Let $m: \mathbb R_+\to [0,1]$ be continuous and decreasing. Consider

$$X_t=1+bt+\int_0^t\frac{\sigma}{1+m(s)}dW_s,\quad \forall t\ge 0,$$

where $b, \sigma>0$ are given and $(W_t)_{t\ge 0}$ is a standard Brownian motion. Do we have an explicit (or semi-explicit) formula for the probability

$$\mathbb P[\inf_{0\le s\le t}X_s>0]?$$

Any answer and comments are highly appreciated.

  • 1
    $\begingroup$ I don't think such a formula can exist. $\endgroup$ Nov 25, 2021 at 21:52
  • $\begingroup$ the Ito integral is a time-changed Brownian motion, so then Xt is time-changed Brownian motion with a drift of different time though. From there maybe you can at least get some bounds. $\endgroup$ Nov 25, 2021 at 23:14
  • $\begingroup$ @ThomasKojar Thanks for the nice observation. I tried to do some calculus following your idea, and it needs the joint law of the stochastic integral and running mininum $\endgroup$
    – user128095
    Nov 26, 2021 at 6:17
  • $\begingroup$ @IosifPinelis the requested formula is a byproduct of mathoverflow.net/a/424512/64449 $\endgroup$ Aug 23, 2022 at 10:23
  • 1
    $\begingroup$ @NawafBou-Rabee : Thank you for your comment. $\endgroup$ Aug 23, 2022 at 11:48

1 Answer 1


I claim this is not an answer but some calculus following Thomas's observation. As I cannot write it as a comment, I write it down here. The stochastic integral

$$\left(M_t:=\int_0^t\frac{\sigma}{1+m(s)}dW_s\right)_{t\ge 0}$$

is a continuous martingale starting from zero. Hence, there exists some Brownian motion, denoted by $B$, s.t. $M_t=B_{\langle M\rangle_t}$ for all $t\ge 0$. Therefore, $X$ is equal in law to the following stochastic process :

$$Y_t=1+bt+W_{f(t)} \quad \mbox{with}\quad f(t):=\langle M\rangle_t=\int_0^t\frac{\sigma^2}{(1+m(s))^2}ds,\quad \forall t\ge 0.$$


$$\mathbb P[\inf_{0\le s\le t}X_s>0]=\mathbb P[\inf_{0\le s\le t}Y_s>0]=\mathbb P\big[\inf_{0\le u\le f(t)}Y_{f^{-1}(u)}>0\big]=\mathbb P\big[\inf_{0\le u\le f(t)}\big(1+bf^{-1}(u)+W_u\big)>0\big].$$

Let $\lambda:=(f^{-1})'$ (more precisely $\lambda(u)=\big(1+m(f^{-1}(u))\big)^2\big/\sigma^2$) and define the change of measure by

$$\frac{d\mathbb Q}{d\mathbb P}:=\exp\left(-b\int_0^{f(t)}\lambda(u)dW_u-\frac{b^2}{2}\int_0^{f(t)}\lambda(u)^2du\right).$$

Then $(Z_u:=bf^{-1}(u)+W_u)_{u\ge 0}$ is a Brownian motion under $\mathbb Q$. Further,

$$\mathbb P\big[\inf_{0\le u\le f(t)}\big(1+bf^{-1}(u)+W_u\big)>0\big]=\mathbb E^{\mathbb Q}\left[\frac{d\mathbb P}{d\mathbb Q}{\bf 1}_{\{\inf_{0\le u\le f(t)}(1+Z_u)>0\}}\right]=\mathbb E^{\mathbb Q}\left[\exp\left(b\int_0^{f(t)}\lambda(u)dZ_u-\frac{b^2}{2}\int_0^{f(t)}\lambda(u)^2du\right){\bf 1}_{\{\inf_{0\le u\le f(t)}Z_u>-1\}}\right].~~~~ (\ast)$$

Here $(\ast)$ provides an upper bound of the probability. To compute the probability, it remains to find out the joint law of

$$\left(\int_0^t \lambda(s)dZ_s, \inf_{0\le s\le t}Z_s\right).$$

Is this known in the literature?


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy