= Solution
For $t>0$, write $\Phi$ for the <standard normal distribution function>. Since the <Brownian running maximum> is nonnegative, $a\leq0$ gives simply $\mathbb P(B_t\leq b)=\Phi(b/\sqrt t)$.
For $a>0$ and $b<a$, reflect the <Brownian motion> after its first hit of $a$. This is a <stopping time>, and the <Strong Markov property> together with symmetry of Brownian increments shows that reflection preserves its law. On paths that hit $a$, it sends the endpoint $B_t$ to $2a-B_t$. Thus the event with endpoint at most $b$ maps to endpoints at least $2a-b$, all of which necessarily hit $a$. This proves
$$
\mathbb P(M_t\geq a,B_t\leq b)=\mathbb P(B_t\geq2a-b)=\Phi((b-2a)/\sqrt t).
$$
The same reflection gives $\mathbb P(M_t\geq a)=2(1-\Phi(a/\sqrt t))$. If $b\geq a$, every path with $B_t>b$ has hit $a$, so subtracting that endpoint tail gives the complete answer:
$$
\boxed{\mathbb P(M_t\geq a,B_t\leq b)=
\begin{cases}
\Phi(b/\sqrt t),&a\leq0,\\
\Phi((b-2a)/\sqrt t),&a>0,\ b<a,\\
1-2\Phi(a/\sqrt t)+\Phi(b/\sqrt t),&a>0,\ b\geq a.
\end{cases}}
$$
The expressions agree at $b=a$. This is the <joint distribution of Brownian motion and its running maximum>. In particular $M_t$ has the distribution of $|B_t|$ and has no atoms for $t>0$. If $t=0$, the pair is $(0,0)$ deterministically, so the requested probability is $\mathbf1_{\{a\leq0,\ b\geq0\}}$.
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