Solution and Analysis of a One-Dimensional First-Passage Problem with a Nonzero Halting Probability

Abstract

This paper treats a kind of a one-dimensional first-passage problem, which seeks the probability that a random walker first hits the origin at a specified time. In addition to a usual random walk which hops either rightwards or leftwards, the present paper introduces the “halt” that the walker does not hop with a nonzero probability. The solution to the problem is expressed using a Gauss hypergeometric function. The moment generating function of the hitting time is also calculated, and a calculation technique of the moments is developed. The author derives the long-time behavior of the hitting-time distribution, which exhibits power-law behavior if the walker hops to the right and left with equal probability.

1. Introduction

The first-passage problem is a useful and fundamental problem in statistical physics [1]. In this problem we seek the probability that a diffusion particle or a random walker first reaches a specific set of sites at a specified time. The first-passage problem has many applications in statistical physics, such as reaction-rate theory [2], neuron dynamics [3], and economic analysis [4].

The simplest form of the first-passage problem is a random walker hopping on a one-dimensional lattice until it hits the origin. This problem is called the gambler's ruin problem in probability theory [5]—the motion of the random walker is regarded as the dynamics of a gambler's bankroll. Consider a random walker on a one-dimensional lattice hopping to the right with probability 𝑝_→ and left with 𝑝_←(=1−𝑝_→) in a single time step. The problem seeks statistical properties of duration 𝑇_𝑥, the time at which the walker from the position 𝑥 first hits the origin (of course, 𝑇_𝑥 is a random variable.) The central quantity is the probability 𝑃(𝑥,𝑡) that the walker at 𝑥 first hits the origin after time 𝑡. According to [5], the solution is obtained as 𝑃()=𝑥𝑥,𝑡𝑡𝑡𝑡+𝑥2𝑝_→^{(𝑡−𝑥)/2}𝑝_←^{(𝑡+𝑥)/2}0𝑡and𝑥areofsameparity,𝑡and𝑥areoftheoppositeparity, (1) by solving the equation 𝑃(𝑥,𝑡+1)=𝑝_→𝑃(𝑥+1,𝑡)+𝑝_←𝑃(𝑥−1,𝑡) with initial and boundary conditions 𝑃(0,0)=1 and 𝑃(𝑥,0)=𝑃(0,𝑡)=0 (𝑥,𝑡≥1). The coefficient before 𝑝_→^{(𝑡−𝑥)/2}𝑝_←^{(𝑡+𝑥)/2} in (1) is the number of different paths from 𝑥 hitting the origin first at time 𝑡, and it is connected with the reflection principle of a random walk [6]. Obviously, 𝑃(𝑥,𝑡)=0 holds when 𝑡<𝑥, because ^𝑡_{(𝑡+𝑥)/2}=0.

The present paper analyzes an extended form of the above one-dimensional first-passage problem; a random walker hops to the right with probability 𝑝_→, to the left with 𝑝_←, and it does not hop with probability 𝑝₀=1−𝑝_→−𝑝_←. Figure 1 schematically shows the problem. The only difference from the original ruin problem is that the halting probability 𝑝₀ is introduced, but the results change greatly. In fact, the solution (1) of nonhalting case (𝑝₀=0) is superseded by a formula involving a Gauss hypergeometric function in halting case 𝑝₀≠0. Moreover, moment analysis and asymptotic (long-time) behavior are developed.

Figure 1 

An illustration of the problem. The random walker hops to the right with probability 𝑝→ and to the left with 𝑝←; hopping does not occur with probability 𝑝0. (𝑝→+𝑝←+𝑝0=1). The problem focuses on statistical properties of the duration 𝑇𝑥, which is the time the walker first hits the origin.

2. Hypergeometric Solution

As in the classical problem described in the previous section, let 𝑃(𝑥,𝑡) be the probability that the walker starting from 𝑥 has duration 𝑇_𝑥=𝑡. 𝑃(𝑥,𝑡) satisfies the equation 𝑃(𝑥,𝑡+1)=𝑝_→𝑃(𝑥+1,𝑡)+𝑝_←𝑃(𝑥−1,𝑡)+𝑝₀𝑃(𝑥,𝑡) with initial and boundary conditions 𝑃(0,0)=1 and 𝑃(𝑥,0)=𝑃(0,𝑡)=0 (𝑥,𝑡≥1). However, it is hard to solve this equation directly, and we take another way by employing a classical result (1) effectively.

