Brownian bridge interpreted as Brownian motion on the circle - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T06:56:23Z http://mathoverflow.net/feeds/question/46991 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/46991/brownian-bridge-interpreted-as-brownian-motion-on-the-circle Brownian bridge interpreted as Brownian motion on the circle Simon Lyons 2010-11-22T20:09:45Z 2012-05-02T09:18:11Z <p>Is it reasonable to view the Brownian bridge as a kind of Brownian motion indexed by points on the circle? The Brownian bridge has some strange connections with the Riemann zeta function (see Williams' article <a href="http://www.statslab.cam.ac.uk/~grg/books/hammfest/22-dw.ps" rel="nofollow">http://www.statslab.cam.ac.uk/~grg/books/hammfest/22-dw.ps</a>, for example). </p> <p>I'm looking for a heuristic explanation of why this might be the case. If one could interpret the Brownian bridge as described above, then the heuristic would be that Brownian motion is naturally associated with heat flow, which goes hand in hand with theta functions, which goes some way toward explaining the appearance of the zeta function.</p> <p>I don't know where to begin reading about Brownian motion indexed by anything other than $\mathbb{R}^{+}$. The standard stochastic analysis texts don't really address the idea.</p> <p>Many thanks.</p> http://mathoverflow.net/questions/46991/brownian-bridge-interpreted-as-brownian-motion-on-the-circle/46995#46995 Answer by Michael Hardy for Brownian bridge interpreted as Brownian motion on the circle Michael Hardy 2010-11-22T20:26:20Z 2010-11-22T20:26:20Z <p>The circle qua circle has no distinguished point. The Brownian bridge is tied down at the two endpoints, and its variance at any point $x \in [0,1]$ is $x(1-x)$, which equals 0 at the endpoints. So is there a way to get rid of the special point and make it into a process with the same variance at all points, while having those endpoints match up? It seems to me that's the question you'd have to answer.</p> <p>Added a minute or so later: But you could single out another point to be the distinguished point, subtract the value of the BB at that point from its value at every point, and get another BB, not independent of the first one, with the same probability distribution.</p> <p>So I think probably the answer is yes.</p> http://mathoverflow.net/questions/46991/brownian-bridge-interpreted-as-brownian-motion-on-the-circle/47004#47004 Answer by Louigi Addario-Berry for Brownian bridge interpreted as Brownian motion on the circle Louigi Addario-Berry 2010-11-22T21:21:24Z 2010-11-22T21:47:46Z <p><a href="http://www3.interscience.wiley.com/journal/113393695/abstract" rel="nofollow">Aldous and Pitman</a> have a paper on "Brownian bridge asymptotics for random mappings", which describes a setting in which Brownian bridge shows up as a limit object and is most naturally thought of as indexed by a circle rather than by an interval. There are two follow-up papers (<a href="http://www.math.washington.edu/~ejpecp/viewarticle.php?id=1429&amp;layout=abstract" rel="nofollow">one</a>, <a href="http://www.springerlink.com/content/g8134863lt734607/?p=ed2ec1d76a7f4250990539b6c19a321c&amp;pi=4" rel="nofollow">two</a>) by Aldous, Miermont and Pitman, the first of which "give[s] a conceptually straightforward argument which both proves convergence and more directly identifies the limit" (as well as extending the results to more general kinds of random mappings). </p> <p>The basic idea is that mapping, i.e. a function $f$ from <code>$[n] = \{1,\ldots,n\}$</code> to $[n]$ can be represented in terms of "basins of attraction". Create a digraph by joining $i$ to $j$ if $f(i)=j$. In this digraph, each connected component will contain a unique directed cycle, and each vertex $i$ of the cycle will be the root of a tree all of whose edges are oriented towards $i$. It is then possible to code the structure of the cycle-plus-trees in terms of a lattice path, with height corresponding to distance from the cycle.