$\textbf{Background:}$ 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.



A 2d recurrence equation representing a step-free continuous-time Markov process on N^2 with frequency-dependent rates
Some experimenting reveals that the hitting time $t\_1(x)$ of the 1D process started at $x\in\mathbb{N}$ (in the notation of OP's comment) has a simple cdf,
$$\mathbb{P}(t\_1(x) < t ) = \left(\frac{t}{t+1}\right)^x.$$
I don't know an easy combinatorial proof, but one may check explicitly that it solves
$$\mathbb{P}(t\_1(x) < t ) = \int\_0^t \mathrm{d}s\, 2x\,e^{-2x s} \tfrac12\left[\mathbb{P}(t\_1(x+1) < t-s )+\mathbb{P}(t\_1(x-1) < t-s )\right].$$


**Edit:** The fact that $\mathbb{P}(t\_1(x) < t ) = \mathbb{P}(t\_1(1) < t )^x$ can easily be understood in the light of [James Martin's answer](https://mathoverflow.net/a/315733). For the whole population to die out before time $t$, the descendants of each of the $x$ initial individuals have to die out before time $t$. 


It follows immediately that
$$\mathbb{P}(\min(t\_1(x),t\_2(y))>t) = \left[1-\left(\frac{t}{t+1}\right)^{x}\right]\left[1-\left(\frac{t}{t+1}\right)^{y}\right].$$
Hence
\begin{align\*}v(x,y) &= \mathbb{E}(\min(t\_1(x),t\_2(y))) \\
&= \int\_0^\infty \mathrm{d}t\left[1-\left(\frac{t}{t+1}\right)^{x}\right]\left[1-\left(\frac{t}{t+1}\right)^{y}\right]\\
&=1 - x\,H\_x-y\, H\_y+(x+y)H\_{x+y-1} < \infty,
\end{align\*}
where $H\_n$ is the $n$th harmonic number.


