This is a problem given by a professor that has been perplexing me.
Suppose a particle takes a random walk on a $100 \times 100$ checkerboard in the following way. After an exponential time with rate 1, it attempts to move up, down, left, or right -- each with probability 1/4. If the attempted move would take the particle off the board, it stays put instead. Then, after an exponential time with rate 1, it tries to move again, and on and on. What is the stationary distribution of the particle's position.
Now suppose there are 1278 such particles on the board moving independently, and multiple particles can occupy the same squares. What is the stationary distribution for the number of particles on each square? You might want to think of your state space as consisting of all the $100 \times 100$ arrays, where the number in the $(i, j)$ position in the array corresponds to the number of particles there.
Finally answer the previous question when the 1278 particles are only allowed to move to empty squares. That is, each square can only accomodate one particle. Now the state space would be all the $100 \times 100$ arrays of 1s and 0s with exactly 1278 1s.
Now, I'm certain that if I understand how to solve the first part (the first paragraph that is, then I'll be able to use similar logic to solve the latter paragraphs. I believe that the principle I need to use because this process is clearly time reversible is $$P_i^A = \frac{P_i}{\sum_{j \in A}P_j}$$ where $A$ is a truncated state space of the whole CTMC. I can envision truncating the space to just the neighboring checkerboard squares, but I'm still uncertain of what to even do with that information.
A clever approach for the first part is to consider an infinite board ruled off in $100 \times 100$ sections. When the particle moves off the edge of your board, it moves to a space of the same type (corner or edge) of a neighboring board, but on the infinite grid all the cells are equivalent as there are no reflections. This shows the chance of the particle being on any given cell on your board is the same. For the second part, you have $1278$ particles each with a chance of $10^{-4}$ to be in your given cell.