$\textbf{Background:}$ Consider a particle in $\mathbb{N}^2$ starting at a point $(x,y)$ that can only move one step at a time along the lattice. The particle moves each up or down in continuous time with rate $y$, and each left or right with rate $x$. It will therefore be moving faster, the further away from the origin that it is. Furthermore, once it hits an axis, it cannot leave that axis, so the axes act as a kind of absorbing boundary. I am wondering about the expected time (or, in an ideal world, the distribution) for the particle to hit either axis for the first time.

$\textbf{Question:}$ Write $v(x,y)$ for the expected time for the particle to hit either axis given it began at $(x,y)$. What is $v(x,y)$? We can write the recurrence equation $$ v(x,y) = \frac{1}{2(x+y)} + \frac{x}{2(x+y)} (v(x-1,y) + v(x+1,y)) + \frac{y}{2(x+y)} (v(x,y-1) + v(x,y+1))$$ for $x,y>0$, and with boundary conditions $v(x,0) = v(0,y) = 0 \; \; \; \forall \; x,y>0$.

In particular, if we let $x=y$, the equation can be written as $$ \frac{1}{2x} + (v(x+1,x) - v(x,x)) = (v(x,x) - v(x-1,x))$$ which feels, to me, like it might be a good place to start.

$\textbf{What I've got so far}:$

- A poor upper bound on $v(x,y)$ that shows it's finite.
- The particle must hit one of the axes eventually, and can't hit both at the same time. The probability that the particle hits the x-axis before the y-axis, starting at the point $(x,y)$, is $x/(x+y)$ .
- If one defines the same process in 1d, the expected time to hit the origin given a start point of $x$ is clearly infinite. But the expected hitting time of either $0$ or $A$, starting at $0<x<A$, is $x(H_A - H_x)$, where $H_m$ is the $m$-th harmonic number.