User yuri bakhtin - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-19T08:44:31Z http://mathoverflow.net/feeds/user/2968 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/121922/a-strictly-positively-correlated-e-valued-random-variables/122008#122008 Answer by Yuri Bakhtin for A (strictly) positively correlated E-valued random variables. Yuri Bakhtin 2013-02-16T18:35:38Z 2013-02-16T18:35:38Z <p>There are various concepts of dependence. I think, what you are trying to describe is close to the standard notion of associated or positively associated random variables, but maybe not quite. I found the original paper where these were introduced here: <a href="http://www.stat.cmu.edu/~brian/720/week02/esary-proschan-walkup-1967.pdf" rel="nofollow">http://www.stat.cmu.edu/~brian/720/week02/esary-proschan-walkup-1967.pdf</a> This is also tightly related to the so called FKG inequalities.</p> http://mathoverflow.net/questions/121976/derivative-of-a-random-process/122005#122005 Answer by Yuri Bakhtin for Derivative of a random process Yuri Bakhtin 2013-02-16T18:15:06Z 2013-02-16T18:15:06Z <p>Stochastic processes with i.i.d. values in continuous time are not nice. Such a process $X$ with nondegenerate marginals (e.g. Gaussian with nonzero variance) satisfies $\limsup_{t\to t_0} X(t)\ne\liminf_{t\to t_0}X(t)$ for any time $t_0$, so the trajectories cannot even be continuous, leave alone differentiable in a reasonable sense.</p> http://mathoverflow.net/questions/121238/uniform-law-of-large-numbers-for-martingale-difference/121535#121535 Answer by Yuri Bakhtin for Uniform law of large numbers for martingale difference Yuri Bakhtin 2013-02-11T21:57:08Z 2013-02-11T21:57:08Z <p>Even if $\xi$'s are independent (a specific case of your martingale-difference stitation), uniform LLN sometimes holds and sometimes it does not. This type of questions has been studied in Machine learning and, specifically, in Vapnik-Chervonenkis theory. The Glivenko-Cantelli theorem (see wikipedia) describes one situation where the convergence is uniform. More general results can be formulated in terms of VC (Vapnik-Chervonenkis) classes. Perhaps some literature is available for martingales, too.</p> http://mathoverflow.net/questions/121495/do-there-exist-almost-surely-c-infty-smooth-gaussian-random-fields/121531#121531 Answer by Yuri Bakhtin for Do there exist (almost surely) $C^{\infty}$-smooth Gaussian random fields? Yuri Bakhtin 2013-02-11T21:31:46Z 2013-02-11T21:31:46Z <p>I think, you get what you want if you convolve nonsmooth trajectories (like those of a Wiener process) with a $C^\infty$ kernel that is not analytic.</p> http://mathoverflow.net/questions/120512/exit-probability-of-a-brownian-particle/120526#120526 Answer by Yuri Bakhtin for Exit probability of a Brownian particle. Yuri Bakhtin 2013-02-01T15:49:18Z 2013-02-01T22:07:32Z <p>This probability satisfies the heat equation on your interval with zero boundary condition and initial condition being identical 1. Solve it using Fourier series and separation of variables and you will obtain that the probability decays exponentially with exponent being the leading eigenvalue of the problem.</p> <p>Also, see <a href="http://en.wikipedia.org/wiki/Doob" rel="nofollow">http://en.wikipedia.org/wiki/Doob</a>'s_martingale_inequality#Application:_Brownian_motion</p> http://mathoverflow.net/questions/120438/limit-of-a-wiener-integral/120537#120537 Answer by Yuri Bakhtin for Limit of a Wiener integral Yuri Bakhtin 2013-02-01T19:42:32Z 2013-02-01T19:42:32Z <p>Here's one way of dealing with it. Integrate by parts to see that the expression under the sup is $$\Bigl|B(t)-\alpha\int_0^t e^{\alpha(s-t)}B(s)ds\Bigr|\le\alpha\int_0^t e^{\alpha(s-t)}|B(t)-B(s)|ds +e^{-\alpha t}|B(t)|.