1,for j=1,…,k-1, with d0=dk=0; proceeding as above by letting dj+1-dj=qj, we can rewrite this second-order linear, nonhomogeneous difference equation as follows: (54)qj+1-1+jΔr-aΔr1+jΔr+aΔrqj=-2A1+jΔr1+jΔr+aΔr.We deduce from the preceding subsection that (55)Rj∗=-2AΓ1/Δr+j-aΓ1/Δr+j+a∑n=0∞1Δr+n+kΓ1/Δr+k+n+aΓ1/Δr+k+n-a+1.To complete the work, we now show that dj×Δt converges to the function mr when we choose Δt=Δr2/A. To do so, we express the general solution of (53) in terms of r=1+jΔr and write dj as dr: (56)d′r≃α0ΔrΓr/Δr-a+1Γ1/Δr+a+1Γ1/Δr+a+1Γr/Δr-a+1-2AΓr/Δr-aΓr/Δr+a∑n=0∞r+nΔr/ΔrΓr+nΔr/Δr+aΓr+nΔr/Δr-a+1,since for z→∞ and argz+a<π (see [9]) (57)Γz+aΓz+b≃za-b1+O1z.Then for Δr small enough, we may write that (58)d′r≃α0Δrr-2a1+OΔrr1Δr-2a×1+OΔr-2ArΔr-2a1+OΔrr×∑n=0∞1Δrr+nΔrr+nΔrΔr2a-11+OΔrr.Next, let(59)Sr=∑n=0∞1Δrr+nΔrr+nΔrΔr2a-11+OΔrr=1+OΔrr1Δr2a+1∑n=0∞r+nΔr2aΔr.Now, let l=N-r/Δr, where [x] is the least integer greater than or equal to x. Then, for small Δr, we may write that (60)Sr=liml→∞⁡1+OΔrr1Δr2a+1∑n=0lr+nΔr2aΔr=1+OΔrr1Δr2a+1∫rNy2ady.Thus, from what precedes, we deduce that (61)limΔr→0⁡d′rΔr22A=α0ΔrΔr2Ar-2a-2r-2a∫rNy2ady,and we get (62)limΔr→0⁡d′rΔt=r-2aα0ΔrΔrΔt-2r-2a∫rNy2ady.Hence, if we let Du=limΔr→0⁡duΔt, we can write (by making use of the boundary condition D1=0) that (63)Du=∫1ur-2alimΔr→0⁡α0ΔrΔrΔt-2r-2a∫rNy2adydr.Finally, to satisfy the boundary condition DN=0, we find that the constant α0Δr must be chosen so that Du=mu. That is, we retrieve formula (9) for the function mu.


## 3.2.1. The CaseΔr=1

IfΔr=1, the state of the Markov chain is the set 1,2,…,N. Conditioning on the result of the first transition of the chain and taking this first transition into account, we obtain that the quantity dj=ERTj=N satisfies the second-order linear, nonhomogeneous difference equation (45)dj=pj,j+1dj+1+pj,j-1dj-1+pj,jdj+1forj=2,…,N-1,where the probabilities pj,j+1, pj,j-1, and pj,j are defined, respectively, in (17), with the boundary conditions:(46)d1=dN=0.We have(47)qj-j-aj+aqj-1=-2Ajj+awhereqj=dj+1-dj.Let r1j=j-a/j+a, r2j=-2Aj/j+a.Now, by making use of the formula of Batchelder [10] (5, p.62), we obtain that a formal particular rational solution Rj∗ of (47) is given by (48)Rj∗=-∑k=0∞r2j+k∏i=0kr1j+i=2AΓj-aΓj+a∑k=0∞j+kΓj+a+kj+k-aΓj-a+k=2AΓj-aΓj+a∑k=0∞Γj+a+k+1Γj-a+k+1-a∑k=0∞Γj+a+kΓj-a+k+1.Then (49)Rj∗=2AΓj-aΓj+aΓj+a+1Γj-a+1Γ1Fj+a+1,1;j-a+1,1-aΓj+aΓ1Γj-aFj+a,1;j-1,1,where Fa,b;c;z=Γc/ΓaΓb∑n=0∞Γa+nΓb+n/Γc+nzn/n!, hypergeometric function.Proposition 8.
In the caseΔr=1, the mean number dj of transitions needed by the Markov chain, starting from j, to reach either 1 or N can be expressed as follows: (50)dj=∑k=1j-1α0Γk+1-aΓk+1+a+Rk∗,for j=1,…,N, where the constant α0 is given by (51)α0=-∑k=1N-1Rk∗∑k=1N-1Γk+1-a/Γk+1+a.


