Random walk on $\mathbb{Z}$/L $\mathbb{Z}$

Question

Consider a random walk on $\mathbb{Z}$/L $\mathbb{Z}$ where:

$$ Z_{0}=0 $$ and if d (i,j) = 1 then:

$$\mathbb{P}\left(Z_{n+1}=i \mid Z_{n}=j\right) = \dfrac{1}{2}\text{.}$$

else: $$\mathbb{P}\left(Z_{n+1}=i \mid Z_{n}=j\right) = 0\text{.}$$

We define $\beta$=$\beta_{L}$ $\in$ $\mathbb{Z}$/L $\mathbb{Z}$ as the last non visited point.

1) What is the (probability law of $\beta$ ?

2) Is $\beta$ more likely to be closed to to $\mathbb{0}$ or far from $\mathbb{0}$?

How would you plot this random walk for Z/LZ I find it easy for a simple random walk for integers, but in the Z/LZ I don't know. — , Apr 23 '18 at 15:58
@RobertLewis The (quotient) ring of integers modulo L, where L is an integer — , Apr 23 '18 at 16:04
You'd plot it as a random walk on a circle with $L$ lattice points distributed around it. The larger problem here is the issue brought up by @Servaes; how are you defining "last point visited"? — Aaron Montgomery, Apr 23 '18 at 16:25
The transition matrix is double stochastic, so should not "last point visited" be any point with the same probability? — dEmigOd, Apr 23 '18 at 16:34
https://math.stackexchange.com/questions/116446/random-walk-on-n-cycle — Mike Earnest, Apr 02 '19 at 16:23

score 1 · Accepted Answer · answered Apr 23 '18 at 16:54

This is a classic probability puzzle, whose result is counterintuitive.

For each $0<i< L$, let $p_i$ be the probability that $i$ is the last non-visited point. I claim that for any $1<i<L-1$, we have $$ p_i=\frac12p_{i-1}+\frac12p_{i+1}\tag{1} $$ To see this, suppose your first step is in the clockwise direction. Now, point $i$ is $i-1$ steps clockwise of you, so conditioned on this first step, the probability of seeing $i$ last is the same as the probability of seeing $i-1$ last. The other summand comes from conditioning on an anti-clockwise step.

The equation $(1)$ implies that in the list $$ p_1,p_2,p_3,\dots,p_{L-1}, $$ each entry is the average of its neighbors, implying this list is an arithmetic progression. Furthermore, the symmetry of this problem implies $p_1=p_{L-1}$. Therefore, the arithmetic progression is constant, so each $p_i=\frac1{L-1}$; that is, no particular point is more likely to be visited last!

Random walk on $\mathbb{Z}$/L $\mathbb{Z}$

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