Diffusion sample paths as deformed Brownian sample paths - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T11:07:36Z http://mathoverflow.net/feeds/question/53073 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/53073/diffusion-sample-paths-as-deformed-brownian-sample-paths Diffusion sample paths as deformed Brownian sample paths Simon Lyons 2011-01-24T15:47:00Z 2011-07-02T12:14:31Z <p>Suppose $X$ is a non-explosive diffusion with dynamics</p> <p>$dX_t = \mu(X_t)dt + \sigma(X_t)dW_t$, </p> <p>where $W$ is a standard Brownian motion. My intuition about $X$ is that if $\mu$ and $\sigma$ are sufficiently nice, then the sample paths of $X$ are in some sense "deformed" sample paths of $W$. Is there any way to formalise this idea? For example, is it possible to define a suitable topology on sample paths of $W$ and construct diffusion sample paths $X(\omega)$ as homeomorphisms of $W(\omega)$?</p> <p>Part of the motivation for this question comes from the observation that it's possible to do something very similar in the discrete-time case. Given the Euler approximation</p> <p>$\Delta X_{t+1} = \mu(X_t)\Delta t + \sigma(X_t)\sqrt{\Delta t} W_t$</p> <p>with $W_t \sim N(0,1)$, then if one knows the values of $\Delta X_t$, then one can unambiguously recover the driving noise $W$. In that sense, one can view $X$ as a transformed version of $W$.</p> http://mathoverflow.net/questions/53073/diffusion-sample-paths-as-deformed-brownian-sample-paths/53077#53077 Answer by Leonid Petrov for Diffusion sample paths as deformed Brownian sample paths Leonid Petrov 2011-01-24T16:24:53Z 2011-01-24T16:24:53Z <p>You should probably look at the Girsanov's theorem <a href="http://en.wikipedia.org/wiki/Girsanov_theorem" rel="nofollow">http://en.wikipedia.org/wiki/Girsanov_theorem</a></p> <p>The process $X$ is a probability distrubution on the space of continuous functions, so is the Wiener process $W$. Girsanov' theorem states that the distribuitons of $X$ and $W$ are equivalent, and gives explicitly the Radon-Nykodim derivative. </p> http://mathoverflow.net/questions/53073/diffusion-sample-paths-as-deformed-brownian-sample-paths/53203#53203 Answer by The Bridge for Diffusion sample paths as deformed Brownian sample paths The Bridge 2011-01-25T08:12:35Z 2011-01-25T08:12:35Z <p>Hi,</p> <p>This is an interesting question</p> <p>I don't have the complete answer to your question rather some leads about it, but it seems to me that what you need to show for solutions of your SDE, is that the natural filtrations of $X$ is the same as the natural filtration of $W$. </p> <p>If you have this done, then by some kind of Doob's Lemma you should be able to write your Brownian path with respect to a "measurable" functional of the path of $X$ (i.e. $W_t=f((X_s,s\le t))$ for some functional $f$). This not a constructive way of showing the result though (i.e. you only have existence) </p> <p>Anyway I think this is not the case for very broad class of SDEs even if I don't have a counterexample at hand, but there must be some litterature about this (maybe in Revuz and Yor's book).</p> <p>You can also look at the Lamperti's Transform (beware I think that there two Lamperti transform in the litterature), which says that under some conditions you can transform a SDE of the form : $dX_t=\sigma(X_t,t)dW_t$ into some SDE of the form $dX_t=\mu_{\sigma}(X_t,t)dt+dB_t$ but I can't remember if this is done in a path-by-path way (i.e. if B_t =f(W.) where f is a function over the path space). You should have a look at the proof yourself. Here is a paper where the Lamperti's Transform is disccussed "Moller, Madsen - From State Dependent Diffusion to Constant Diffusion in SDEs by the Lamperti Transformation" and the references therein.