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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 $T_a=\min\{t\geq0\mid X_t^{\max}\geq a\}$, 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 $T_b=\min\{t\geq0\mid X_t^{\min}\leq b\}$ 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 $b$.

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$\newcommand{\vpi}{\varphi}\newcommand\Z{\mathbb Z}$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 $b=2$:

enter image description here

enter image description here

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 $\sum_{k=-1}^1$.

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  • $\begingroup$ Dear Iosif, thank you very much for your reply and the reference you posted. I have set up simulations (on R) to check the accuracy of the predictions but they are not accurate (they are -instead- for the first passage times that i previously computed). Would you be available to discuss it in the chat so that i can close the question? $\endgroup$
    – DreDev
    Jan 3, 2022 at 21:02
  • $\begingroup$ @DreDev : We can try to discuss this. However, I do not know R. I now suspect that the expression for $\text{ss}_y$ on p. 641 of the handbook is incorrect, with the factor $(-1)^k$ missing in the $k$th summand -- cf. the formula for $\text{cc}_y$ on p. 641 there and the expression in terms of $\text{cc}_t$ in formula 3.0.2 on p. 212 of the handbook. $\endgroup$ Jan 3, 2022 at 22:03
  • $\begingroup$ @DreDev : I am going to re-derive the distribution by hand. $\endgroup$ Jan 3, 2022 at 23:01
  • $\begingroup$ Thank you a lot, Iosif! However, with your allowance, I can write you an email to the address specified in your mathoverflow profile $\endgroup$
    – DreDev
    Jan 3, 2022 at 23:11
  • $\begingroup$ @DreDev : Sure, you are welcome to do so. $\endgroup$ Jan 4, 2022 at 0:42

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