Stopping time problem - Show that T is bounded

Question

Let $a< 0 < b$ and $W_t$ is Brownian motion

$T_a$=inf{$t\ge$0|$W_t\le a$}

$T_b$=inf{$t\ge$0|$W_t\ge b$}

T=min{$T_a$,$T_b$}

$1)$ Show that $T <\infty$

My attempt :

P(T<$\infty$)=P($T_a ,\infty$ or $T_b < \infty$ )

= P($T_a$<$\infty$)+P(T_b)<\infty) - P(T_a<\infty \cap T_b<\infty)$

I stuck here , it seems $P(T_a<\infty$) and $P(T_b<\infty$) bounded somehow but I dont know how to put it.

$2)$ Show that P($T_a<T_b) = P(W_T=a)=b/(b-a)$

This one I actually have very little clue, it seems when $W_T=a$ ,T= min( $T_a;T_b$ ) = a.

When $W_T = a$ , it can't be b, so $W_T$ should have reach $a$ before $b$, so $T=T_a= a$.

Answer 1

To show that $T$ is finite almost surely, note that $$ [T\gt t]\subset[a\lt W_t\lt b], $$ and that, for every Borel subset $B$, since the density of $W_t$ is uniformly bounded by you-know-what, $$ P[W_t\in B]\leqslant\frac{\mathrm{Leb}(B)}{\sqrt{2\pi t}}. $$ Thus, for every $t$, $$ P[T=+\infty]\leqslant P[T\gt t]\leqslant\frac{b-a}{\sqrt{2\pi t}}, $$ that is, $P[T=+\infty]=0$.