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Setup

In an $n$-dimensional hypercube $C_n = \{0,1\}^n$, we define the Hamming distance of two vertices $d(A,B)$ to be the number of coordinates in which they differ. (e.g. $d((0,0,1), (1,0,1)) = 2$.)

A simple random walk on the vertices of $C_n$ has $1/n$ chance of moving to each of the $n$ adjacent vertices.

Problem

I am trying to find some expected hitting times $t_d$, where

A and B are given (fixed) vertices of $C_n$,

$d = d(A, B)$, and

$$t_d = \mathbb{E}(\text{time to hit B starting from A}).$$

For example, $t_0$ is the expected return time.

My attempts

Using the inverse of the invariant distribution, we get $$t_0 = 2^n.$$ To find $t_1$, we try to express $t_0$ in another way, by conditioning on the first step: $$t_0 = \underbrace{\frac1n \times (1+t_1) + \ldots + \frac1n \times (1+t_1)}_{n\text{ terms}} $$ $$ \Rightarrow t_1 = 2^n - 1.$$ Similarly, we can find $t_2$. Note that the only outcomes after two steps are i) returning, and ii) being $d = 2$ away from the start. $$t_0 = \frac1n \times 2 + \frac{n-1}n \times (2+t_2) $$ $$\Rightarrow t_2 = \frac{n2^n-2}{n-1}-2 = \frac{n(2^n - 2)}{n-1}.$$

Why do I need help

To find $t_3$, I try to do the same trick. However, I suspect I have made a mistake.

$$t_0 = \frac1n \times 2 + \frac{n-1}{n}\frac1n \times (3+t_1) + \frac{n-1}{n}\frac{n-1}{n}\times (3+t_3)$$

This gives $t_3 = 8.5$ when $n = 3$ (a cube), contradicting this question, which says that $t_3 = 10$ in this case.

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  • $\begingroup$ An alternative way to compute this once you know that $t_1 = 2^n - 1$ and $t_2 = \frac{n(2^n - 2)}{n-1}$ is to write $t_2 = 1 + \frac{2}{n} t_1 + \frac{n-2}{n} t_3$ and solve for $t_3.$ $\endgroup$
    – Roy
    Commented Sep 5, 2023 at 12:01

1 Answer 1

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Suppose you’re at distance 2 at 2nd step. Then, there’s two (not one) way of getting to a vertex at distance 1. (Distances are relative to the starting vertex).

Now, $$t_0 = \frac1n \times 2 + \frac{n-1}{n}\frac2n \times (3+t_1) + \frac{n-1}{n}\frac{n-2}{n}\times (3+t_3)$$

which gives

$$t_3= \frac{(n^2-2n+2)2^n-(6n-4)}{(n-1)(n-2)} - 3.$$

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