To calculate 𝑃(𝑥,𝑡), we classify the walker’s paths according to the number of hopping. We count the paths consisting of 𝑗 hops and 𝑡−𝑗 halts. First, the different patterns of putting 𝑡−𝑗 halts into 𝑡 steps are given by 𝑡−1𝑡−𝑗 in total, where “𝑡−1” (not 𝑡) comes from the fact that a halting step never comes to the last 𝑡th step. Next, if we focus on only the hopping steps (and ignore the halting steps), the paths are reduced to those of classical ruin problem; the probability that the walker from 𝑥 hits the origin after 𝑗 hops is given by 𝑃(𝑥,𝑗). Thus, the total occurring probability of a path with 𝑗 hops and 𝑡−𝑗 halts is 𝑝𝑡−1𝑡−𝑗^0𝑡−𝑗𝑃𝑥,𝑗. (2) Summing up all 𝑗, we get 𝑃()=𝑥,𝑡_{𝑡𝑗=𝑥}𝑝𝑡−1𝑡−𝑗^0𝑡−𝑗𝑃=𝑥,𝑗_{𝑡𝑗=𝑥𝑗+𝑥even}𝑝_→^{(𝑗−𝑥)/2}𝑝_←^{(𝑗+𝑥)/2}𝑝₀^𝑡−𝑗𝑥×(𝑡−1)!!!!=𝑗+𝑥/2𝑗−𝑥/2𝑡−𝑗_{∞𝑗=𝑥𝑗+𝑥even}𝑝_→^{(𝑗−𝑥)/2}𝑝_←^{(𝑗+𝑥)/2}𝑝₀^𝑡−𝑗𝑥×(𝑡−1)!!!!=𝑗+𝑥/2𝑗−𝑥/2𝑡−𝑗_∞𝑘=0𝑝^𝑘_→𝑝_←^𝑥+𝑘𝑝₀^{𝑡−𝑥−2𝑘}𝑥(𝑡−1)!()()!.𝑥+𝑘!𝑘!𝑡−𝑥−2𝑘 (3) In the third equality, the range of summation is changed; the terms corresponding to 𝑗>𝑡 have no contribution because 1/((𝑡−𝑗)!)=0. In the last equality we change the summation variable as 𝑗=𝑥+2𝑘. This result can be also obtained by solving directly the equation 𝑃(𝑥,𝑡+1)=𝑝_→𝑃(𝑥+1,𝑡)+𝑝_←𝑃(𝑥−1,𝑡)+𝑝₀𝑃(𝑥,𝑡) with conditions 𝑃(0,0)=1 and 𝑃(𝑥,0)=𝑃(0,𝑡)=0 (𝑥≥1, 𝑡≥1).

We further proceed with calculation; we rewrite factorials using gamma functions, which is a preparatory step toward a hypergeometric function. Employing some formulas of the gamma function, one obtains 1()!=1𝑡−𝑥−2𝑘Γ()×Γ))+())Γ)))𝑡+1−𝑥(((𝑥−𝑡/21/2+𝑘(((𝑥−𝑡/2+𝑘Γ))+(Γ))2(((𝑥−𝑡/21/2))((𝑥−𝑡/2^2𝑘. (4) See Appendix A for the calculation in detail. The probability 𝑃(𝑥,𝑡) is expressed as 𝑃(𝑥,𝑡)=𝑝^𝑥_←𝑝₀^𝑡−𝑥𝑥(𝑡−1)!Γ()Γ()×Γ()𝑥+1𝑡−𝑥+1𝑥+1Γ))+(Γ))×(((𝑥−𝑡/21/2))((𝑥−𝑡/2_∞𝑘=0Γ(((𝑥−𝑡)/2)+𝑘+(1/2))Γ(((𝑥−𝑡)/2)+𝑘)Γ()×1𝑥+𝑘+1𝑘!4𝑝_→𝑝_←𝑝²₀^𝑘=𝑝^𝑥_←𝑝₀^𝑡−𝑥(𝑡−1)!(𝑥−1)!(𝑡−𝑥)!×𝐹𝑥−𝑡2,𝑥−𝑡+12;𝑥+1;4𝑝_→𝑝_←𝑝²₀, (5) where 𝐹(𝛼,𝛽;𝛾;𝑧) is the Gauss hypergeometric function. This is an explicit form of the solution of our problem. This solution cannot be deduced from the nonhalting solution (1). The hypergeometric function in (5) is a genuine hypergeometric function, in the sense that it cannot be expressed using simpler functions. (It has been studied that some hypergeometric functions have tractable expressions, e.g., 𝐹(1/2,1;3/2;−𝑧²)=𝑧⁻¹arctan𝑧 [7].)

We comment here on the convergence of the sum in (5)—at first sight, it seems to diverge when 4𝑝_→𝑝_←/𝑝²₀>1. By using the Pochhammer symbol (𝛼)_𝑘∶=𝛼(𝛼+1)⋯(𝛼+𝑘−1) instead of gamma functions, 𝑃(𝑥,𝑡)=𝑝^𝑥_←𝑝₀^𝑡−𝑥(𝑡−1)!()!()!×𝑥−1𝑡−𝑥_∞𝑘=0((𝑥−𝑡+1)/2)_𝑘((𝑥−𝑡)/2)_𝑘(𝑥+1)_𝑘1𝑘!4𝑝_→𝑝_←𝑝²₀^𝑘. (6) If 𝑥>𝑡, 𝑃(𝑥,𝑡)=0 holds automatically because 1/(𝑡−𝑥)!=0. On the other hand, if 𝑥≤𝑡, either ((𝑥−𝑡)/2)_𝑘 or ((𝑥−𝑡+1)/2)_𝑘 becomes zero for 𝑘>⌊(𝑡−𝑥)/2⌋. (More precisely, the former becomes zero when 𝑥 and 𝑡 are of same parity, and the latter becomes zero otherwise.) The sum consists of a finite number of terms in reality, so one does not need to worry about the convergence.

