7
$\begingroup$

I am interested in the distribution of the $\text{argmax}_{t \in [0,1]} \{B(t) + f(t)\}$, where $B$ is a Brownian motion (or Brownian bridge) and $f:[0,1] \to \mathbb{R}$ is a continuous function. There are good results when the function $f$ is constant, linear or parabolic. However, I am particular interested in the case when $f$ is a sample path of another Brownian motion.

Consider the following problem: Let $B(t)$ and $\tilde B,(t)$, $t \in [0,1]$ be two independent standard Brownian motions (or Brownian bridges). Define $I := \text{argmax}_{t \in [0,1]} \{ B(t) + \tilde B(t) \}$. Consider the following two questions about the conditional distribution of $I$ given $\tilde B$:

(1) Is this conditional distribution absolutely continuous with respect to the Lebesgue measure on $[0,1]$, for almost every sample path of $\tilde B$? In another words, is there a regular conditional density function $g( \cdot \mid \tilde B)$ ?

(2) If the conditional density function $g( \cdot \mid \tilde B)$ exists, do we have $E[ \int_0^1 g^2(t\mid \tilde B) dt ] = \infty$, where the expectation $E[\cdot]$ is with respect to the law of $\tilde B$ ?

$\endgroup$
  • $\begingroup$ The answer to your question is that it's almost surely singular. The basic idea is to view the Brownian motion locally at the point where it achieves its local maximum. By the Williams' decomposition theorem, its distribution in the neighborhood of such a point is, up to absolute continuity, that of a Bessel(3) process. From this point of view your question is equivalent to the following: show that for a Brownian motion $B$ and an independent Bessel(3) process $U$, both starting from $0$, the distribution of $(B + U) / \sqrt{2}$ is singular to that of the Brownian motion... $\endgroup$ – Alexander Shamov Feb 16 '15 at 22:38
  • $\begingroup$ ... And this can be done using various invariants of the equivalence class of the distribution of Bessel(3). For instance, the stochastic integrals $\intop f(t) d (U_t - \mathsf{E} U_t)$ converge for all deterministic $f \in L^2$, so also the stochastic integrals $\intop f(t) d ((B_t + U_t) / \sqrt 2 - \mathsf{E} U_t / \sqrt 2)$ converge for such $f$. For the Brownian motion (and any other semimartingale with distribution equivalent to that of a Brownian motion) such integrals would converge without the $\mathsf{E} U_t$ recentering. Note that $\mathsf{E} U_t = \mathsf{const} \cdot t^{1/2}$... $\endgroup$ – Alexander Shamov Feb 16 '15 at 22:41
  • $\begingroup$ ... which is not in the $W^{1,2}$ Sobolev space, so $\intop f d \mathsf{E} U$ doesn't converge for some $f \in L^2$. So the distributions of $B$ and $(B + U) / \sqrt 2$ must be singular to each other: one needs the recentering and the other one doesn't. $\endgroup$ – Alexander Shamov Feb 16 '15 at 22:42
  • $\begingroup$ I'll try to cook up a readable answer later... $\endgroup$ – Alexander Shamov Feb 16 '15 at 22:46
  • $\begingroup$ @AlexanderShamov Thanks for the explanation! I am not familiar with the Bessel(3) process, but I will look it up and try to understand your argument. It seems to me that your argument shows that the processes $B$ and $(B+U)/\sqrt{2}$ are singular to each other, however, could you explain what this implies for the location of the maximum? $\endgroup$ – X. Wang Feb 18 '15 at 17:23
2
$\begingroup$

They are almost surely singular to each other.

First of all, here is a basic observation from measure theory that is useful in the proof. Consider two random measures, $M_1$ and $M_2$ on the same space - say, $[0,1]$, depending on some randomness $\omega \in \Omega$. With these measures one associates measures $\mathsf{Q}_1$, $\mathsf{Q}_2$ on $\Omega \times [0,1]$, defined as

$$\mathsf{Q}_i(d\omega, dt) := \mathsf{P}(d\omega) M_i(\omega, dt)$$

In other words, $M_i$ can be obtained from $\mathsf{Q}_i$ by disintegration w.r.t. the $t \in [0,1]$ coordinate. On the other hand, one can do the disintegration w.r.t. the other coordinate, i.e. $\omega$, and define the corresponding functions $t \mapsto \mathsf{Q}_i(t)$ with values in measures on $\Omega$, as follows:

$$\mathsf{Q}_i(d\omega, dt) = \mathsf{Q}_i(t, d\omega) \mu(dt),$$

where $\mu$ is a measure on $[0,1]$, such that $\mathsf{E} M_i \ll \mu$. The basic observation that I'm talking about is this:

