Consider a 2D Brownian Motion $(X_1(t),X_2(t))$ starting at $(x_1,x_2) \in \mathbb{R}^2$. For every $s\geq0$, let $$\tau_s = \inf \left\{t \geq 0 \mid X_1(t) - x_1 > s \right\}\qquad Y_s = X_2(\tau_s)$$ The problem is to show that $Y=(Y_s)$ has the stationary, independence property.
Here's what I've tried:
Stationary: $Y_t - Y_s = X_2(\tau_t) - X_2(\tau_s) = X_2(\tau_t - \tau_s) = X_2(\tau_{t-s}) = Y_{t-s}$
This uses the fact that $\tau$ is stationary and the strong Markov property on $X_2$
Independence: Know $P(\tau_s < \infty) = 1$ for every $s$. By simple Markov property, I can reshift $X_2$ to a new process $\tilde{X}_2$ so that $\tilde{X}_2(0) = 0$. So by Strong Markov Property, $ E^{x_2}[Y_{\tau_s} \circ \theta_{\tau_s} | F_{\tau_s}] = E^{0}[Y_{\tau_s}] = 0$.
But I get stuck at the independence part here since I'm not sure if I'm using the strong Markov property correctly. I want to essentially argue that because I know $X_2$ has the independent increment properties, that $Y$ should have it as well but I'm struggling to make it work.