$$</p> <p>Now the result follows since $B$ is a.s.-bounded and a.s.-Holder on [0,T].</p> http://mathoverflow.net/questions/120363/is-there-a-good-concept-of-a-measurable-fibration/120533#120533 Answer by Yuri Bakhtin for Is there a good concept of a measurable fibration? Yuri Bakhtin 2013-02-01T18:34:57Z 2013-02-01T18:34:57Z <p>This seems to be similar to the concept of measurable partition. One related result is the Rokhlin theorem, see <a href="http://w3.impa.br/~viana/out/rokhlin.pdf" rel="nofollow">http://w3.impa.br/~viana/out/rokhlin.pdf</a> This notion is extensively used in ergodic theory, see books by Sinai (and coauthors).</p> http://mathoverflow.net/questions/118339/is-there-a-general-process-for-conditioning-a-stochastic-process-above-a-boundary/120527#120527 Answer by Yuri Bakhtin for Is there a general process for conditioning a stochastic process above a boundary? Yuri Bakhtin 2013-02-01T16:00:10Z 2013-02-01T16:00:10Z <p>www.math.harvard.edu/~alexb/rm/Doob.pdf</p> http://mathoverflow.net/questions/105176/markov-chains-invariant-measures-and-explosion/105342#105342 Answer by Yuri Bakhtin for Markov chains: invariant measures and explosion Yuri Bakhtin 2012-08-23T18:44:58Z 2012-08-23T18:44:58Z <p>It is incorrect to call the solution of the detailed balance equation that you found an invariant measure. No invariant measure exists for this process since it is transient as you have mentioned.</p> http://mathoverflow.net/questions/85137/how-to-find-critical-points/85142#85142 Answer by Yuri Bakhtin for How to find critical points? Yuri Bakhtin 2012-01-07T17:35:09Z 2012-01-07T17:35:09Z <p>I see the following plan of proof. The trajectories of your system are approximately circles, for large $x^2+y^2$. So an orbit of a point $(x_0,0)$ with large $x_0$ intersects $x$-axis at some point $(x_1,0)$ again after the orbit makes an approximate circle. Then the compact set with the boundary composed of the segment connecting $(x_0,0)$ to $(x_1,0)$ and the piece of the orbit between these points is either forward or backward invariant (depending on whether $x_1>x_0$ or $x_1&lt; x_0$). Therefore, it has a fixed point.</p> http://mathoverflow.net/questions/84216/a-non-degenerate-martingale/84290#84290 Answer by Yuri Bakhtin for A non-degenerate martingale Yuri Bakhtin 2011-12-26T01:07:15Z 2011-12-26T01:07:15Z <p>I have not thought about such questions for a while, but I do not see an immediate mistake in the following reasoning:</p> <p>Let $Y_t=E[\mathrm{sign}(W_1)| \mathcal{F}_t]$. Then it is a bounded martingale, and it also has a continuous modification since $Y_t=E[\mathrm{sign}(W_1) | W_t]$ and $E[\mathrm{sign}(W_1) | W_t=x]$ depends on $t$ and $x$ continuously. Therefore, it admits a representation via stochastic integral, but $P(Y_t=\pm1 )=1/2$.</p> <p>This does not contradict the stochastic representation since $\sigma_t$ in it becomes increasingly large if $t$ is close to 1, but the process is far from $\pm1$. This diffusion pushes the process closer to the boundaries of $[-1,1]$.</p> <p>Is this true, or am I missing something?</p> http://mathoverflow.net/questions/83176/poisson-equationwhy-the-boundary-regularity-of-the-domain-is-important-for-the-r/83191#83191 Answer by Yuri Bakhtin for Poisson Equation:Why the boundary regularity of the domain is important for the regularity of the solution? Yuri Bakhtin 2011-12-11T17:45:47Z 2011-12-11T17:45:47Z <p>I am not sure if this helps when teaching a basic PDE class, but this is certainly a useful understanding:</p> <p>Elliptic problems can be interpreted via diffusion processes. The solution at a point $x$ can be written as expectation of the boundary condition at the (random) exit point for the diffusion emitted from $x$ and associated to the elliptic operator.