</p> <p>When the underlying mapping $f$ is a uniformly random mapping, the resulting lattice path, suitably interpreted and after rescaling, converges to (the absolute value of) Brownian bridge. </p> http://mathoverflow.net/questions/46991/brownian-bridge-interpreted-as-brownian-motion-on-the-circle/47008#47008 Answer by Shai Covo for Brownian bridge interpreted as Brownian motion on the circle Shai Covo 2010-11-22T22:19:22Z 2010-11-22T22:53:15Z <p>First of all, a Brownian motion can be indexed by a quite arbitrary parameter set (so-called set-indexed Brownian motion). </p> <p>Now to your question. There is a natural extension to two-parameters of the ordinary Brownian motion on <code>$\mathbb{R}_+$</code>, namely the Brownian sheet, commonly denoted by <code>$W$</code> and indexed by <code>$\mathbb{R}_ + ^2 = \{ (s,t):s \ge 0,t \ge 0\}$</code>. A very good textbook on this topic (in a general setting) is "An Introduction to Continuity, Extrema, and Related Topics for General Gaussian Processes"; the relevant pages (6-7) are available online. Now, restricting a Brownian sheet to parametrized curves in <code>$\mathbb{R}^2_+$</code> results in one-parameter processes, which are continuous centered Gaussian processes. A prominent example is the one-parameter process <code>$A = \{A_t: 0 \leq t \leq 1\}$</code> defined by <code>$A_t = W_{t,1-t}$</code>. The process <code>$A$</code> is in fact a Brownian bridge. So, this is an example of a Brownian bridge which is "a kind of Brownian motion indexed by points on" the segment connecting <code>$(0,1)$</code> and <code>$(1,0)$</code>. However, this is essentially the only way to get a Brownian bridge by restricting a Brownian sheet to some parametrized curve in <code>$\mathbb{R}^2_+$</code>. Nevertheless, a Brownian bridge which is "a kind of Brownian motion indexed by points on the circle" might be obtained as follows. The Brownian sheet is defined, for <code>$s,t \geq 0$</code>, by <code>$W_{s,t} = W((0,s] \times (0,t])$</code>, where <code>$W(A)$</code>, <code>$A \in \mathcal{B}(\mathbb{R}_ + ^2 )$</code> is the Gaussian white noise based on Lebesgue measure; see pages 6-7 in the aforementioned book. Now, letting the underlying measure <code>$\nu$</code> in <code>$(\mathbb{R}_ + ^2,\mathcal{B}(\mathbb{R}_ + ^2 ),\nu)$</code> be a (<code>$\sigma$</code>-finite) measure different from Lebesgue measure, the process <code>$\tilde W$</code> defined by <code>$\tilde W_{s,t} = W((0,s] \times (0,t])$</code>, where <code>$W$</code> is a Gaussian white noise based on <code>$\nu$</code>, is no longer a Brownian sheet; it is very likely that one can find <code>$\nu$</code> such that the process <code>$\tilde A = \{\tilde A_t: 0 \leq t \leq \pi/2\}$</code> defined by <code>$\tilde A_t = \tilde W_{\sin t, \cos t}$</code> is a Brownian bridge (which is "a kind of Brownian motion indexed by points on the circle"). I can guide you how to find the suitable $\nu$, if you wish.</p> http://mathoverflow.net/questions/46991/brownian-bridge-interpreted-as-brownian-motion-on-the-circle/47051#47051 Answer by Tom LaGatta for Brownian bridge interpreted as Brownian motion on the circle Tom LaGatta 2010-11-23T04:40:20Z 2010-11-23T04:40:20Z <p>Let $X_t$ be a continuous stochastic process on the circle $S^1$. For any reasonable application, you probably want that the distribution of $X_t$ is invariant under rotations of the circle. If you instead consider the process as defined on the interval $[0,1]$ with $X_0 = X_1$, then this means that the distribution must be invariant by translations $\operatorname{mod} \mathbb Z$. The Brownian bridge fails this.</p> <p>A more natural approach, in my opinion, would be to consider a <a href="http://en.wikipedia.org/wiki/Stationary_stochastic_process" rel="nofollow">stationary stochastic process</a> on the interval $[0,1]$, along with the bridge'' condition $X_0 = X_1$. Note that, unlike Brownian motion, the value of $X_0$ need not be concentrated at a single value.