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Suppose $M$ is a $2D$ random walk on a grid of squares where it travels in any direction uniformly, but if the direction traveled is out of bounds, it stays in the same location.

How would one prove that the uniform distribution is the stationary distribution for this case?

I found,

$$\pi_i = \frac{1}{4} (\sum_{j \in N(i)} \pi_j) + \frac{4-n}{4}(\pi_i)$$

where $N(i) =$ the set of neighbors of square i and $n =$ size of $N(i)$, but I'm not sure how to proceed.

edit:

$M$ is clearly irreducible and aperiodic

how does one prove its reversible as well?

deah12
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  • Is there a compelling reason to try to describe the transition matrix? You haven't given the size of the "bounds", but if they're large (a $100 \times 100$ grid) or variable (a $k \times k$ grid) I see nothing to be gained from seeing the matrix. – Misha Lavrov Mar 10 '23 at 17:05
  • Maybe I was mistaken, it should be possible to find the stationary distribution without using the matrix. – deah12 Mar 10 '23 at 17:32
  • I think the assumption if finite but too big to be calculated – deah12 Mar 10 '23 at 17:40
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    It should be clear the chain is irreducible. Supposing your finding is correct, try the guess $\pi_:= p$ and it reads $p = \frac{1}{4}\big(n\cdot p\big) +\frac{4-n}{4}p=\frac{1}{4}np + \frac{4}{4}p -\frac{1}{4}np= p$. The detailed balance equations are satisfied so you are done. (I tacitly used the the fact that there are $N\lt \infty$ states here; you directly solve for $p$ by setting $p:=\frac{1}{N}$.) – user8675309 Mar 10 '23 at 18:36

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