We have a random walk in 2D. In this many dimensions, we return to the origin with probability $1$. However, the number of steps it takes to do so seems to vary greatly from computer simulations I've ran.
Thus, I'm curious about the distribution concerning the number of steps required for one to return to the origin in a 2D random walk. An explicit probability for each $n$ steps would be fantastic, but an expected value will do, too. I'm just curious about the topic in general.
For what it's worth, I have looked into the 1D case, which seems a quite bit easier.
Best Answer
For a symmetric random walk in two dimensions the probability that you are back at the origin after the $2n$th step is:
$$p_0(2n) = \left(\frac{1}{4}\right)^{2n} \sum_{m=0}^{n}\frac{(2n)!}{(m!)^2[(n-m)!]^2} = \left(\frac{1}{2}\right)^{4n} \:{2n\choose n}^{\!\!2}$$