The exact solution (5) is difficult to understand intuitively. We show numerical evaluation of 𝑃(𝑥,𝑡) in Figure 2. The initial position of the walker is fixed as 𝑥=50. We separate graphs according to the parameter Δ𝑝∶=𝑝_←−𝑝_→; Figures 2(a), 2(b), and 2(c), respectively, correspond to Δ𝑝=0.2, 0.1, and 0. (Δ𝑝 is a key parameter for the average duration, as discussed in the following section.) 𝑃(𝑥,𝑡) is a unimodal function of 𝑡 for each parameter value. Power-law behavior 𝑃(𝑥,𝑡)∝𝑡^−3/2 is suggested in large 𝑡 when Δ𝑝=0 (see Figure 2(d)), which is further discussed in Section 4.

(a)

(b)

(c)

(d)

Figure 2 

Numerical results of 𝑃(𝑥,𝑡) as a function of 𝑡 (𝑥=50). The graphs are separated according to the value of Δ𝑝=𝑝←−𝑝→: (a) Δ𝑝=0.2, (b) Δ𝑝=0.1, and (c) Δ𝑝=0. The two curves in each panel represent 𝑝→=0.1 and 0.3. Long-time behavior of 𝑃(𝑥,𝑡) in Δ𝑝=0 is shown in (d) on logarithmic scales, which suggests a power law 𝑃(𝑥,𝑡)∝𝑡−3/2.

3. Analysis of Moment

In order to see properties of the random variable 𝑇_𝑥, moment analysis is developed in this section.

First, we calculate the moment generating function of 𝑇_𝑥, defined as 𝑀_𝑥(𝑠)=^∞_𝑡=0𝑒^𝑠𝑡𝑃(𝑥,𝑡). (7) Calculation process is summarized in Appendix B, and the result is 𝑀_𝑥(𝑠)=𝑒^−𝑠−𝑝₀2𝑝_→−𝑒^−𝑠−𝑝₀2𝑝_→²−𝑝_←𝑝_→^𝑥. (8) Remarkably, 𝑀_𝑥(𝑠) contains no special functions, though the infinite summation in (7) involves a genuine hypergeometric function via 𝑃(𝑥,𝑡).

Repeatedly differentiating 𝑀_𝑥(𝑠) and putting 𝑠=0, we have the first and second moments of the duration 𝑇_𝑥 as 𝑇_𝑥=𝑥𝑝_←−𝑝_→,𝑇²_𝑥=𝑥²𝑝_←−𝑝_→²+𝑝_←+𝑝_→𝑝_←−𝑝_→³−1𝑝_←−𝑝_→𝑥. (9) The average and the other moments remain finite when 𝑝_←>𝑝_→, and they diverge to infinity as 𝑝_←−𝑝_→↘0. The variance of 𝑇_𝑥 is given by 𝑉𝑇_𝑥∶=𝑇²_𝑥−𝑇_𝑥²=𝑝_←+𝑝_→𝑝_←−𝑝_→³−1𝑝_←−𝑝_→𝑥. (10)

We can give a plain explanation for the average ⟨𝑇_𝑥⟩. The walker moves to the left by a length Δ𝑝=𝑝_←−𝑝_→ on average in a single hop. In other words, Δ𝑝 is a mean velocity of the walker toward the origin. Hence, it takes 𝑥/Δ𝑝 time steps on average to reach the origin.

The moment generating function (8) looks too complicated to calculate higher moments by differentiation. Alternatively, we provide another calculation method for moments of the duration 𝑇_𝑥.

