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 $b$.


1 Answer 1


$\newcommand{\vpi}{\varphi}\newcommand\Z{\mathbb Z}$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 $\sum_{k=-1}^1$.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.