Randomly chosen walk of fixed length

Question

Let $G=(V, E)$ be the graph on vertices $V = \{0, \cdots, k\}^n$, where vertices $(v_1, \cdots, v_n)$ and $(w_1, \cdots, w_n)$ share an edge iff $\lvert v_i - w_i\rvert \leq 1$ for all $i$.

A walk of length t is a sequence $X_0, \cdots, X_t\in V$ where $(X_i, X_{i+1})\in E$. Among these walks of length $t$, choose $X_0, \cdots, X_t$ uniformly at random. As far as I see, this is not simply a Markov process, as we choose the $(X_0, \cdots, X_k)$ globally at random, rather than performing a random walk.

Is there a way to prove some of the following intuitive statements?

1. The distribution of $X_t$ given $X_0$ converges to something independent of $X_0$ for $t\to\infty$?

2. Say, the weight of a vertex is given by $$w(v_1, \cdots, v_n) := v_1 + \cdots + v_n$$ and the weight of the entire walk by $$w(X_0, \cdots, X_t) := w(X_0) + \cdots + w(X_t).$$ Then, $\mathbb{E}[w(X_0, \cdots, X_t)\vert X_0 = x, X_t = y]$ is asymptotically independent of $x$ and $y$ when $t\to\infty$.

Answer 1

The right setup here is that of topological Markov chains. This is essentially the same as a directed graph, i.e., a (finite, for simplicity) set of states (vertices) $A$ endowed with a $\{0,1\}$-valued $A\times A$ admissibility matrix $\Sigma$. Then one looks at the sequences (finite or infinite) of states $x_0,x_1,\dots$ subject to the condition $\Sigma_{x_i x_{i+1}}=1$ for any pair of adjacent states.

A natural mixing condition on the matrix $\Sigma$ (satisfied in your situation) is that there exists a power $n$ with the property that all entries of $\Sigma^n$ are positive (i.e., any two states can be joined by a length $n$ admissible path). Under this condition the behaviour of various uniform distributions assocaited with the topological Markov chain is described by the "ordinary" Markov chain on $X$ whose transition probabilities are determined by the Perron - Frobenius eigenvectors of the admissibility matrix. In particular, the answers to both your questions are "yes" (and the "something" in question 1 is precisely the stationary distribution of the aforementioned Markov chain). For instance, see Chapter 3 of the Brin - Stuck textbook.