We start from the difference equation for 𝑃(𝑥,𝑡): 𝑃(𝑥,𝑡+1)=𝑝_→𝑃(𝑥+1,𝑡)+𝑝_←𝑃(𝑥−1,𝑡)+𝑝₀𝑃(𝑥,𝑡). (11) Multiply 𝑡 and take summation for 𝑡 to get the following: ^∞_𝑡=0()=𝑡𝑃𝑥,𝑡+1_∞𝑡=0𝑡𝑝_→𝑃(𝑥+1,𝑡)+𝑡𝑝_←𝑃(𝑥−1,𝑡)+𝑡𝑝₀𝑃(𝑥,𝑡)=𝑝_→⟨𝑇_𝑥+1⟩+𝑝_←⟨𝑇_𝑥−1⟩+𝑝₀𝑇_𝑥. (12) The left-hand side can be expressed as 𝑡𝑃(^∞_𝑡=0𝑥,𝑡+1)=^∞_𝑡=0()𝑃()−𝑡+1𝑥,𝑡+1_∞𝑡=0𝑃(𝑥,𝑡+1)=𝑇_𝑥−1. (13) The last equality uses ∑^∞_𝑡=0𝑃(𝑥,𝑡+1)=1, which means that a random walker surely hits the origin. (This normalization breaks if 𝑝_←<𝑝_→—see the discussion of total probability of hitting the origin in Section 5 in more detail.) We have a difference equation: 𝑝_→𝑇_𝑥+1−𝑝_→+𝑝_←𝑇_𝑥+𝑝_←𝑇_𝑥−1=−1. (14) A particular solution is given by ⟨𝑇_𝑥⟩=𝑥/(𝑝_←−𝑝_→), and the characteristic equation 𝑝_→𝜆²−(𝑝_→+𝑝_←)𝜆+𝑝_←=0 admits the two roots 1 and 𝑝_←/𝑝_→. Thus, the general solution is 𝑇_𝑥=𝑥𝑝_←−𝑝_→𝑝+𝐴+𝐵^←𝑝_→^𝑥, (15) which has two constants 𝐴 and 𝐵. In a limiting case 𝑝_→→0, the problem is still well-defined, and ⟨𝑇_𝑥⟩ should be finite, so 𝐵=0. Also, it is obvious that ⟨𝑇₀⟩=0, so 𝐴=0. The appropriate solution is therefore 𝑇_𝑥=𝑥𝑝_←−𝑝_→. (16)

Next we calculate the second moment. Multiplying 𝑡² and taking summation, we derive a difference equation: 𝑇²_𝑥−2𝑇_𝑥+1=𝑝_→𝑇²_𝑥+1+𝑝_←𝑇²_𝑥−1+𝑝₀𝑇²_𝑥, (17) where we use 𝑡²=(𝑡+1)²−2(𝑡+1)+1 for the calculation on the left-hand side. We assume a particular solution in the form ⟨𝑇²_𝑥⟩=𝐶₂𝑥²+𝐶₁𝑥, and determine two constants as 𝐶₁=𝑝_←+𝑝_→𝑝_←−𝑝_→³−1𝑝_←−𝑝_→,𝐶₂=1𝑝_←−𝑝_→². (18) The second moment is 𝑇²_𝑥=𝑥²𝑝_←−𝑝_→²+𝑝_←+𝑝_→𝑝_←−𝑝_→³−1𝑝_←−𝑝_→𝑥=𝑥²𝑝_←−𝑝_→²+𝑝₀1−𝑝₀+4𝑝_→𝑝_←𝑝_←−𝑝_→³𝑥. (19) The homogeneous-solution part vanishes for the same reason as the average.

In short, this “bottom-up” method calculates higher moments inductively and generally ⟨𝑇^𝑘_𝑥⟩ becomes a polynomial of degree 𝑘. For instance, the third moment is calculated as 𝑇³_𝑥=𝑥³𝑝_←−𝑝_→³+𝑝_←+𝑝_→𝑝_←−𝑝_→³−1𝑝_←−𝑝_→3𝑥²+2𝑝_←+𝑝_→²+4𝑝_←𝑝_→𝑝_←−𝑝_→⁵−3𝑝_←+𝑝_→𝑝_←−𝑝_→³+1𝑝_←−𝑝_→𝑥. (20)

4. Asymptotic Behavior

We look at long-time behavior of 𝑃(𝑥,𝑡) in this section. In particular, we confirm the power-law behavior 𝑃(𝑥,𝑡)∝𝑡^−3/2 mentioned in Section 2 when 𝑝_→=𝑝_← (see Figure 2(d) for reference).

