X and Y are two independent Brownian motions both equal to $1$ at time $0$. Consider the first time, T, that Y process hits $0$. What is the probability that $X_T>0$?
My intuition: I think that at any time $t>0$, $P(X_t>Y_t)=0.5$. The question asks about the random time T and at T we know $Y_T=0$. So $P(X_T>0) = P(X_T>Y_T) = 0.5$.
I would like to know if my intuition is correct? I appreciate it if anyone can provide a hint for a more formal solution. Thanks, guys!
Hint: Let $S$ be the first time that $X$ hits $0$.
If $S>T$, then certainly $X_T>0$.
If $S<T$, then $Z_r:= X_{S+r}$ is a Brownian motion independent of $Y$ with $Z_0=0$, so $P(Z_{T-S}>0\mid S<T)=\frac12$.
Furthermore, $P(T>S)=\frac12$.