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?


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 oppinion 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 $$\mathbb{P}(T_1<T_4<T_{-1}) = \frac{-1}{-1-4} = \frac{1}{5}.$$


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