A summation and factorial, as well as a hypergeometric function, are not suitable to study limiting behavior, so we first reexpress the probability 𝑃(𝑥,𝑡) into a tractable form (it may be also possible to estimate factorials directly by Stirling’s formula). According to [5], the solution (1) of the classic ruin problem has another expression: 𝑃(𝑥,𝑡)=2^𝑡𝑝_→^{(𝑡−𝑥)/2}𝑝_←^{(𝑡+𝑥)/21}₀cos^𝑡−1𝜋𝜙⋅sin𝜋𝜙⋅sin𝜋𝑥𝜙𝑑𝜙. (21) This form is convenient in that one does not need to care about the parities of 𝑥 and 𝑡. In fact, the integral holds the information of parities: ¹₀cos^𝑡−1=𝑥𝜋𝜙⋅sin𝜋𝜙⋅sin𝜋𝑥𝜙𝑑𝜙2^𝑡(𝑡−1)!))!))!0((𝑡+𝑥/2((𝑡−𝑥/2𝑡and𝑥areofsameparity,otherwise. (22) In Appendix C, we prove this formula. Applying (21) to (3), we have the integral form of the probability 𝑃(𝑥,𝑡): 𝑃()=𝑥,𝑡_𝑡𝑗=1𝑝𝑡−1𝑡−𝑗^0𝑡−𝑗𝑃𝑥,𝑗=2𝑝_→^(1−𝑥)/2𝑝_←^(1+𝑥)/2×¹₀^𝑡_𝑗=1𝑝𝑡−1𝑡−𝑗^0𝑡−𝑗×2𝑝_→𝑝_←cos𝜋𝜙^𝑗−1sin𝜋𝜙⋅sin𝜋𝑥𝜙𝑑𝜙=2𝑝_→^(1−𝑥)/2𝑝_←^(1+𝑥)/21₀𝑝₀+2𝑝_→𝑝_←cos𝜋𝜙^𝑡−1×sin𝜋𝜙⋅sin𝜋𝑥𝜙𝑑𝜙=2𝑝_→^(1−𝑥)/2𝑝_←^(1+𝑥)/2𝑝₀+2𝑝_→𝑝_←^𝑡−1×¹₀𝑝₀+2𝑝_→𝑝_←cos𝜋𝜙𝑝₀+2𝑝_→𝑝_←^𝑡−1sin𝜋𝜙⋅sin𝜋𝑥𝜙𝑑𝜙. (23) Here we estimate the integral 𝐼∶=¹₀𝑝₀+2𝑝_→𝑝_←cos𝜋𝜙𝑝₀+2𝑝_→𝑝_←^𝑡−1=sin𝜋𝜙⋅sin𝜋𝑥𝜙𝑑𝜙_0𝜋𝑡𝑝₀+2𝑝_→𝑝_←cos(𝜈/𝑡)𝑝₀+2𝑝_→𝑝_←^𝑡−1𝜈sin𝑡⋅sin𝑥𝜈𝑡⋅1𝜋𝑡𝑑𝜈, (24) where the integral variable is changed as 𝜈=𝜋𝑡𝜙 in the second equality. For large 𝑡, consider the following: 𝑝₀+2𝑝_→𝑝_←cos(𝜈/𝑡)𝑝₀+2𝑝_→𝑝_←^𝑡−1≃1−𝑝_→𝑝_←𝑝₀+2𝑝_→𝑝_←𝜈²𝑡^2𝑡−1≃exp−𝑝_→𝑝_←𝑝₀+2𝑝_→𝑝_←𝜈²𝑡. (25) Hence, the integrand is a rapidly decreasing function of 𝜈, and we can extend the upper limit of the integration to infinity. Together with the approximation sin(𝜈/𝑡)≃𝜈/𝑡, we can carry out the integration as 𝑥𝐼≃𝜋𝑡^3∞₀exp−𝑝_→𝑝_←𝑝₀+2𝑝_→𝑝_←𝜈²𝑡𝜈²=𝑥𝑑𝜈4√𝜋𝑝₀+2𝑝_→𝑝_←𝑝_→𝑝_←^3/2𝑡^−3/2. (26)

Therefore, the asymptotic form is expressed as 𝑃()≃𝑥𝑥,𝑡2√𝜋𝑝_→^(1−𝑥)/2𝑝_←^(1+𝑥)/2𝑝₀+2𝑝_→𝑝_←^𝑡−1×𝑝₀+2𝑝_→𝑝_←𝑝_→𝑝_←^3/2𝑡^−3/2. (27) By the arithmetic mean-geometric mean inequality, we note that 𝑝₀+2𝑝_→𝑝_←≤𝑝₀+𝑝_→+𝑝_←=1, (28) with equality holding if and only if 𝑝_→=𝑝_← (or Δ𝑝=0 by the notation in Section 2). Thus, the long-time behavior of 𝑃(𝑥,𝑡) is completely different according to whether Δ𝑝 equals zero, or not as follows.(i) If Δ𝑝≠0 (i.e., 𝑝_→≠𝑝_←), 𝑝₀+2𝑝_→𝑝_←<1, and hence 𝑃(𝑥,𝑡) asymptotically exhibits exponential decay due to (𝑝₀+2𝑝_→𝑝_←)^𝑡. (ii) If Δ𝑝=0 (i.e., 𝑝_→=𝑝_←=∶𝑝), a power law with exponent −3/2 is concluded: 𝑃()≃𝑥𝑥,𝑡2√𝑡𝜋𝑝^−3/2. (29) See Figure 2(d) for reference.

5. Concluding Remarks and Discussions

In the present paper, we have solved a one-dimensional first-passage problem where a random walker hops to the right with probability 𝑝_→, left with 𝑝_←, and does not hop with 𝑝₀. We have calculated exactly the probability 𝑃(𝑥,𝑡) that the walker starting from position 𝑥 has duration 𝑇_𝑥=𝑡. The average and higher moments of 𝑇_𝑥 are calculated in two ways: by the moment generating function, and by the bottom-up calculation. The asymptotic form of 𝑃(𝑥,𝑡) is derived, and a power law 𝑃(𝑥,𝑡)∼𝑡^−3/2 is obtained when 𝑝_→=𝑝_←.

