TITLE: Random walk on cubic lattice
QUESTION [7 upvotes]: Suppose at every point of the cubic grid in n dimensions is a particle, and at every timestep every particle moves at random to one of its 2n neighbours. As time goes to infinity, what is the probability distribution for the number of particles at a chosen point?
And for hexagonal grid in 2d?

REPLY [2 votes]: Let $X_p(T)$ be the number of particles at point $p$ after $T$ time steps.
Let $Y_{pq}(T)$ be the indicator random variable for the event  that the particle starting at $p$ ends up at $q$ after $T$ time steps.  Then $X_p(t) = \sum_q Y_{qp}(t)$.  These are independent, and since by symmetry 
$E[Y_{pq}(t)] = E[Y_{qp}(t)]$ we have $E[X_p(t)] = E \left[\sum_q Y_{qp}(t)\right] = 1$.  Also for each $q$ we have $E[Y_{qp}(t)] \to 0$ as $t \to \infty$.   Then the limiting distribution should be Poisson with mean $1$.
I believe this can be proven using Le Cam's theorem.
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