</p> <p>If the boundary is smooth, then as $x\to x_0\in\partial \Omega$ the exit distribution converges to the Dirac measure at $x_0$, hence regularity of the solution.</p> <p>If the boundary is bad, then the diffusion initiated at a boundary point $x_0$ can, with positive probability, hit the boundary next time at a completely different place, and the exit distribution can be very far from the Dirac measure, hence there is a problem.</p> http://mathoverflow.net/questions/81265/getting-b-t-from-its-local-times-lx-t/81305#81305 Answer by Yuri Bakhtin for Getting $B_t$ from its local times $L^x_t$ Yuri Bakhtin 2011-11-19T04:06:01Z 2011-11-19T04:06:01Z <p>Knowing local times you can derive if the path $\gamma={(t,B_t): t\in[0,T]}$ passes through any rectangle of the following form: $[k/2^n,(k+1)/2^n]\times[j/2^n,(j+1)/2^n]$. For fixed $n$, denote by $G_n$ the union of all these visited rectangles.</p> <p>Since $B_t$ is uniformly continuous on $[0,T]$, we have $\gamma=\bigcap_n G_n$.</p> http://mathoverflow.net/questions/81278/sum-of-a-gaussian-and-an-independent-second-moment-constrained-random-variable/81303#81303 Answer by Yuri Bakhtin for Sum of a Gaussian and an independent second moment constrained random variable Yuri Bakhtin 2011-11-19T03:54:01Z 2011-11-19T03:54:01Z <p>If $X$ has polynomial tails, then so does $Y$.</p> http://mathoverflow.net/questions/79869/is-the-hausdorff-metric-on-sub-sigma-fields-separable/79895#79895 Answer by Yuri Bakhtin for Is the Hausdorff metric on sub-$\sigma$-fields separable? Yuri Bakhtin 2011-11-03T04:51:20Z 2011-11-03T04:51:20Z <p>I suspect that the following is a dense set:</p> <p>For each $n\in\mathbb{N}$ take all sub-algebras of the finite sigma-algebra generated by intervals of the form $[i/2^n,(i+1)/2^n)$, $i=0,\ldots,2^n-1$.</p> http://mathoverflow.net/questions/78849/processes-approximating-a-reflected-brownian-motion/78896#78896 Answer by Yuri Bakhtin for Processes approximating a reflected brownian motion. Yuri Bakhtin 2011-10-23T14:07:54Z 2011-10-23T14:07:54Z <p>It looks like it should converge in distribution in Skorokhod space D. Martingale problem approach (see the book by Ethier &amp; Kurtz on Markov processes) should work.</p> http://mathoverflow.net/questions/76727/what-are-two-independent-uniformly-distributed-random-variables-on-the-unit-inte/76893#76893 Answer by Yuri Bakhtin for What are two independent, uniformly distributed random variables on the unit interval? Yuri Bakhtin 2011-09-30T22:58:18Z 2011-09-30T22:58:18Z <p>Using binary digits of a uniformly distributed random variable on [0,1] is the first step in obtaining a variety of distributions in various spaces.</p> <p>In fact, distributions on almost any reasonable measurable space you can think of are pushforwards of Lebesgue measure on [0,1], see e.g., <a href="http://en.wikipedia.org/wiki/Standard_probability_space" rel="nofollow">http://en.wikipedia.org/wiki/Standard_probability_space</a></p> http://mathoverflow.net/questions/74784/exchangeable-normal-r-v-s/74802#74802 Answer by Yuri Bakhtin for exchangeable normal r.v.s Yuri Bakhtin 2011-09-07T22:11:53Z 2011-09-07T22:11:53Z <p>i.i.d. standard normals seem to work, don't they?