I was just wondering about $2D$ random walks when I got the idea of a position dependent $2D$ random walk:-
A man is initially at $(x,y)$ and can move in a line parallel to the X and Y-axis only. The probability that the man takes a step in the X-direction is $\frac {|x|}{|x|+|y|}$ and the probability that the man takes a step along the Y-direction is $\frac {|y|}{|x|+|y|}$. Given that the man takes a step in the X-direction the probability that he takes a step in $+$ve x-direction is $\frac {|x|}{1+|x|}$ and that he takes a step along the negative X-direction is $\frac 1{1+|x|}$. Given that he takes a step in the y-direction probability that he takes a step along the positive Y-direction is $\frac 1{1+|y|}$ and probability that he takes a step in the negative Y-direction is $\frac {|y|}{1+|y|}$. Find the probability that his motion along a direction stops after $n$ steps. Also find probability that his motion stops along a direction at some point of time. (Consider that he doesn't reach the origin at any point of time.) I have no idea about how to do the question, I think some kind of recurrence relation would help but which one?
Here's something I've been thinking about, there may be some errors though so you probably want to check to see if it all makes sense.
This guy, Rick Durrett (p164 remark), uses this cool method in his book to make the x and y steps mutually independent, meaning the x and y can move at the same time, and you can transform this walk into a conventional one. Instead of each step being $(x \pm 1,y)$ or $(x, y \pm 1)$ , he rotates the graph 45 degrees and multiplies each step by $\sqrt{2}$, so that each step is now $(x \pm 1, y \pm 1)$ .
Using this, you first take the point $(x,y)$ the man originally wanted to walk to, rotate it 45 degrees (counterclockwise), and make it $\sqrt{2}$ times farther from the origin. This can be done simply by taking the vector sum between this point and the point rotated 90 degrees counterclockwise, since the "parallelogram" formed during the sum is always a square, so the diagonal will be rotated 45 degrees and $\sqrt{2}$ times longer.
Note that when rotating a point by 90 degrees gives
$x' = -y$ and $y' = x$.
You can try this out for each of the four quadrants.
Now take the vector sum of $(x,y) + (-y,x)$ , giving the resultant vector $(x-y, x+y)$ . And from this we get the new, transformed point to be the same: $(x-y, x+y)$ .
All we have to find now is the probability that a random walk on the x-axis will end up at $x-y$ and the probability that a random walk on the y-axis will end up at $x+y$.
You can do some algebra and see that for a one-dimensional random walk to end up at point $q$ in $n$ steps, the probability is
$\dbinom{n}{\frac{n+q}{2}}p_1^{\frac{n+q}{2}}p_2^{\frac{n-q}{2}}$
(this also means that for some values of $n$ the man can never reach $(x,y)$ )
The same is true for both the x- and y-axis random walks, and we can replace $q$ with its actual value in the problem, giving the final probability (you make the distinction between x and y the coordinates and x and y the probabilities difficult, so I'm going to make the probability of moving in the x direction p, and in the y direction q):
$\dbinom{n}{\frac{n+x-y}{2}}\dbinom{n}{\frac{n+x+y}{2}}p^{\frac{n+x-y}{2}}(1-p)^{\frac{n-x+y}{2}}q^{\frac{n+x+y}{2}}(1-q)^{\frac{n-x-y}{2}}$
or
$\dfrac{(n!)^2 p^{\frac{n+x-y}{2}}(1-p)^{\frac{n-x+y}{2}}q^{\frac{n+x+y}{2}}(1-q)^{\frac{n-x-y}{2}}}{\Big(\dfrac{n+x-y}{2}\Big)!\Big(\dfrac{n-x+y}{2}\Big)!\Big(\dfrac{n+x+y}{2}\Big)!\Big(\dfrac{n-x-y}{2}\Big)!}$
Hopefully that all makes sense, ask questions if its unclear and point out any potential holes or flaws!