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.

Thanks in advance.

  • 1
    $\begingroup$ 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. $\endgroup$ – Nap D. Lover May 8 at 0:11
  • 2
    $\begingroup$ @NapD.Lover If $\mu>0$ and $a>0$ then of course there will be a last time because $X_t \to \infty$ as $t \to \infty$. $\endgroup$ – 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.

  • $\begingroup$ 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? $\endgroup$ – Zak Laberg May 9 at 12:41
  • 1
    $\begingroup$ @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. $\endgroup$ – Shalop May 10 at 1:04
  • $\begingroup$ 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.. $\endgroup$ – Zak Laberg May 10 at 8:40
  • $\begingroup$ 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 $t^{-3/2}$.. $\endgroup$ – Zak Laberg May 10 at 14:14
  • 1
    $\begingroup$ @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. $\endgroup$ – 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.


Your Answer

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

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