</p> http://mathoverflow.net/questions/74545/approximation-to-the-ratio-of-a-gaussian-cdf-to-pdf/74617#74617 Answer by Yuri Bakhtin for Approximation to the ratio of a Gaussian CDF to PDF Yuri Bakhtin 2011-09-05T20:31:39Z 2011-09-05T20:31:39Z <p>Reproducing a lemma from the classic Feller book, first we can write</p> <p>$$(1-3x^{-4})\phi(x)&lt;\phi(x)&lt;(1+x^{-2})\phi(x).$$</p> <p>Integrating this from $x$ to $+\infty$, we obtain</p> <p>$$(x^{-1}-x^{-3})\phi(x)&lt;1-\Phi(x)&lt; x^{-1}\phi(x),$$ </p> <p>so you easily get an approximation rate $x^{-3}\phi(x)$, too.</p> http://mathoverflow.net/questions/74552/a-formal-definition-of-scaling-limits/74608#74608 Answer by Yuri Bakhtin for A formal definition of Scaling Limits? Yuri Bakhtin 2011-09-05T18:52:14Z 2011-09-05T18:52:14Z <p>This is the idea. Suppose that you have a family or sequence of structures of growing complexity (long random walks, realizations of a random field on a large piece of a lattice or in a large continuous domain, large random trees, etc.). You want to understand the behavior of the large structures of your family. Often you want to say that your large random object is similar to a simpler object that you can describe precisely. Since the random walk consisting of 1000 steps is quite different from the "same" random walk consisting of 100000 steps, but you still want to find similarities between them, it makes sense to normalize or rescale your objects appropriately. If you manage to find the right rescaling (it is given by shrinking the time by $n$ and space by $n^{1/2}$ for a standard simple symmetric random walk), then you might discover that thus rescaled (and appropriately embedded into the space of continuous functions or the Skorokhod space) random walk converges in distribution to the Wiener process.</p> <p>So, scaling limits provide approximative descriptions of what your objects look like when "you look at them from a large distance" or "zoom out".</p> <p>At a more formal level, suppose $\xi_n$ is a sequence of random objects in some space $X$ . Suppose $\phi_n$ is a (carefully chosen) sequence of scaling tansformations in $X$. It is hard to say precisely what a scaling transformation is, often it is a linear map depending on $n$ with coefficients decaying in $n$. Often, a (time)-reparametrization of the random objects involved is a part of $\phi_n$. A scaling limit is the distributional limit of $\phi_n(\xi_n)$.</p> <p>Some more comments:</p> <ol> <li><p>One point of view is understanding scaling limits as limiting points for renormalization group.</p></li> <li><p>Papers by P.Major from around 1980 on self-semilarity and renormalization are useful in understanding the concept.</p></li> <li><p>I will take this chance to advertise my own paper on scaling limits for random trees, where I describe what large random trees look like if drawn on the plane and looked at from a large distance. It appeared this year in Markov Processes and Related Fields and is also available at <a href="http://arxiv.org/abs/0909.2283" rel="nofollow">http://arxiv.org/abs/0909.2283</a> The construction and scaling used is different from Aldous's continuum trees (and there are strong connections to superprocesses).</p></li> </ol> http://mathoverflow.net/questions/71837/lyapunov-exponent-and-degree-of-chaos/71854#71854 Answer by Yuri Bakhtin for Lyapunov Exponent and degree of chaos Yuri Bakhtin 2011-08-02T03:30:47Z 2011-08-02T03:30:47Z <p>The (Kolmogorov--Sinai metric) entropy is a measure of chaos in a dynamical system w.r.t. an invariant measure. For a broad class of dynamical systems it is equal to the sum of all positive Lyapunov exponents. I would recommend reading <a href="http://www.scholarpedia.org/article/Pesin_entropy_formula" rel="nofollow">http://www.scholarpedia.org/article/Pesin_entropy_formula</a></p> <p>Quoting this article, </p> <p><i>The content of this formula is that the entropy of a measure is given exactly by the total expansion in the system.