We make a brief discussion on the continuum limit, which is an appropriate scaling limit where the step width and time interval of the walker tend to zero. Let the step width be 𝛿 and time interval be 𝜖, and position and time are scaled as 𝜉=𝛿𝑥 and 𝜏=𝜖𝑡. In the continuum limit we take that 𝛿,𝜖→0, together with 𝑝_→−𝑝_←𝛿𝜖𝛿⟶𝑣,²2𝜖⟶𝐷, (30) are kept finite. 𝑣 is the mean displacement per time unit, and 𝐷 is the diffusion coefficient corresponding to a nonhalting (or formally called a simple) random walk. The probability density function 𝜌()1𝜉,𝜏=lim𝜖𝑃𝑥𝛿,𝑡𝜖 (31) is suitable in the continuum limit, rather than the probability distribution 𝑃(𝑥,𝑡). (“lim” represents the continuum limit.) Carrying out calculation similar to that in [5], 𝜌()=𝜉𝜉,𝜏4𝜋1−𝑝₀𝐷𝜏³−()exp𝜉+𝑣𝜏²41−𝑝₀𝐷𝜏 (32) is obtained. The probability density 𝜌 is called the inverse Gaussian distribution [8]. In the continuum limit, the halting effect 𝑝₀ is reflected only upon the diffusion coefficient as (1−𝑝₀)𝐷. Comparing 𝑃(𝑥,𝑡) of a discrete problem in (5) and above 𝜌(𝜉,𝜏) of a continuum limit, we conclude that the discrete random walk is far more difficult than the continuous diffusion and that the results about the discrete random walk in this paper cannot be attained from continuous diffusion problem.

Combining (5), (7), and (8), we get ^∞_𝑡=0𝑒^𝑠𝑡𝑝^𝑥_←𝑝₀^𝑡−𝑥𝐹𝑡−1𝑥−1𝑥−𝑡2,𝑥−𝑡+12;𝑥+1;4𝑝_→𝑝_←𝑝²₀=𝑒^−𝑠−𝑝₀2𝑝_→−𝑒^−𝑠−𝑝₀2𝑝_→²−𝑝_←𝑝_→^𝑥. (33) We have not used the condition 𝑝_→+𝑝_←+𝑝₀=1 in the derivation of (5) and (8), so 𝑝_→, 𝑝_←, and 𝑝₀ can take any value independently. Setting 𝑝_→=𝑧/4 and 𝑝_←=𝑝₀=1, ^∞_𝑡=0𝑒^𝑠𝑡𝐹𝑡−1𝑥−1𝑥−𝑡2,𝑥−𝑡+12=2𝑒;𝑥+1;𝑧_−𝑠−1𝑧−2𝑒^−𝑠−1𝑧²−4𝑧^𝑥. (34) Furthermore, setting 𝑠=−ln2 (i.e., 𝑒^−𝑠=2), ^∞_𝑡=012^𝑡𝐹𝑡−1𝑥−1𝑥−𝑡2,𝑥−𝑡+12=√;𝑥+1;𝑧2−21−𝑧𝑧^𝑥. (35) These formulas are nontrivial, but the author cannot tell whether they are useful in practice. We stress that an infinite series of which each term involves a genuine hypergeometric function is hardly known.

We comment on the total probability of hitting the origin. For the discrete problem, putting 𝑠=0 the moment generating function 𝑀_𝑥(𝑠), ^∞_𝑡=0𝑃(𝑥,𝑡)=𝑀_𝑥(0)=1−𝑝₀2𝑝_→−1−𝑝₀2𝑝_→²−𝑝_←𝑝_→^𝑥=𝑝_→+𝑝_←2𝑝_→−𝑝_←−𝑝_→²2𝑝_→^𝑥=1𝑝_←≥𝑝_→,𝑝_←𝑝_→^𝑥𝑝_←<𝑝_→. (36) If 𝑝_←>𝑝_→, the walker surely comes to the origin eventually, but if 𝑝_←<𝑝_→, the walker manages to avoid hitting the origin with a nonzero probability.

The equation 𝑃(𝑥,𝑡+1)=𝑝_→𝑃(𝑥+1,𝑡)+𝑝_←𝑃(𝑥−1,𝑡)+𝑝₀𝑃(𝑥,𝑡) describes a traffic jam, where 𝑃(𝑥,𝑡) stands for the probability distribution that the jam consists of 𝑥 cars at time 𝑡, and 𝑝_→ and 𝑝_← are involved in the rates at which cars enter and leave the jam. The scaling behavior with exponent −3/2 has been observed in the lifetime distribution of jams [9]. We think that results obtained in this paper can become theoretical bases; in particular, our discrete problem is comparable to a microscopic description of traffic, where discreteness is not negligible.

