bio | website | statslab.cam.ac.uk/~beresty |
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location | ||
age | 34 | |
visits | member for | 3 years |
seen | Oct 24 '12 at 19:15 | |
stats | profile views | 344 |
Nov 30 |
awarded | Yearling |
Aug 25 |
awarded | Nice Question |
Aug 25 |
comment |
Markov chains: invariant measures and explosion
It is a classical (and surprising) feature of continuous Markov chains that they can have an invariant measure while being transient. See, for instance, section 3.5 in James Norris' book on Markov chains. (Notice that the definition of invariant measure is the usual one and does not require the process to be recurrent or non-explosive). In any case, no matter how you would call such a measure, I hope you'll agree it is interesting to know what it means for the process... |
Aug 25 |
comment |
Markov chains: invariant measures and explosion
Dear Robert, Thanks for your comments and sorry for long time in response. I am not totally comfortable with the derivation of your equation. I always thought this process would satisfy the Kolmogorov backward and forward equations - without being the minimal solution. See, for instance, section 2.9 in James Norris' book on Markov chains. But you may well be right - in which case the question of "what is this invariant measure" is even more puzzling to me ! |
Aug 21 |
asked | Markov chains: invariant measures and explosion |
Dec 31 |
answered | Is the maximum tree-path length distributed lognormally (in the limit) ? |
Dec 20 |
comment |
Green's formula for a Markov process
I don't know if this is what you are looking for, but the left-hand side in your last identity is called the Dirichlet form $\mathcal{E}(f,g)$. For a Markov chain on a countable space and with invariant measure $\pi$, (not necessarily reversible), it is always true that $$\mathcal{E}(f,g) = \sum_{x,y} \pi(x) P(x,y) g(x) \nabla_{x,y} f.$$ But this is an easy calculation, so presumably you are aware of it. |
Dec 19 |
comment |
Correlations in last-passage percolation
Hi James ! Thanks, very helpful. I guess the story about competing interface in FPP (which results in random slope) shows indeed that there is a nonzero probability that the geodesics have no edge in common at all. That is somehow slightly counterintuitive to me, you'd expect the geodesics to go get the same goodies for a while, before they diverge... So is the covariance $O(1)$ as well? |
Dec 19 |
awarded | Scholar |
Dec 19 |
accepted | Correlations in last-passage percolation |
Dec 19 |
comment |
Correlations in last-passage percolation
very cool picture ! what did you use to generate it? |
Dec 18 |
awarded | Student |
Dec 18 |
asked | Correlations in last-passage percolation |
Dec 17 |
answered | Comparing two measures on trees on $n$ vertices |
Dec 17 |
answered | Stopping time of a Markov chain |
Dec 5 |
comment |
A percolation problem
Yes, I agree ! Very interesting... |
Dec 3 |
comment |
A percolation problem
@Peter: yes, I agree that my comment above is not correct: the model is not strictly equivalent to finding monotone paths. But the argument I outlined initially to show $p_c \ge 1/2$ is still valid, do you agree? ps. sorry to be answering here, but this is the only place I am allowed to put comments... |
Dec 3 |
comment |
A percolation problem
Come to think of it, the question can be rephrased in terms of standard percolation. It's fairly easy to check that the question is equivalent to asking for the existence of monotone paths (by which I mean, paths for which the x and y coordinates are monotone functions of time, e.g. that travel only in the North and East direction). I think the problem is clearer this way. Phrased this way, it is clear that the problem is monotone in $p$ so $p_c$ is well-defined, and moreover it is obvious that $p_c \ge 1/2$. |
Dec 2 |
awarded | Supporter |
Dec 2 |
answered | A percolation problem |