</i></p> http://mathoverflow.net/questions/71803/billiard-dynamics-under-gravity/71827#71827 Answer by Yuri Bakhtin for Billiard dynamics under gravity Yuri Bakhtin 2011-08-01T20:39:15Z 2011-08-01T21:26:11Z <p>This is not an answer to the specific question asked, but I cannot resist sharing this with you.</p> <p>There is a beautiful result of Chernov and Dolgopyat, see <a href="http://www-users.math.umd.edu/~dmitry/galton11new.pdf" rel="nofollow">http://www-users.math.umd.edu/~dmitry/galton11new.pdf</a> , on the billiard in Galton's board in the presence of gravity. The Galton board is a well-known device indended to reproduce the binomial distribution (see, e.g., <a href="http://mathworld.wolfram.com/GaltonBoard.html" rel="nofollow">http://mathworld.wolfram.com/GaltonBoard.html</a>). </p> <p>Chernov and Dolgopyat studied the situation when the board is extended periodically and indefinitely. Collisions with the obstacles are assumed elastic, so that no loss of energy occurs. The result is that despite the fact that gravity is directed down, the ball or particle will eventually reach the same horizontal level as it is at initially. Moreover, it reaches a small neighborhood of its initial state infinitely many times. This happens for a.e. initial condition.</p> http://mathoverflow.net/questions/68680/counterexamples-in-pde/68783#68783 Answer by Yuri Bakhtin for Counterexamples in PDE Yuri Bakhtin 2011-06-25T11:39:46Z 2011-06-25T11:39:46Z <p>In his <a href="http://www.mathnet.ru/php/getFT.phtml?jrnid=sm&amp;paperid=6410&amp;volume=42&amp;year=1935&amp;issue=2&amp;fpage=199&amp;what=fullt&amp;option_lang=eng" rel="nofollow"> classic 1935 work</a> Tikhonov showed that the Cauchy problem for the heat equation with 0 initial data has nonzero solutions. He also identified uniqueness classes, of course.</p> http://mathoverflow.net/questions/66990/shortest-path-in-plane/66992#66992 Answer by Yuri Bakhtin for Shortest Path in Plane Yuri Bakhtin 2011-06-05T22:19:55Z 2011-06-05T22:47:14Z <p>In this generality, the shortest path is not necessarily well defined, i.e., it can happen that the minimum is not attained, approximations to the minimizer being paths going further and further away to infinity.</p> <p>Hence, no algorithm is guaranteed to terminate correctly in finite time.</p> http://mathoverflow.net/questions/66892/is-the-laplacian-on-a-manifold-the-limit-of-graph-laplacians/66993#66993 Answer by Yuri Bakhtin for Is the Laplacian on a manifold the limit of graph Laplacians? Yuri Bakhtin 2011-06-05T22:26:40Z 2011-06-05T22:26:40Z <p>Koltchinskii an Gine study how discrete Laplacians constructed for randomly sampled points on a manifold approximate the true Laplacian:</p> <p><a href="http://projecteuclid.org/DPubS?service=UI&amp;version=1.0&amp;verb=Display&amp;handle=euclid.lnms/1196284116" rel="nofollow">http://projecteuclid.org/DPubS?service=UI&amp;version=1.0&amp;verb=Display&amp;handle=euclid.lnms/1196284116</a></p> http://mathoverflow.net/questions/66663/covariance-sign/66836#66836 Answer by Yuri Bakhtin for Covariance sign Yuri Bakhtin 2011-06-03T17:17:11Z 2011-06-03T17:47:40Z <p>There is a short proof based on coupling. Let $Y$ be a r.v. with the same distribution as $X$ and independent of $X$. Then $f(X)-f(Y)$ and $g(X)-g(Y)$ are centered r.v.'s.</p> <p>On the one hand, $Cov(f(X)-f(Y), g(X)-g(Y))= \mathsf{E} (f(X)-f(Y)) (g(X)-g(Y))\ge 0$. (actually $>0$ unless $g(X)$ or $f(X)$ is constant a.s.)