</p> http://mathoverflow.net/questions/46991/brownian-bridge-interpreted-as-brownian-motion-on-the-circle/47058#47058 Answer by Gideon Schechtman for Brownian bridge interpreted as Brownian motion on the circle Gideon Schechtman 2010-11-23T06:17:09Z 2010-11-23T10:43:27Z <p>What is wrong with the following simple definition? Let $g_1$ and $g_2$ be two independent standard Gaussian variables. For $t=(a,b)\in S^1$ (so $a^2+b^2=1$), let $B_t=ag_1+bg_2$. You get a Gaussian process whose distribution is invariant under rotation. Each $B_t$ is a standard Gaussian variable and the variance of $B_t-B_s$ is $\|t-s\|^2$. </p> <p>Following Ori's remark below, concerning the 2-dimensionality, maybe a better simple suggestion is the following: Take two independent Brownian motions on $(-\infty, \infty)$, $C_a,D_a$ ($C_0=D_0=0$) and for $t=(a,b)\in S^1$ define $B_t=C_a+D_b$.</p> <p>On third thought, this is probably just a Brownian motion on $R^2$ (zero at the origin) restricted to $S^1$.</p> http://mathoverflow.net/questions/46991/brownian-bridge-interpreted-as-brownian-motion-on-the-circle/47391#47391 Answer by Omer for Brownian bridge interpreted as Brownian motion on the circle Omer 2010-11-26T01:38:33Z 2010-11-26T01:38:33Z <p>You could consider an Ornstein-Uhlenbeck bridge, i.e. a O-U process conditioned on $X_1=X_0$. You can no longer scale away the length of the circle.</p> http://mathoverflow.net/questions/46991/brownian-bridge-interpreted-as-brownian-motion-on-the-circle/95706#95706 Answer by AlexArvanitakis for Brownian bridge interpreted as Brownian motion on the circle AlexArvanitakis 2012-05-01T23:48:45Z 2012-05-02T00:03:07Z <p>Well, Brownian motion on $R^n$ is a diffusion process; that is, the coefficients of the SDE of which Brownian motion is a solution are time-independent, and I think that this is also the case for generalisations of Brownian motion. However, the SDE solved by a Brownian bridge is time dependent, so I would think that the Brownian bridge is a different kind of object.</p> http://mathoverflow.net/questions/46991/brownian-bridge-interpreted-as-brownian-motion-on-the-circle/95737#95737 Answer by Vincent Beffara for Brownian bridge interpreted as Brownian motion on the circle Vincent Beffara 2012-05-02T09:18:11Z 2012-05-02T09:18:11Z <p>If you want an Ornstein-Uhlenbeck process (i.e. Brownian motion confined by a quadratic potential), there is no difficulty because this has a stationary distribution: start from that definition at time $0$, condition to be back at the same point at time $1$, and then extend by periodicity / wrap this on the circle. This gives a well-defined process indexed by the circle, and no point is special because we are always at equilibrium.</p> <p>If you insist on Brownian motion, but do not want a special point on the circle, then there is a problem because there is no invariant probability measure. The right way to define things goes through <em>white noise</em> i.e. through the <em>increments</em> of Brownian motion. Start with a Brownian bridge $B$ on the interval $[0,1]$, continue it periodically / wrap it on the circle, and let $X_{s,t} = B_t-B_s$. This makes sense for any $s,t \in \mathbb R / \mathbb Z$, and it is easy to see that the distribution of $(X_{s,t})$ does not depend on the choice of the origin. So $(X_{s,t})$ is somehow "the" right object to look at.</p> <p>From $(X_{s,t})$ you can recover $B_t$ if you want (because $B_t = X_{0,t}$), but then this depends on the choice of the origin. A better way is to say that you will identify two functions if they differ by a constant; then you will get the same equivalence class wherever you choose the origin, and that is as good a definition as any of Brownian motion on the circle. In particular, if what you want to do is stochastic calculus, the increments of the process are usually the relevant thing.</p> <p>Another way to break the indeterminacy is to choose the path with average value $0$, which then coincides with the Fourier expansion version of Brownian motion (i.e., the Fourier series with independent Gaussian coefficients of appropriate variance).</p>