# Distribution of last exit time of Brownian motion with drift

If $$X_t = \mu t + \sigma W_t$$ with $$W_t$$ a Wiener process, I would like to know if the distribution for the last time $$X_t = a$$ is known - and if so, what it is. My googling has turned up a bunch of results for first exit time, first hitting time etc but these are not too useful to me. Any references/derivations are also highly appreciated, but I am a physicist without much training in probability, unfortunately.

• Brownian motion is recurrent in dimension $1$ so there is no last time $W_t=a$. I suspect this can be extended to BM with drift, $X_t$, but do not have a proof at the moment. – Nap D. Lover May 8 at 0:11
• @NapD.Lover If $\mu>0$ and $a>0$ then of course there will be a last time because $X_t \to \infty$ as $t \to \infty$. – Shalop May 9 at 8:05

Yes you can compute the distribution of the last hitting time.

Assume $$\mu,a>0$$ so the last hitting time is a.s. finite. Basically let $$B_t = tW_{1/t}$$. which is also a brownian motion. This time inversion allows us to "convert" the last hitting time into a first hitting time.

Specifically, if $$t_a = \sup\{t \ge 0 : \mu t + \sigma W_t \le a\}$$ then $$t_a^{-1} = \inf\{u \ge 0: au-\sigma B_u \ge \mu\}$$. And the latter is something whose distribution you know how to compute, because it is a first hitting time.

• Thank you, that makes sense. One follow-up question: as far as I can tell, $t_a^{-1}$ will follow an inverse gaussian distribution. But then $t_a$ does not, as the density picks up a factor 1/t^2 when I compute the distribution of $t_a$ from that of $t_a^{-1}$. But this goes against my numerical simulations, where it appears that $t_a$ does follow an inverse gaussian distribution. Do you know which is correct, e.g. should I expect $t_a$ to follow an IG or not? – Zak Laberg May 9 at 12:41
• @ZakLaberg yes $t_a$ should be inverse Gaussian. I don’t know why you expect $t_a^{-1}$ to be as well, as there’s no reason for that to be true. – Shalop May 10 at 1:04
• Now I'm confused; wikipedia at least states that first hitting times of Brownian motion are inverse gaussian distributed, so that was why I thought $t_a^{-1}$ should be IG.. – Zak Laberg May 10 at 8:40
• To elaborate a bit, I thought the distribution of $t_a^{-1}$ expressed in terms of $u$ would look like an IG, inserting $u = 1/t$ would give something different. But then switching to the distribution of $t_a$ I still pick up a factor 1/t^2 such that the distribution I get is something like $t^{-1/2}\cdot exp(\dots)$, whereas IG has $t^{-3/2}$.. – Zak Laberg May 10 at 14:14
• @ZakLaberg Sorry i misspoke. What I meant was that $t_a^{-1}$ will have an inverse Gaussian distribution (by definition of the inverse Gaussian distribution) however there is no reason to expect $t_a$ to have an inverse Gaussian distribution as well. I got confused by my own notation. – Shalop May 10 at 15:11

$$\def\si{\sigma}$$ $$\def\cF{\mathcal{F}}$$ $$\def\ol{\overline}$$

Here is an alternative derivation. Let $$B_t=(\mu/\si)W_{\si^2t/\mu^2}$$, $$Y_t=t+B_t$$, and $$b=a\mu/\si^2$$. Then $$X_t=a$$ if and only if $$Y_{\mu^2t/\si^2}=b$$. Hence, if $$T$$ is the last time $$X$$ hits $$a$$ and $$\tau$$ is the last time $$Y$$ hits $$b$$, then $$T=\si^2\tau/\mu^2$$.

By the Markov property, \begin{align} P(\tau < t) &= P(Y_t > b, Y_s > b \text{ for all s\ge t})\\ &= E[P(Y_t > b, Y_s > b \text{ for all s\ge t} \mid \cF_t)]\\ &= E[1_{\{Y_t > b\}}P^{Y_t}(\tau_b = \infty)], \end{align} where $$\tau_b$$ is the first hitting time of $$b$$. For $$y>b$$, $$P^y(\tau_b = \infty) = P(\tau_{b-y} = \infty) = 1 - e^{-2(y-b)}.$$ Thus, \begin{align} P(\tau < t) &= P(Y_t > b) - \frac1{\sqrt{2\pi t}}\int_b^\infty\exp\left({ -2(y - b) - \frac{(y - t)^2}{2t} }\right)\,dy\\ &= P(Y_t > b) - \frac1{\sqrt{2\pi t}}\int_b^\infty\exp\left({ 2b - \frac{(y + t)^2}{2t} }\right)\,dy\\ &= P(t + B_t > b) - e^{2b}P(-t + B_t > b)\\ &= \ol\Phi\left({\frac{b-t}{\sqrt t}}\right) - e^{2b}\ol\Phi\left({\frac{b+t}{\sqrt t}}\right), \end{align} where $$\ol\Phi=1-\Phi$$, and $$\Phi$$ is the standard normal distribution function. Differentiating and doing a little algebra gives the density of $$\tau$$, $$f_\tau(t) = \frac1{\sqrt{2\pi t}}\exp\left({-\frac{(b-t)^2}{2t}}\right),$$ and from here, we get the density of $$T$$, $$f_T(t) = \frac{\mu}{\si\sqrt{2\pi t}}\exp\left({ -\frac{(\si^2b-\mu^2t)^2}{2\mu^2\si^2t} }\right).$$ Not that you needed another derivation to tell you this, but there must be something wrong with your numerical simulations.