Suppose we have a random walker starting at $(0,0)$ at $t_0$ and the walker takes one step randomly per second. We have a square absorbing boundary going through the points $(2,2)$, $(2,-2)$, $(-2,2)$ and $(-2,-2)$, which means the game is over if the random walker lands on any of the points spanned by the square. How long does it take on average to end the game?
My first attempt:
I generalized the problem to a 1-d walker with 2 absorbing points at $x=n$ and $x=-n$. For this problem the average game time is:
$T=n^2$
(special case of Eq 9 on page 74 from "Probability and Random Processes", 3rd ed, by Grimmett and Stirzake).
However when trying to generalize this result to 2-d I failed. Does anyone maybe know how to do this?
My second attempt was successful but inelegant:
We view the lattice as a Markov chain this time and we setup the probability transition matrix in the following way:
$$ P = \begin{bmatrix} Q & R\\ 0 & I_r \end{bmatrix} $$ , where Q are the transion states. Apparently for a the hitting time for a Markov process is:
$T = (Q-I)^{-1}*\vec{1}$
I have checked the 1d case and it works but doing the 2d case will require a $25\times 25$ matrix. I know it can be done but I was wondering if anyone has a better solution in mind? Perhaps a generalization of my first attempt?
Let $M_{ij}$ be the expected absorption time starting at $(i,j)$. Then by conditioning on the first step and using symmetry, $$M_{00}=1+M_{10}$$ $$M_{10}=1+M_{00}/4+M_{11}/2$$ $$M_{11}=1+M_{10}/2 \,.$$ The solution is $M_{00}=9/2,\;\, M_{10}=7/2,\;\, M_{11}=11/4 .$