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Suppose $X$ is a non-explosive diffusion with dynamics

$dX_t = \mu(X_t)dt + \sigma(X_t)dW_t$,

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 $W(\omega)$?

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

$\Delta X_{t+1} = \mu(X_t)\Delta t + \sigma(X_t)\sqrt{\Delta t} W_t $

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 $W$.

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A (one-dimensional) diffusion is expressed fairly explicitly in terms of so-called "scale measure" and "speed measure", you can easily find the formula in old classical textbooks. – zhoraster Jan 24 '11 at 18:30
@Zhoraster Thanks - I'll work through the material in Karatzas & Shreve. – Simon Lyons Jan 24 '11 at 18:32
I mean, expressed as a transform of a Wiener process path. – zhoraster Jan 24 '11 at 18:33
Just give a shout if you don't find the formula, I'll find a reference. – zhoraster Jan 24 '11 at 18:34


This is an interesting question

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 $W$.

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)

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).

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.

Best Regards and let us know if you find something interesting

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Thanks, this is the type of idea I was looking for. Yes, I can see that this kind of strategy will not work for many SDEs. Perhaps one could construct a counterexample based on Tanaka's formula. I'll take a look at the Lamperti transform. – Simon Lyons Jan 25 '11 at 15:56
The pleasure is mine if I was of any help – The Bridge Jan 25 '11 at 16:10
Simon, you are right, for $X_t=|B_t|$ it seems difficult to rebuilt $B_t$ from $X_t$'s paths. – The Bridge Jan 25 '11 at 17:49

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.

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You should probably look at the Girsanov's theorem

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.

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Hm, Girsanov's theorem isn't really what I'm looking for. It tells you how to transform the laws of $X$ and $W$, whereas I'm looking for a transform on the paths themselves, if such a thing exists. That said, it wasn't me who voted your answer down. – Simon Lyons Jan 24 '11 at 20:15
  1. 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).

  2. 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).

  3. 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.

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Another way to approach the problem is as follows.

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

$$M_t = B_{< M,M>_t},$$ where $B$ is some Brownian motion. This is know as "Dambis, Dubins-Schwarz Theorem" (e.g. see Chapter V of Revuz, Yor - "Continous martingales ...").

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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.

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