</p> <p>Best Regards and let us know if you find something interesting</p> http://mathoverflow.net/questions/53073/diffusion-sample-paths-as-deformed-brownian-sample-paths/53212#53212 Answer by Ilya for Diffusion sample paths as deformed Brownian sample paths Ilya 2011-01-25T09:53:01Z 2011-01-25T09:53:01Z <ol> <li><p>If you want to have $X_t$ as a "deformed" $W_t$ - at first I advise to assume $\sigma\neq 0$ a.s. Otherwise you will have some problems (really in such points you may have almost deterministic dynamics).</p></li> <li><p>If $\mu = 0$ then you can just change the time since all continuous martingales are time-transformed Brownian motion (it seems to me that zhoraster talked about something closer).</p></li> <li><p>If $\mu \neq 0$ then you can either apply change of measure +change of time, but since you do not want to apply the change of measure - please make your question more precise. What do you want? If we have the function $t^2$ and $t$ are their paths "homeomorphic"? The same question I would like to ask you about all functions of bounded variation.</p></li> </ol> http://mathoverflow.net/questions/53073/diffusion-sample-paths-as-deformed-brownian-sample-paths/69241#69241 Answer by Piotr Miłoś for Diffusion sample paths as deformed Brownian sample paths Piotr Miłoś 2011-07-01T10:00:05Z 2011-07-01T10:00:05Z <p>Another way to approach the problem is as follows.</p> <p>One can notice that $X$ is a semimartingale (probably under some mild assumptions on $\sigma,\mu$). The martingale $M$ part of $X$ can be represented as</p> <p>$$M_t = B_{&lt; M,M>_t},$$ where $B$ is some Brownian motion. This is know as "Dambis, Dubins-Schwarz Theorem" (e.g. see Chapter V of Revuz, Yor - "Continous martingales ..."). </p> http://mathoverflow.net/questions/53073/diffusion-sample-paths-as-deformed-brownian-sample-paths/69261#69261 Answer by Alekk for Diffusion sample paths as deformed Brownian sample paths Alekk 2011-07-01T13:58:40Z 2011-07-02T12:14:31Z <p>As other have said, in the one dimensional case at least, you can suppose that the volatility is constant. Then the solution of the SDE is nothing else than a solution of the integral equation $$X(t) = \int_0^t \mu(X_s) ds + \sigma W_t \qquad \forall t \in [0,T].$$ You can then check that if $\mu(\cdot)$ is a Lipschitz function, say, then the function $\Psi$ that maps $(W_t)_{t \in [0,T]}$ to the solution of the above integral equation is continuous (Gronwall Lemma) on $C([0,T],\mathbb{R})$ with the supremum norm. Hence you can indeed write $X = \Psi(W)$ and see the path $(X_t)_{t \in [0,T]}$ as a 'deformation' of the Brownian path $(W_t)_{t \in [0,T]}$. The function $\Psi: C([0,T],\mathbb{R}) \to C([0,T],\mathbb{R})$ is sometimes called the 'Ito map' in the literature.</p> http://mathoverflow.net/questions/53073/diffusion-sample-paths-as-deformed-brownian-sample-paths/69299#69299 Answer by Jason Swanson for Diffusion sample paths as deformed Brownian sample paths Jason Swanson 2011-07-01T22:36:15Z 2011-07-01T22:36:15Z <p>Suppose $\mu$ and $\sigma$ are sufficiently well-behaved so that we may define $$B_t := - \int_0^t\frac{\mu(X_s)}{\sigma(X_s)}\,ds + \int_0^t \frac1{\sigma(X_s)}\,dX_s.$$ This should be the case, for example, if $\mu$ and $\sigma$ are globally Lipschitz with linear growth and $|\sigma|$ is bounded below. We then obtain $$B_t = - \int_0^t\frac{\mu(X_s)}{\sigma(X_s)}\,ds + \int_0^t\frac{\mu(X_s)}{\sigma(X_s)}\,ds + \int_0^t \,dW_s = W_t.$$ If $\sigma$ is continuous, then the stochastic integral in the definition of $B$ can be realized as a limit in probability of left-endpoint Riemann sums, and so it is evident that $B$ (i.e. $W$) is adapted to the natural filtration of $X$. Since $X$ is also adapted to the natural filtration of $W$, these two filtrations must be the same.</p>