When can a continuous path (Markovian) stochastic process in one dimension be represented as an Ito or a diffusion process? What are the examples when it can not be?
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3 Answers
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A continuous strong Markov process is a semimartingale if and only if it can be represented as a time-changed Ito diffusion. More generally, a Hunt process is a semimartingale if and only if it is a time-changed jump diffusion. This is a result of Cinlar, Jacod, Protter, and Sharpe, from their truly amazing paper Semimartingales and Markov processes.
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$\begingroup$ Thank you, Dan. I am trying to get a pdf copy of the mentioned paper off the web, but it seems hard... $\endgroup$– HansCommented Dec 15, 2013 at 4:32
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These types of questions are treated in great generality in the book
Rogers-Williams: Diffusions, Markov processes and martingales, Volume 1
One of the great results is due to Dynkin. Let $(X_t)_{t \ge 0}$ be a continuous Markov process whose semigroup is Feller-Dynkin (that is restricts to a strongly continuous semigroup on continuous functions vanishing at $\infty$). Then if the generator of $X$ contains the space of smooth and compactly supported functions, this generator is necessarily a second-order semi-elliptic differential operator and so $X$ is a diffusion process.
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A simple counterexample is a process that, starting at zero moves with constant velocity 1 (with probability $1/2$) or with constant velocity -1 (with probability $1/2$).
There are various subtleties here, and the understanding of what a diffusion process is has varied in time and from one author to another.
In my favorite introductory book on stochastic processes by Wentzell (A Course In The Theory Of Stochastic Processes, originally in Russian), with his choice of definitions, a sufficient condition for a Markov process to be a diffusion is essentially a combination of stochastic continuity and nice behavior of (truncated) mean and variance of transition probabilities.
Update A better example is a strictly increasing nonrandom trajectory that is not absolutely continuous with respect to time.
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$\begingroup$ Thank you, Yuri, for the explanation and reference. The counter example however does not seem to be correct. I suppose you want the two trajectory to be deterministic except at time 0? But ODE is just a special case of SDE, or $dx = \mbox{sign}(x)dt$, then assign velocity -1 and 1 to $x=0$ with equal probability. Why can't this be classified as diffusion? Regarding classification of diffusion, I suppose you are saying the Ito process is but one type of diffusions, the general definition of which, as you say, varies. Also, Dan has provided a reference for one sufficient and necessary condition. $\endgroup$– HansCommented Dec 15, 2013 at 13:38
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$\begingroup$ Right, i should have taken care of the infinite diffusion coefficient at zero. To that end, instead of constant speed motion, take $x(t)=t^a$ with $a$ small. You still can say that away from $0$ these trajectories solve an ODE and in the integral sense even including $0$. And here we again need a definition of a diffusion process. If it includes linear behavior of variance of the increment then this is not a diffusion. Of course, a time change also cures this singularity, so, unfortunately, this question is partially about definitions. $\endgroup$ Commented Dec 15, 2013 at 17:19
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$\begingroup$ Thank you for providing these examples and going along with me, Yuri. As you said, I don't like the example $x(t)=t^\alpha$ with $\alpha$, since I can still call it ODE and thus SDE. Your Update example is better. Still it is not a strictly stochastic example which I seek and would prefer. $\endgroup$– HansCommented Dec 16, 2013 at 18:38
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$\begingroup$ By the way, do you understand Martin Hairer's strong Markov derivation, specifically about the law of $X(t)$ and $Y(t)$ being equal, in the 1-d continuous path Markov process increasing variance problem? Can you explain it to me there if you don't mind? $\endgroup$– HansCommented Dec 16, 2013 at 20:02
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$\begingroup$ You want a "stochastic" example of a process that is not a "diffusion"? Don't you think it's getting strange? Well, I don't know if this is gonna satisfy you: take that increasing non-absolutely continuous deterministic trajectory $f(t)$ and instead of time substitute a Wiener process obtaining $f(W(t))$. As for Martin's comment, I thought I understood what he was saying, especially because I had a similar generalization in mind (not quite the same), but why don't you ask him. Anyway, I have not thought much about the nonsymmetric case. $\endgroup$ Commented Dec 17, 2013 at 0:18