Markov random field with continuous index set - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-19T17:46:27Z http://mathoverflow.net/feeds/question/58298 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/58298/markov-random-field-with-continuous-index-set Markov random field with continuous index set Mahdiyar 2011-03-12T22:50:11Z 2011-03-13T23:40:43Z <p>Hi</p> <p>There's Markov random field (MRF) which, by my Wikipedia-based knowledge, is an extension of Markov chain. I'd like to think of it as going from 1D to higher dimensional spaces. Inherent in its definition though, the space a MRF lives on (i.e., the index set of the stochastic process) is a discrete graph. So it's actually a lattice and not the continuum of a Euclidean space (or some manifold for that matter). I'm wondering if there exists a MRF of the latter form, in other words, an extension of the Markov process (as opposed to the Markov chain) to higher dimensions.</p> <p>I know that a similar extension exists for the Poisson process, namely the spatial Poisson process. But for MRF, I'm not even sure how I'd define the Markov property.</p> <p>Any references or remarks are appreciated.</p> http://mathoverflow.net/questions/58298/markov-random-field-with-continuous-index-set/58314#58314 Answer by John Jiang for Markov random field with continuous index set John Jiang 2011-03-13T04:23:05Z 2011-03-13T04:28:23Z <p>There is the continuum 2-dimensional Gaussian free field, which is a higher dimensional generalization of Brownian motion. The continuum GFF satisfies the so-called domain Markov property (meaning if you condition on the value of the GFF on a subset of $\mathbb{R}^2$, its value outside that subset depends only on the value on the boundary of that subset), and can be viewed as weak limit of GFF defined on 2-dimensional infinite lattices. The discrete GFF is basically a probability distribution on $\mathbb{R}^{\mathbb{Z}^2}$, i.e., the space of real-valued functions on $\mathbb{Z}^2$. The distribution is given in terms of a Hamiltonian, $$\displaystyle H(x) = \sum_{(i,j) \in E} x_i x_j + \sum_{i \in \mathbb{Z}^2} x_i^2$$ In other words, the probability density is proportional to $\exp(-H(x))$. Thus it's the most natural Gaussian measure on $\mathbb{R}^{\mathbb{Z}^2}$ that takes into account the underlying graph structure.</p> <p>As an analogy, the Brownian motion at discrete time points, say $\mathbb{Z}$ is the Gaussian free field on $\mathbb{Z}$. Another perspective is to view Gaussian free field as the standard Gaussian random variable on the $\mathbb{Z}^2$ or $\mathbb{Z}$ but with the Dirichlet inner product instead of the usual Euclidean inner product; since Dirichlet product is basically the $\mathcal{L}^2$ inner product of the gradient, one needs to impose boundary condition or define certain equivalence classes of functions in order for the inner product to be nondegenerate. </p> <p>When we take finer and finer grid in $\mathbb{Z}$ or $\mathbb{Z}^2$, we obtain in the weak limit the Brownian motion and the continuum 2-dimensional Gaussian free field. The only thing weird about the continuum GFF in 2 dimensions is that it's no longer a probability law on a function space, but rather on the space of distributions, or generalized functions on $\mathbb{R}^2$; Brownian motion on the other hand can still be viewed as a distribution on continuous functions on $\mathbb{R}$. For more details on how to define such a limiting object, see the lecture notes by Scott Sheffield: Gaussian free fields for mathematicians. </p> http://mathoverflow.net/questions/58298/markov-random-field-with-continuous-index-set/58350#58350 Answer by Yuri Bakhtin for Markov random field with continuous index set Yuri Bakhtin 2011-03-13T19:08:37Z 2011-03-13T19:08:37Z <p>Essentially the Markov property in higher dimensions means that for any index set $D$ with nice boundary, the conditional distribution of the restriction of the field to indices in $D$ conditioned on the realization outside of $D$ coincides with the same thing conditioned on the realization on the boundary of $D$.</p> <p>From Rozanov's book <a href="http://www.amazon.com/Markov-Random-Fields-Applications-Mathematics/dp/0387907084/ref=sr_1_1?ie=UTF8&amp;s=books&amp;qid=1300042943&amp;sr=8-1" rel="nofollow">http://www.amazon.com/Markov-Random-Fields-Applications-Mathematics/dp/0387907084/ref=sr_1_1?ie=UTF8&amp;s=books&amp;qid=1300042943&amp;sr=8-1</a> that I looked into about 10 years ago just for fun, I vaguely remember that this property sometimes has to be altered a little, but maybe this is only because Rozanov wanted to consider generalized Markov fields ("generalized" means in Sobolev-Schwartz sense).</p> http://mathoverflow.net/questions/58298/markov-random-field-with-continuous-index-set/58373#58373 Answer by Shai Covo for Markov random field with continuous index set Shai Covo 2011-03-13T23:35:07Z 2011-03-13T23:40:43Z <p>In general, the simplest extension of (real-valued) one-parameter processes is to (real-valued) multi-parameter processes, where the index set is $\mathbb{R}^n_+$ (usually $n=2$). This includes, for example, the $n$-parameter Brownian sheet, and more generally, $n$-parameter L\'evy sheets (or processes). However, such processes are much simpler than general set-indexed processes. </p> <p>You'll probably find the following paper very useful (especially the introduction, since the set-indexed framework might be too heavy): <a href="http://arxiv.org/PS_cache/math/pdf/0412/0412350v1.pdf" rel="nofollow">A Markov Property For Set-Indexed Processes</a>.</p>