Appendices

A. Calculation of (4)

Here we follow the calculation of (4) in detail. Employing the duplication formula Γ(2𝑧)=𝜋^−1/22^2𝑧−1Γ(𝑧)Γ(𝑧+(1/2)), we rewrite as follows: 1()!=1𝑡−𝑥−2𝑘Γ()=√𝑡−𝑥−2𝑘+1𝜋2^{𝑡−𝑥−2𝑘}Γ))(Γ))).(((𝑡−𝑥/2−𝑘+1/2))(((𝑡−𝑥/2−𝑘+1 (A.1) By Euler’s reflection formula Γ(𝑧)Γ(1−𝑧)=𝜋/sin𝜋𝑧, Γ𝑡−𝑥21−𝑘+2=1Γ())+(×𝜋𝑘−((𝑡−𝑥/21/2))))(=1sin𝜋(((𝑡−𝑥/2−𝑘+1/2))Γ())+(×()𝑘−((𝑡−𝑥/21/2))−1^𝑘𝜋))+(=()sin𝜋(((𝑡−𝑥/21/2))−1^𝑘Γ(𝑘−((𝑡−𝑥)/2)+(1/2))×Γ𝑡−𝑥2+12Γ12−𝑡−𝑥2. (A.2) Note that sin𝜋(((𝑡−𝑥)/2)−𝑘+(1/2))=(−1)^𝑘sin𝜋(((𝑡−𝑥)/2)+(1/2)) because 𝑘 is an integer. Similarly, Γ𝑡−𝑥2=()−𝑘+1−1^𝑘Γ(𝑘+((𝑥−𝑡)/2))×Γ𝑡−𝑥2+1Γ𝑥−𝑡2 (A.3) is obtained. Substituting (A.2) and (A.3) into (A.1), then using the duplication formula again as Γ𝑡−𝑥2+12Γ𝑡−𝑥2=√+1𝜋2^𝑡−𝑥Γ(𝑡−𝑥+1), (A.4) we eventually come to the following result: 1()!=1𝑡−𝑥−2𝑘Γ()×Γ))+())Γ)))𝑡+1−𝑥(((𝑥−𝑡/21/2+𝑘(((𝑥−𝑡/2+𝑘Γ))+(Γ))2(((𝑥−𝑡/21/2))((𝑥−𝑡/2^2𝑘. (A.5)

B. Calculation of the Moment Generating Function (8)

We follow here the calculation of (8). Substituting (3) into (7), 𝑀_𝑥(𝑠)=^∞_𝑡=0𝑒^∞𝑠𝑡_𝑘=0𝑝^𝑘_→𝑝_←^𝑥+𝑘𝑝₀^{𝑡−𝑥−2𝑘}𝑥(𝑡−1)!()()!=𝑥+𝑘!𝑘!𝑡−𝑥−2𝑘_∞𝑘=0𝑝^𝑘_→𝑝_←^𝑥+𝑘𝑝₀^{−𝑥−2𝑘}𝑥1()×𝑥+𝑘!𝑘!_∞𝑡=0(𝑡−1)!()!𝑝𝑡−𝑥−2𝑘⁰𝑒^𝑠𝑡=^∞_𝑘=0𝑝^𝑘_→𝑝_←^𝑥+𝑘𝑝₀^{−𝑥−2𝑘}𝑥1()𝑝𝑥+𝑘!𝑘!⁰𝑒^{𝑠𝑥+2𝑘}×^∞_𝜏=0(𝜏+𝑥+2𝑘−1)!𝑝𝜏!⁰𝑒^𝑠𝜏, (B.1) where a summation variable is changed as 𝜏=𝑡−𝑥−2𝑘 in the last equality. Applying the negative binomial expansion 1−𝑦^−𝑚=^∞_𝜏=0(𝜏+𝑚−1)!()!𝑦𝜏!𝑚−1^𝜏 (B.2) with 𝑦=𝑝₀𝑒^𝑠 and 𝑚=𝑥+2𝑘, one obtains ^∞_𝜏=0(𝜏+𝑥+2𝑘−1)!𝑝𝜏!⁰𝑒^𝑠𝜏=(𝑥+2𝑘−1)!1−𝑝₀𝑒^{𝑠−𝑥−2𝑘} (B.3) and simplifies (B.1) into 𝑀_𝑥(𝑠)=𝑝_←𝑒^−𝑠−𝑝₀^𝑥𝑥^∞_𝑘=0(𝑥+2𝑘−1)!(𝑥+𝑘)!𝑘!𝑝_→𝑝_←𝑒^−𝑠−𝑝₀^2𝑘. (B.4) We can transform as follows using the duplication formula Γ(2𝑧)=𝜋^−1/22^2𝑧−1Γ(𝑧)Γ(𝑧+(1/2)): ()()=2𝑥+2𝑘−1!=Γ𝑥+2𝑘_{𝑥−1+2𝑘}√𝜋Γ𝑥2+𝑘Γ𝑥+12=2+𝑘_𝑥−1√𝜋Γ𝑥2Γ𝑥+12×Γ((𝑥/2)+𝑘)Γ(((𝑥+1)/2)+𝑘)Γ()Γ))2𝑥/2((𝑥+1/2^2𝑘(𝑥)Γ))Γ)))=Γ((𝑥/2+𝑘(((𝑥+1/2+𝑘Γ()Γ))2𝑥/2((𝑥+1/2^2𝑘. (B.5) The moment generating function can be expressed by a Gauss hypergeometric function as 𝑀_𝑥(𝑠)=𝑝_←𝑒^−𝑠−𝑝₀^𝑥𝑥Γ(𝑥)Γ()Γ()𝑥+1𝑥+1Γ()Γ))×𝑥/2((𝑥+1/2_∞𝑘=0Γ((𝑥/2)+𝑘)Γ(((𝑥+1)/2)+𝑘)Γ()×1𝑥+1+𝑘2𝑘!𝑝_→𝑝_←𝑒^−𝑠−𝑝₀^2𝑘=𝑝_←𝑒^−𝑠−𝑝₀^𝑥𝐹𝑥2,𝑥+12;𝑥+1;4𝑝_→𝑝_←𝑒^−𝑠−𝑝₀². (B.6) It can be expressed by an elementary function as follows (see [7]): 𝐹𝑥2,𝑥+12=2;𝑥+1;𝑧1+√1−𝑧^𝑥=2−2√1−𝑧𝑧^𝑥. (B.7) This leads to the conclusion.

