$P(T_2=T_{-3}), P(T_1<T_4<T_{-1})$ and $P(T_3<2)-P(T_{-3}<2)$

Question

Let $W(t)$ be a Brownian motion and $T_x=\inf\{t:W(t)=x\}$.

I need to calculate $P(T_2=T_{-3}), P(T_1<T_4<T_{-1})$ and $P(T_3<2)-P(T_{-3}<2)$.

I'm not sure if I understand these correctly. I assume the first and the last one are $0$?

The second one I guess I can write as $$P(T_1<T_4<T_{-1})=\\P(T_1=\min(T_1,T_4,T_{-1}))\cdot P(T_4<T_{-1}|T_1=\min(T_1,T_4,T_{-1}))=\\\frac12\cdot P(T_3<T_{-2})=\\\frac12\frac2{2+3}=\frac15$$, but I'm not really confident about this either.

Can anyone confirm if these are correct?

Answer 1

The first and the third probability are zero, yes.

Your computations for the second probability are a bit difficult to follow, without any explanation. The final result is correct. In my opinion it's a bit clearer to argue as follows:

It follows directly from the definition of $T_x$ and the continuity of the sample paths of Brownian motion that $T_1<T_4$ almost surely. Hence, $$\mathbb{P}(T_1<T_4<T_{-1}) = \mathbb{P}(T_4<T_{-1}).$$ Applying the formula $$\mathbb{P}(T_x<T_y) = \frac{y}{y-x}, \qquad y<0<x$$ (...which is a consequence of the fact that $\mathbb{E}(W_{\tau})=0$ for $\tau:=\inf\{T_x,T_y\}$...) we get $$\mathbb{P}(T_1<T_4<T_{-1}) = \frac{-1}{-1-4} = \frac{1}{5}.$$