Regular Conditional Probability given a natural filtration of a stochastic process - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-19T06:28:22Z http://mathoverflow.net/feeds/question/71137 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/71137/regular-conditional-probability-given-a-natural-filtration-of-a-stochastic-proces Regular Conditional Probability given a natural filtration of a stochastic process Pierre 2011-07-24T17:27:32Z 2011-07-24T18:22:22Z <p>OK, this is kind of re-posting, but I think I can clarify the question more, so it's worth a shot.</p> <p>Consider a real valued process $(X_t)_{t \leq T}$, cadlag on a probability space $(\Omega, (\mathcal{F}^\circ_t)_{t \leq T}, \mathbb{P}). \mathcal{F}^\circ_t=\sigma(X_s;s\leq t)$ is the uncompleted, natural filtration generated by $X_t$. Unfortunately $X_t$ neither has independent increments, nor is it markov. Since $\Omega$ is a Polish space, $\mathcal{F}^\circ_T$ and also $\mathcal{F}^\circ_t$ are countably generated, so we know, there exists a regular version of the conditional probability of $\mathbb{P}$ for any fixed $t$ for $\mathbb{P}$-a.a. $\omega$, i.e. for fixed $t$, $\mathbb{P}(\cdot|\mathcal{F}_t)(\omega)$ is a prob. measure f.a.a. $\omega$.</p> <p>Hence we know, that for all $t\in [0,T]\cup \mathbb{Q}$, we find a regular conditional probability f.a.a. $\omega$, depending on $t$. In words: Given almost any path of the process up to time $t$, we can deduce the probablity of events, taking that information into account.</p> <p>On the remaining $\omega$'s, define some meaningless measure, so we have a measure $\forall \omega$. How can I extend this to all $t$ in a reasonable way? Reasonable means: There is one Null set $N$, so that $\forall t$ $\mathbb{P}(\cdot|\mathcal{F}_t)(\omega)$, $\omega\in N^c$, is a measure Anybody seen anything like this?</p> <p>I read something like this only for Markov and Feller processes using infinitesimal generators, but this cannot be carried over one to one, because we do not have a transition semigroup.</p> <p>Maybe I have a deep misunderstanding here. Grateful for any objections, hints and comments.</p> http://mathoverflow.net/questions/71137/regular-conditional-probability-given-a-natural-filtration-of-a-stochastic-proces/71141#71141 Answer by Tom LaGatta for Regular Conditional Probability given a natural filtration of a stochastic process Tom LaGatta 2011-07-24T18:22:22Z 2011-07-24T18:22:22Z <p>Let's assume that we are working with the canonical probability space $\Omega = D(\mathbb R)$ of càdlàg functions, and $\mathbb P$ is the law of the process. I would doubt that there is a satisfactory answer at the level of maximal generality you've stated. At the very least, the measure $\mathbb P$ should be Radon. There are extremely general results on the existences of RCPs for Radon measures (cf. Leão, Fragoso and Ruffino, <i>Regular conditional probability, disintegration of probability and Radon spaces</i>). </p> <p>The RCP is a measure-valued function $P : [0,T] \times \Omega \to \mathcal M(\Omega)$ such that for $\mathbb P$-almost every $\omega$, the measure $P(t,\omega, \cdot)$ is a version of $\mathbb P(\cdot|\mathcal F_t)$. Do you want the function $(t, \omega) \mapsto P(t,\omega,\cdot)$ to simply exist and be measurable? If so, this can be done in the wide generality stated above; see Leão et al.</p> <p>Recently, I have needed more regularity properties for RCPs, namely, continuity. Consider the space $\mathcal M(\Omega)$ of Radon measures on $\Omega$ equipped with the topology of weak convergence of measures. We say that the RCP is a <i>continuous disintegration</i> (or continuous RCP) when it satisfies the following property: $$\mbox{if \omega_n \to \omega, then the measures P(t,\omega_n,\cdot) converge weakly to P(t,\omega,\cdot).}$$</p> <p>If the law is Gaussian, then my preprint <a href="http://arxiv.org/abs/1003.0975" rel="nofollow">Continuous Disintegrations of Gaussian Processes</a> gives a necessary and sufficient condition for the law $\mathbb P$ to have a continuous disintegration. I haven't thought about this in the case of càdlàg functions, but I'm pretty sure that this will extend easily. Note that this is just for fixed $t$.</p> <p>To show that the map $(t, \omega) \mapsto P(t,\omega,\cdot)$ jointly continuous, a little more work is needed. As part of a larger project, Janek Wehr and I have general results in this direction for stationary, Gaussian processes. If this is what you need, I'm happy to discuss this with you further.</p> <p><b>Open Question:</b> If the law $\mathbb P$ is not Gaussian but at least is log-Sobolev, then all the same results should hold. This is because log-Sobolev measures satisfy very strong concentration-of-measure properties. I have some ideas how to do this, but I haven't worked out the details because I've been busy with other projects. If anybody is interested in collaborating on extending this work to the log-Sobolev case, please contact me.</p>