4
$\begingroup$

I've come across the following resampling identity and was wondering if this is known since it seems rather natural. Take $X$ a two-sided Brownian motion conditioned to always stay below $1$. (So if you take $Y$ and $\tilde Y$ two independent Bessel(3) processes with initial condition $1$, one can set $X_t = 1-Y_t$ for $t \ge 0$ and $X_t = 1-\tilde Y_{-t}$ for $t \le 0$.) Let now $A = \{t\in\mathbb{R} : X_t \ge \sup_s X_s - 1\}$, pick a point $t^\ast$ uniformly at random in $A$ (according to normalised Lebesgue measure), and set $X^\ast_t = X_{t+t^\ast} - X_{t^\ast}$. The claim is that $X^\ast$ then has the same law as $X$.

Is this known and does it have a name? It's presumably known to someone like Pitman and there are various well known decomposition results about Bessel(3) processes and Brownian motion, but I wasn't able to find a reference for this one...

Edit: A similar identity is true if one gives $X$ a linear drift $-\gamma |t|$, but then $t^\ast$ has to be picked according to the (random) probability measure on $A$ with weight proportional to $e^{\gamma X_t}$.

Improve this question
$\endgroup$
5
  • $\begingroup$ Hi, I am confused: why is ${\rm Leb}(A)\in (0,\infty)$? $\endgroup$
    – ofer zeitouni
    Commented Jul 18 at 20:15
  • $\begingroup$ @oferzeitouni $X_t \to -\infty$ for $t \to \pm \infty$ so $A$ is bounded and $\sup_s X_s < 1$, so $A$ contains a neighbourhood of the origin. $\endgroup$
    – Martin Hairer
    Commented Jul 18 at 20:27
  • $\begingroup$ Thanks! I misread. Did you ask Pitman? $\endgroup$
    – ofer zeitouni
    Commented Jul 18 at 21:34
  • $\begingroup$ @oferzeitouni I hadn't, but just sent him an email. $\endgroup$
    – Martin Hairer
    Commented Jul 18 at 21:49
  • $\begingroup$ @oferzeitouni Surprisingly Pitman wasn't aware of it either... $\endgroup$
    – Martin Hairer
    Commented Jul 20 at 12:55

0

Reset to default

You must log in to answer this question.