In a simple random walk on $\mathbb Z ^2$, starting in the origin, I'm trying to find out the probability of visiting the point (1,0) before the walk returns to the origin. Means, if $O$ is the origin point, and $p$ is the point (1,0), The question is what is the probability of $\mathbb P _O(\tau_p<\tau_O^+)$, when $\tau$ is the stopping time of a point, and $\tau^+$ is the stopping time for $t>0$. I tried to apply mirroring strategies simply by counting symmetric paths etc., but it doesn't work. I also tried to project the walk on each axis and analyze it separately (playing separately with the axes made me guess that the answer should be 0.5, but I still can't see how to prove it rigorously, and I'm not sure it's indeed the right answer).
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Projecting the walk to $\Bbb Z$ doesn't work at all, as that just checks whether the first step is left or right. There are many ways to avoid $(1,0)$ but hit $1$ in the projection, and there are ways to get to $(1,0)$ without passing the origin even if the first step is to the left, but you can't get to $1$ without hitting $0$. – Arthur Sep 23 '20 at 07:42
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Maybe the description was confusing - I'm aware of the problems in just projecting the walk on a spesific axis. I tried something a bit more complicated, but it also failed, so I didn't write the entire idea. – S. R Sep 23 '20 at 07:53
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1The question is not stated well: if you are only considering closed walks that return to the origin, the probablility if returning from (1,0) is the same as the probability of returning from (0,1), (-1,0) or (0,-1), so it's 1/4 by symmetry. If multiple returns to the origin are allowed some sort of geometric series for the n'th return to the origin results. If the question is not for closed walks, but asking for a probability of a length-n walk to step (1,0) (exactly once, at least once) , things get more complicated. – R. J. Mathar Sep 23 '20 at 11:13
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@R.J.Mathar Why it's not stated well? the walk is recurrent So it must reach the origin somewhen. The question is what is the probability that the walk will reach the points (1,0) before the first return to the origin. Note that the walk can start by going to (1,0) and return to the origin through the point (-1,0) for example. So anyway it's not simply 1/4 as you described – S. R Sep 23 '20 at 11:22
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@R.J.Mathar I edited the question. Is it clear now? – S. R Sep 23 '20 at 11:32
1 Answers
The generating function for the number of RW's that return to the origin after $N$ steps is given by $a0 = 1+4z^2+36z^4+... ={}_2F_1(1/2,1/2;1;(4z)^2)$, see A002894. The generating function of RW's that are at point (1,0) after $N$ steps is $a1=z+9z^3+100z^5+\cdots = z {}_3F_2(1,3/2,3/2;2,2;(4z)^2)$, see A060150. Plug these two into the expression $\frac{1}{2(a0-a1)}-\frac{1}{2(a0+a1)}$ by Rubin and Weiss. Insert $z=1/4$ into the generating functions to obtain probabilities, which means divide the number of walks by $4^N$ to weigh them with the $1/4$ probability of the 2D square lattice. Then $a0+a1\to \infty$, and with a little bit of Gamma-function manipulation one sees that $a0-a1\to 1$ as $z\to 1/4$. This is the conjectured result, i.e., probability of $1/2$. The full analysis is in A337869.
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