First hitting time for a drifted Brownian motion While the solution for a first hitting time for a drifted Brownian Motion is well known, I want to post a different question.
Take a continuous-time stochastic process $X_t$ and define the the stopping time $T_a=\min\{t\geq0\,|\,X_t^{max}\geq a\}$, where $X_t^{max}=\max_{\tau\in[0,t]}(X_{\tau})$ and $a\in\mathbb{R}$. The well known result I was mentioning before is given by $\mathbb{P}(X_t^{max}\geq a)=\mathbb{P}(T_a\leq t)$ which leads to an Inverse Gaussian distribution with respect to time $t$.
Now take the aforementioned upper-bound limit $a$ and take a lower bound limit $b$; consider the stopping time $T_b=\min\{t\geq0\,|\,X_t^{min}\leq b\}$ with $b<a$, which is the first time the process goes below the limit $b$. Now the question is: what is the probability $\mathbb{P}(X_t^{max}\geq a \,\,\cap \,\, X_t^{min}\leq b)$? I think the procedure is, on the same line of the previous problem, to consider it in terms of stopping times, that is $\mathbb{P}(T_a\leq t \,\,\cap \,\, T_b\leq t)$, but do you have any idea on how to derive the distribution? To put it in words, I am looking for the probability that the process, in a time period $[0,t]$, both goes up $a$ and shrinks beneath $b$.
 A: $\newcommand{\vpi}{\varphi}\newcommand\Z{\mathbb Z}$For real $t\ge0$, let $X_t:=mt+W_t$, where $m$ is a real number and $W_\cdot$ is a standard Brownian motion. So, $X_\cdot$ is a drifted Brownian motion starting at $0$ with the constant drift coefficient $m$.
For real $c$, let $T_c:=\min\{t\ge0\colon X_t=c\}$. The probability in question is
\begin{equation*}
    P_{t,a,b,m}:=P(T_a\le t,T_b\le t)=1+P(T_a>t,T_b>t)-P(T_a>t)-P(T_b>t), 
\end{equation*}
where $-\infty<a<0<b<\infty$ and $t>0$.
By rescaling, without loss of generality $t=1$, because $P_{t,a,b,m}=P_{1,\,a/\sqrt t,\,b\sqrt t,\,m\sqrt t}$. So, it is enough to find
\begin{equation*}
    P_{a,b,m}:=P_{1,a,b,m}=1+Q_{a,b}-Q_{a,\infty-}-Q_{-\infty+,b}, \tag{1}
\end{equation*}
where
\begin{equation*}
\begin{aligned}
    &Q_{a,b} \\
    &:=P(T_a>1,T_b>1) \\
    &=P(a<X_s<b\ \forall s\in[0,1]) \\
    &=P(a-ms<W_s<b-ms\ \forall s\in[0,1]) \\ 
    &=\int\limits_{a-m}^{b-m}P(a-ms<W_s<b-ms\ \forall s\in[0,1],W_1\in[x,x+dx])) \\ 
    &=\int\limits_{a-m}^{b-m}P(a-(m+x)s<W_s-sW_1<b-(m+x)s\ \forall s\in[0,1], \\
    &\, \qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad W_1\in[x,x+dx])) \\ 
    &=\int\limits_{a-m}^{b-m}P(a-(m+x)s<W_s-sW_1<b-(m+x)s\ \forall s\in[0,1]) \\
    &\, \qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad P(W_1\in[x,x+dx]) \\ 
    &=\int\limits_{a-m}^{b-m}P(a-(m+x)s<W_s-sW_1<b-(m+x)s\ \forall s\in[0,1], \\
    &\, \qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad W_1\in[m+x,m+x+dx]) \\ 
    &\, \qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\times\frac{P(W_1\in[x,x+dx])}{P(W_1\in[m+x,m+x+dx])} \\ 
    &=\int\limits_{a-m}^{b-m}P(a<W_s<b\ \forall s\in[0,1], \  W_1\in[m+x,m+x+dx])  
\,\frac{\vpi(x)}{\vpi(m+x)}, 
\end{aligned}
\end{equation*}
where $\vpi$ is the standard normal pdf; the third and second equalities from the end of the above multiline display follow by the independence of the Brownian bridge $(W_s-sW_1)_{s\in[0,1]}$ from $W_1$.
By multiple reflection (Lévy's triple law -- see e.g. Proposition 6.10.6 or Theorem 6.18),
\begin{equation*}
    P(a<W_s<b\ \forall s\in[0,1], \  W_1\in[m+x,m+x+dx])=q_{a,b}(m+x)\,dx,
\end{equation*}
where
\begin{equation*}
    q_{a,b}(z):=\sum_{k\in\Z}[\vpi(z+2kh)-\vpi(2b-z+2kh)],
\end{equation*}
\begin{equation*}
    h:=b-a,
\end{equation*}
and $z\in(a,b)$.
Taking now the the latter integral in the above multiline display, we get
\begin{equation*}
\begin{aligned}
    Q_{a,b}&=Q_{a,b;m}:=\sum_{k\in\Z}e^{-2 h k m} (\Phi (a+2 h k+h-m)-\Phi (a+2 h k-m)) \\ 
&   -\sum_{k\in\Z}  e^{2 m (a+h k+h)} (\Phi (a+2 h (k+1)+m)-\Phi (a+2 h k+h+m)),  
\end{aligned}
\tag{2}
\end{equation*}
where $\Phi$ is the standard normal cdf.
Now one can find the limits $P(T_a>1)=Q_{a,\infty-}$ and $P(T_b>1)=Q_{-\infty+,b}$ of $Q_{a,b}$ as $b\to\infty$ and as $a\to-\infty$, respectively. Alternatively, these limits can be found directly -- similarly to (but more simply than) $Q_{a,b}$.
Anyway, we get
\begin{equation*}
Q_{-\infty+,b}=P(T_b>1)=    \Phi (b-m)-e^{2 m b }\Phi (-b -m)  
\end{equation*}
and
\begin{equation*}
Q_{a,\infty-}=P(T_a>1)= \Phi (-a+m)-e^{2 m a }\Phi (a+m);   
\end{equation*}
cf. e.g. formula (10.13), p. 13.
Thus, by (1),
\begin{equation*}
\begin{aligned}
    P_{a,b;m}&=1+\sum_{k\in\Z}e^{-2 h k m} (\Phi (a+2 h k+h-m)-\Phi (a+2 h k-m)) \\ 
&   -\sum_{k\in\Z}  e^{2 m (a+h k+h)} (\Phi (a+2 h (k+1)+m)-\Phi (a+2 h k+h+m)) \\ 
&- \Phi (b -m)+e^{2 m b }\Phi (-b -m) \\ 
&-\Phi (-a +m)+e^{2 m a }\Phi (a +m). 
\end{aligned}
\end{equation*}

Here is the graph $\{(m,Q_{a,b;m})\colon |m|<5\}$ (above) and $\{(m,P_{a,b;m})\colon-3 < m < 6\}$ (below) for $a=-1$ and $b=2$:


We see that here a small enough positive drift, toward the boundary $b=2$ (which is farther away from the starting point $0$ than the boundary $a=-1$) helps the Brownian motion stay within the two boundaries till time $t=1$. Of course, this should be expected.
The two series in (2) converge very fast: for $a=-1$ and $b=2$, the maximum absolute error seems to be $<2\times10^{-12}$ if the two summations $\sum_{k\in\Z}$ there are each replaced by $\sum_{k=-1}^1$.
