Fix $a>0$ and let $(A_t^-)_{t\geq0}$ and $(A_t^+)_{t\geq 0}$ be two independent one-dimensional Brownian motions, starting from $-a$ and $a$, respectively. Set $$T=\inf\{t\geq0:A_t^-=A_t^+\}.$$ I am tasked to show that $T<\infty$ almost surely and to find a density function $T$. I really don't know how to do this – the hint given is to consider an orthogonal transformation of $(A_t^-,A_t^+)_{t\geq0}$ which would also be a Brownian motion, but I'm unsure how to use this. Any help anyone could give would be great, thanks!
Hitting time for 2-dimensional Brownian motion
brownian motionprobability theorystochastic-calculusstochastic-processesstopping-times
Best Answer
We may write the given Brownian motions as: $$\begin{align*} A_t^{-} &= -a + B_t^{(1)} \\ A_t^{+} &= a + B_t^{(2)} \end{align*}$$ where $B_t^{(1)}$ and $B_t^{(2)}$ are independent standard Brownian motions (i.e. they start at zero). Then, $A_t^{-} = A_t^{-}$ if and only if $\frac{1}{2}(B_t^{(1)} - B_t^{(2)}) = a$. Notice that $W_t = \frac{1}{2}(B_t^{(1)} - B_t^{(2)})$ is a standard Brownian motion, so that we may re-write the given stopping time as $$T = T_a^{W} = \inf\{ t \geq 0 \, : \, W_t = a\}$$ From then on, you may conclude that $T<\infty$ almost surely by, for instance, appealing to the fact that $\limsup_{t\to \infty} W_t = \infty$.
For the density of this stopping time, you may consult this other MSE post, or maybe this one.