What is the distribution of revisits to the starting point for an asymmetric simple random walk on the integers?
The answer is geometric distribution which makes sense but I have a hard problem understanding that.
2026-03-25 06:01:43.1774418503
Simple Assymetric Random Walk Confusion
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The asymmetric random walk is transient, so there is some fixed probability $q>0$ that you will never return to your starting point. So the probability of $0$ revisits is $q$.
If you do return to your starting point (probability $1-q$), you can think of the walk as restarting from the same starting point, so there is the same probability $q$ that you will not return to the starting point again after this. So the probability of exactly $1$ revisit is $q(1-q)$.
Similarly, you get probability $q(1-q)^{k-1}$ of $k$ revisits.
This is just a geometric random variable where "success" is never returning to the starting point again, and you are looking for the number of "failures" - number of times you do return - before the first success.