I know the basics of (discrete time) martingales, and I'd appreciate any help and suggestions on how to prove the following using martingale techniques.
Let $Z_n$, $n\ge 0$ be a random walk on the nonnegative quadrant (i.e. $Z_n=(Z^{(1)}_n,Z^{(2)}_n)$ with $Z^{(1)}_n,Z^{(2)}_n\ge 0$) and $$ P(Z_{n+1}=Z_n+(1,0)|\mathcal{F_n})=P(Z_{n+1}=Z_n+(0,1)|\mathcal{F_n})=1/2. $$ Now let $\gamma$ be any path connecting neighboring lattice points and extending form the vertical axis to the horizontal axis. Let $X, Y$ be the number of steps (to the North and East, respectively) before hitting $\gamma$, and $S$ be the total number of steps until $\gamma$ is reached.
My objective is to show, using martingale techniques, that $$ EX=EY=ES/2. $$
EDIT
I followed the helpful hints below, and have solved it, assuming, additionaly, the following:
If $\gamma$ intersects the $x,y$ axes at $(0,b), (a,0)$, respectively, then $Z^{(1)}_0\le a, Z^{(2)}_n \le b$.
In this case, how can I rigorously prove that $S$ is bounded? It is intuitively clear that, if the path $\gamma$ is included in $[0,K]^2$, then $S\le 2K$. Or, as @Did pointed out below, there's the sharper bound of $\max\{x+y:(x,y)\in\gamma\}$. But, still, I'm looking for a rigorous proof that the range of $\{X_n\}$ indeed intersects $\gamma$.