Exact Formula of Time-Headway Distribution for TASEP with Random-Sequential Update

Abstract

The analytical derivation of time-headway distribution for random-sequential totally asymmetric simple exclusion process (TASEP) with periodic boundaries is presented. The finite and periodic nature of the lattice together with the lattice-size-dependent hopping probability related to the random-sequential update does not allow to use common method for the derivation of the time-headway distribution. Another method is presented in this article. The exact derivation of the time-headway distribution leads to several interesting combinatorial tasks. Further, after proper time scaling and using the large L limit we obtain the approximation of the distribution, which can be considered as exact result for TASEP with continuous time.

Appendix

Lemma 1

For n, m ≥ 1 and z ≥ 0 it holds

$$\displaystyle \begin{aligned} \begin{array}{rcl} C_{n,m,z}(j)&\displaystyle =&\displaystyle \textstyle{{j+n-1\choose j}\left[{z+n+m\choose m}-{j+n+m-1\choose m}\right]}+\\ &\displaystyle +&\displaystyle \textstyle{{j+m-1\choose j}\left[{z+n+m\choose n}-{j+n+m-1\choose n}\right]} \end{array} \end{aligned} $$

(11)

Proof

This combinatorial task leads to the expression

$$\displaystyle \begin{aligned} \begin{array}{rcl} C_{n,m,z}(j)&\displaystyle =&\displaystyle \textstyle{\sum_{d=0}^{z-j}\sum_{r_L=0}^{m-1}{z-j-d+r_L\choose r_m}\frac{(j+m+n-r_L-2)!}{k!(n-1)!(m-1-r_L)!}}\\ &\displaystyle +&\displaystyle \textstyle{\sum_{d=0}^{z-j}\sum_{r_F=0}^{n-1}{z-j-d+r_F\choose r_n}\frac{(j+m+n-r_F-2)!}{j!(m-1)!(n-1-r_F)!}}\,, {} \end{array} \end{aligned} $$

(12)

as can be seen from Fig. 6. This can be summed using combinatorial Lemmas 2 and 3 shown below.

Fig. 6

To the derivation of C _n,m,z(j), where s ₀ = j, r ₀ = z − j − d, r _L + s _L = m − 1, s _F = n − 1. Exchanging F and L leads to the second sum of (12)

Lemma 2

For a, b, m ≥ 0 it holds

$$\displaystyle \begin{aligned} \sum_{k=0}^{m}{a+k\choose k}{b+m-k\choose m-k}={a+b+m+1\choose m}\,. \end{aligned} $$

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Proof

$$\displaystyle \begin{aligned} \begin{array}{rcl} \sum_{m=0}^{+\infty}{\textstyle{a+b+m+1\choose m}}z^m=(1-z)^{-(a+1)-(b+1)}=\sum_{m=0}^{+\infty}\sum_{k=0}^{m}{\textstyle{a+k\choose k}{b+m-k\choose m-k}}z^m\,. \end{array} \end{aligned} $$

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Lemma 3

For a > b ≥ 0 and m ≥ 0 it holds

$$\displaystyle \begin{aligned} \sum_{d=0}^{m}{a+d\choose b}={a+m+1\choose b+1}-{a\choose b+1}\,. \end{aligned} $$

(15)

Proof

The relation can be directly derived using

$$\displaystyle \begin{aligned} {a+d\choose b}={a+d+1\choose b+1}-{a+d\choose b+1}\,. \end{aligned} $$

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