C. Proof of (22)

We calculate the integral to prove (22). We first break up the product of circular functions into the sum: cos^𝑡−1𝑒𝜋𝜙=^𝑖𝜋𝜙+𝑒^{−𝑖𝜋𝜙}2^𝑡−1=12^{𝑡−1𝑡−1}_𝑘=0𝑘𝑒𝑡−1_{𝑖(2𝑘−𝑡+1)𝜋𝜙}, (C.1) whose real part is cos^𝑡−11𝜋𝜙=2^{𝑡−1𝑡−1}_𝑘=0𝑘()𝑡−1cos2𝑘−𝑡+1𝜋𝜙. (C.2) This is just a Fourier series expansion of cos^𝑡−1𝜃 and is associated with the Chebyshev polynomial 𝑇_𝑛(cos𝜃)∶=cos𝑛𝜃 [10]. Moreover, by formulas in trigonometry, ()=1cos2𝑘−𝑡+1𝜋𝜙⋅sin𝜋𝜙⋅sin𝜋𝑥𝜙4()()()().cos2𝑘−𝑡+𝑥𝜋𝜙+cos2𝑘+2−𝑡−𝑥𝜋𝜙−cos2𝑘+2−𝑡+𝑥𝜋𝜙−cos2𝑘−𝑡−𝑥𝜋𝜙 (C.3) Thus, ¹₀cos^𝑡−1=1𝜋𝜙⋅sin𝜋𝜙⋅sin𝜋𝑥𝜙𝑑𝜙2^{𝑡+1𝑡−1}_𝑘=0𝑘×𝑡−1₁₀()()()()cos2𝑘−𝑡+𝑥𝜋𝜙+cos2𝑘+2−𝑡−𝑥𝜋𝜙−cos2𝑘+2−𝑡+𝑥𝜋𝜙−cos2𝑘−𝑡−𝑥𝜋𝜙𝑑𝜙. (C.4) Since 𝑡, 𝑥, and 𝑘 are integers, the first term in the integral is ¹₀()=cos2𝑘−𝑡+𝑥𝜋𝜙𝑑𝜙12𝑘−𝑡+𝑥=0,0otherwise, (C.5) which behaves like the Kronecker delta 𝛿(𝑘,(𝑡−𝑥)/2)—a subscript fraction may be unreadable, so we use 𝛿(𝑖,𝑗) instead of the standard symbol 𝛿_𝑖,𝑗. The following terms in the integral are also substituted by 𝛿(𝑘,(𝑡+𝑥−2)/2), 𝛿(𝑘,(𝑡−𝑥−2)/2), and 𝛿(𝑘,(𝑡+𝑥)/2), respectively.

If 𝑡 and 𝑥 have opposite parity (i.e., 𝑡+𝑥 and 𝑡−𝑥 are odd), each Kronecker delta vanishes for all integer 𝑘. Otherwise, if 𝑡 and 𝑥 have the same parity, each Kronecker delta becomes nonzero at some 𝑘. By picking out such 𝑘, ¹₀cos^𝑡−1=1𝜋𝜙⋅sin𝜋𝜙⋅sin𝜋𝑥𝜙𝑑𝜙2^𝑡+1𝑡−1𝑡−𝑥2+𝑡−1𝑡+𝑥−22−𝑡−1𝑡−𝑥−22−𝑡−1𝑡+𝑥2=𝑥2^𝑡(𝑡−1)!))!))!.((𝑡+𝑥/2((𝑡−𝑥/2 (C.6) Therefore, (22) is derived.

Acknowledgments

The author is grateful to Dr. Yoshihiro Yamazaki and Dr. Jun-ichi Wakita for their beneficial comments.