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 $t\to\infty$.



Randomly chosen walk of fixed length
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](https://mathscinet.ams.org/mathscinet-getitem?mr=3558919).