The following are equivalent:

  1. $M_1 \ll M_2$ almost surely
  2. $\mathsf{Q}_1 \ll \mathsf{Q}_2$ (as measures on $\Omega \times [0,1]$)
  3. $\mathsf{Q}_1(t) \ll \mathsf{Q}_2(t)$ (as measures on $\Omega$) for $\mu$-almost all $t$

Now we turn back to our problem. Consider the random measure $M := \delta_I$, and the measure $M_1 := \mathsf{E} \left[ M \middle| \tilde B \right]$. $M_1$ is exactly the conditional distribution of the maximum of $B + \tilde B$ given a realization of $\tilde B$. I'm proving that $M_1$ is almost surely singular, that is, singular to the deterministic measure $M_2 := \mathrm{Lebesgue}$. By the above observation, this is equivalent to saying that the distribution of $(B, \tilde B)$ according to the measure $\mathsf{Q}_1(t, d\omega)$ is singular to $\mathsf{Q}_2(t) = \mathsf{P}$, for Lebesgue-almost all $t$.

Now we describe the distribution of $(B, \tilde B)$ according to $\mathsf{Q}_1(t)$. It turns out to be easier to describe the distribution according to $\mathsf{Q}(t)$ first --- that is, the one that corresponds to $M = \delta_I$. This description (and a little bit more) is given by the Williams' path decomposition theorem (e.g. Theorem 4.9 in Revuz-Yor). It follows from there that the distribution of the process $\frac{1}{\sqrt 2}(B(t) + \tilde B(t) - B(t + \cdot) - \tilde B(t + \cdot))$ according to $\mathsf{Q}(t)$ is absolutely continuous w.r.t. that of the Bessel(3) process started from $0$. $\frac{1}{\sqrt 2} (B - \tilde B)$ is still the Brownian motion according to $\mathsf{Q}_1(t)$, since it was $\mathsf{P}$-independent of $\frac{1}{\sqrt 2} (B + \tilde B)$ and $M_1$ is measurable w.r.t. $\frac{1}{\sqrt 2}(B + \tilde B)$.

Now, another purely measure-theoretic observation: the fact that $M_1 = \mathsf{E} \left[ M \middle| \tilde B \right]$ corresponds, in terms of the $\mathsf{Q}$ measures, to the following: the distribution of $\tilde B$ according to $\mathsf{Q}_1(t)$ is the same as that according to $\mathsf{Q}(t)$. But from the paragraph above it follows that $\tilde B(t + \cdot) - \tilde B(\cdot)$, according to $\mathsf{Q}_1(t)$, can be decomposed as $\frac{1}{\sqrt 2}$ times the sum of a Brownian motion and an independent Bessel(3) process (up to absolute continuity).

It remains to prove that if $B$ is a Brownian motion and $U$ is an independent Bessel(3) process starting from $0$ then the distribution of $\frac{1}{\sqrt 2}(B + U)$ is singular to that of the Brownian motion. There are many ways to do that; the simplest one that I know is the following. The Bessel(3) process is essentially "the Brownian motion conditioned to stay positive", in particular, its distribution is the limit of Brownian motions conditioned on some convex sets. This implies that it's strictly log-concave (w.r.t. the covariance of the Brownian motion) --- i.e. "more log-concave than the Gaussian distribution of the Brownian motion". It follows from the inequality of Harge (Theorem 1.1 in Harge "A convex/log-concave correlation inequality for Gaussian measure and an application to abstract Wiener spaces") that for any linear functional $\psi$ on $C[0,1]$ we have $\mathsf{E} (\psi(U - \mathsf{E} U))^2 \le \mathsf{E} \psi(B)^2$; by approximation, this extends to measurable linear functionals, i.e. stochastic integrals of deterministic $L^2$ functions. In other words, for any deterministic function $f \in L^2[0,1]$ the stochastic integral $\intop f d (U - \mathsf{E} U)$ "converges" (i.e. is well-defined as an appropriate limit...). For the Brownian motion, however, $\intop f d B$ converges without the need to recenter $B$. Since $\mathsf{E} U(t) \sim t^{1/2}$ is not in the $W^{1,2}$ Sobolev space, this means that there are some functionos $f \in L^2$, such that $\intop f d B$ is well-defined bub $\intop f d U$ is not --- thus, $\intop f d ((B + U) / \sqrt 2)$ is also not defined. This proves that the law of $(B + U) / \sqrt 2$ is singular to that of $B$.

$\endgroup$

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

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