# 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 $$b, which is the first time the process goes below the limit $$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$$.


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 $$-\infty and $$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*}$$ where \begin{equation*} \begin{aligned} &Q_{a,b} \\ &:=P(T_a>1,T_b>1) \\ &=P(a where $$\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), $$\begin{equation*} P(a where $$\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 $$\sum_{k=-1}^1$$.

• 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? Jan 3, 2022 at 21:02
• @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. Jan 3, 2022 at 22:03
• @DreDev : I am going to re-derive the distribution by hand. Jan 3, 2022 at 23:01
• Thank you a lot, Iosif! However, with your allowance, I can write you an email to the address specified in your mathoverflow profile Jan 3, 2022 at 23:11
• @DreDev : Sure, you are welcome to do so. Jan 4, 2022 at 0:42