# Random walk question on 2D grid, probability of vertical line vs horizontal line hit

Hi, I have a problem

(1) where I need to compute the ratio of probabilities of hitting and stopping at a positive vertical barrier x vs hitting and stopping at a negative horizontal barrier y after starting from (0,0).

I feel that by symmetry, the answer to this would be the same as

(2) The probability of hitting -y vs hitting +x, horizontal lines in 2d grid,

which looks like being same as

(3) The probability of hitting -y vs x on a real line.

Can someone please tell me if my 1->2 assumption or 2->3 assumption is wrong. In which case, could someone please tell me how to proceed with the solution to 1.

However, if my assumption is right, can someone tell me how to proceed to prove it. Also, what would be a way to solve the case when both x and y are positive.

I shall be grateful for a response/hint/link.

Thanks.

EDIT: I really wish and hope that someone would answer it. Is there some way to transfer credit from superuser.com/stackoverflow.com here and place a bounty on it? I shall be really grateful if someone could give me an answer. Thanks.

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Sorry, just read the faq and realized that this is not an appropriate place for kid questions. Sorry for the trouble and thanks for the help. – user10401 Nov 1 '10 at 9:15

## 1 Answer

Problems 2 and 3 are equivalent by ignoring horizontal steps.

Problems 1 and 2 are not equivalent. I believe the probabilities are known, but I don't know them. However, if the probability of hitting the vertical barrier were really $y/(x+y)$ as in problem 2, then it would not be a martingale. So, whatever symmetry argument you wanted was incorrect. Specifically, consider the function $y/(x+y)$ on the lattice points $(x,y)$ in the first quadrant. The value at $(1,2)$ is $2/3$ but the average of the values at its neighbors is $11/16$.

For a Brownian motion, the angle from the origin is a martingale so you can use problem 3 to determine the answer to problem 1, $\frac{\arctan(y/x)}{\pi/2}$.

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Thanks very much for the answer. – user10401 Nov 1 '10 at 9:15