## 3.2.2. The CaseΔr>0

To obtain the solution to our problem in the case whenΔr>0, we must find a particular solution of (52)dj=12A1+aΔr1+jΔrdj+1+12A1-aΔr1+jΔrdj-1(53)+1+1Adj+1,for j=1,…,k-1, with d0=dk=0; proceeding as above by letting dj+1-dj=qj, we can rewrite this second-order linear, nonhomogeneous difference equation as follows: (54)qj+1-1+jΔr-aΔr1+jΔr+aΔrqj=-2A1+jΔr1+jΔr+aΔr.We deduce from the preceding subsection that (55)Rj∗=-2AΓ1/Δr+j-aΓ1/Δr+j+a∑n=0∞1Δr+n+kΓ1/Δr+k+n+aΓ1/Δr+k+n-a+1.To complete the work, we now show that dj×Δt converges to the function mr when we choose Δt=Δr2/A. To do so, we express the general solution of (53) in terms of r=1+jΔr and write dj as dr: (56)d′r≃α0ΔrΓr/Δr-a+1Γ1/Δr+a+1Γ1/Δr+a+1Γr/Δr-a+1-2AΓr/Δr-aΓr/Δr+a∑n=0∞r+nΔr/ΔrΓr+nΔr/Δr+aΓr+nΔr/Δr-a+1,since for z→∞ and argz+a<π (see [9]) (57)Γz+aΓz+b≃za-b1+O1z.Then for Δr small enough, we may write that (58)d′r≃α0Δrr-2a1+OΔrr1Δr-2a×1+OΔr-2ArΔr-2a1+OΔrr×∑n=0∞1Δrr+nΔrr+nΔrΔr2a-11+OΔrr.Next, let(59)Sr=∑n=0∞1Δrr+nΔrr+nΔrΔr2a-11+OΔrr=1+OΔrr1Δr2a+1∑n=0∞r+nΔr2aΔr.Now, let l=N-r/Δr, where [x] is the least integer greater than or equal to x. Then, for small Δr, we may write that (60)Sr=liml→∞⁡1+OΔrr1Δr2a+1∑n=0lr+nΔr2aΔr=1+OΔrr1Δr2a+1∫rNy2ady.Thus, from what precedes, we deduce that (61)limΔr→0⁡d′rΔr22A=α0ΔrΔr2Ar-2a-2r-2a∫rNy2ady,and we get (62)limΔr→0⁡d′rΔt=r-2aα0ΔrΔrΔt-2r-2a∫rNy2ady.Hence, if we let Du=limΔr→0⁡duΔt, we can write (by making use of the boundary condition D1=0) that (63)Du=∫1ur-2alimΔr→0⁡α0ΔrΔrΔt-2r-2a∫rNy2adydr.Finally, to satisfy the boundary condition DN=0, we find that the constant α0Δr must be chosen so that Du=mu. That is, we retrieve formula (9) for the function mu.


## 4. Conclusion and Future Research

As it is well known, the Bessel process is a very important model in financial mathematics. In practice, stock or commodity prices vary discretely over time. Therefore, it is interesting to derive formulas forpj and dj for Markov chains that are as close as we want to the diffusion process. Next, we will try to extend the result in the case of two-dimensional diffusion process, which had a lot of applications in real life portfolio insurance [11, 12] and hydrology [13]. It is a difficult task, since we have to solve some systems of difference equation in two variables.



---

*Source: 7189826-2017-12-03.xml*