</p> <p>On the other hand, $Cov(f(X)-f(Y), g(X)-g(Y))=$ </p> <p>$= Cov(f(X),g(X))-Cov(f(X),g(Y))-Cov(f(Y),g(X))+Cov(f(Y),g(Y))=2Cov(f(X),g(X))$</p> <p>Combining these two lines we get $2Cov(f(X),g(X))>0$.</p> <p>A couple of relevant remarks: 1.This statement is also due to Chebyshev. 2. The property can be formulated as "One r.v. forms an associated family". In general, r.v.'s $X_1,\ldots, X_n$ are called associated if for any two bounded coordinatewise nondecreasing functions $f$ and $g$, $Cov(f(X_1,\ldots,X_n),g(X_1,\ldots,X_n))\ge 0$. There is a recent book by Bulinskiy and Shashkin on association.</p> http://mathoverflow.net/questions/64314/levy-theorem-for-signed-measures/64564#64564 Answer by Yuri Bakhtin for Levy theorem for signed measures Yuri Bakhtin 2011-05-11T04:01:08Z 2011-05-11T16:43:30Z <p>Thanks to Yemon, I see that my initial counterexample that I posted here was nonsense.</p> <p>Here is a correct counterexample for the proposed statement with no additional assumptions.</p> <p>For any $n\ge 2$, let $\mu_n=\delta_{n+1/n}-\delta_n$ (two atoms, one with positive mass and one with negative mass placed at points $n+1/n$ and $n$ respectively). </p> <p>Then the Fourier transform is $F(t)=e^{it(n+1/n)}-e^{itn}=e^{itn}(e^{it/n}-1)\to 0$. </p> <p>Zero is the Fourier transform of the zero measure, but $\mu_n$ does not converge to zero measure weakly. To see that, take a continuous bounded function $f$ such that $f(n)=-1$ and $f(n+1/n)=1$ for all $n\ge 2$. Then, for all $n\ge 2$, we have $\int fd\mu_n=2$ and $2$ does not converge to $0$, the integral of $f$ w.r.t. zero measure.</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/48971/has-anyone-found-a-way-to-determine-the-invariant-measure-of-a-one-dimensional-ju/49004#49004 Answer by Yuri Bakhtin for Has anyone found a way to determine the invariant measure of a one-dimensional jump-diffusion? Yuri Bakhtin 2010-12-11T01:46:00Z 2010-12-11T01:46:00Z <p>The general procedure is the following. Take the generator $L$ of the semigroup associated to your process. Then find its dual $L^*$. The latter governs the evolution of 1-dim distributions (via forward Kolmogorov equation), and an invariant density $p$ satisfies $L^*p=0$, should be positive and integrate to 1.</p> <p>I suspect all this can be found in the Ethier&amp;Kurtz book on Markov processes, but I do not have it at hand right now.</p> http://mathoverflow.net/questions/43057/looking-for-a-collection-of-entry-level-proofs/43220#43220 Answer by Yuri Bakhtin for Looking for a collection of entry level proofs Yuri Bakhtin 2010-10-22T18:51:42Z 2010-10-22T18:51:42Z <p>The classic <a href="http://books.google.com/books?id=7_Z27SvIGKAC&amp;printsec=frontcover&amp;dq=isbn%3A082182693X&amp;cd=1#v=onepage&amp;q&amp;f=false" rel="nofollow">"Foundations of Analysis" by Edmund Landau</a>. It pedantically and very carefully derives elementary properties of integers, rationals, etc., from Peano axioms. Or is it too formal?</p> http://mathoverflow.net/questions/123538/convergence-of-a-weakly-dependent-point-process Comment by Yuri Bakhtin Yuri Bakhtin 2013-03-04T15:51:45Z 2013-03-04T15:51:45Z You must be talking to Kostya Khanin. He should have a text with my solution of this problem written in 2009. http://mathoverflow.net/questions/120438/limit-of-a-wiener-integral/120537#120537 Comment by Yuri Bakhtin Yuri Bakhtin 2013-02-18T04:44:41Z 2013-02-18T04:44:41Z @Paul: You are welcome. http://mathoverflow.net/questions/105176/markov-chains-invariant-measures-and-explosion/105342#105342 Comment by Yuri Bakhtin Yuri Bakhtin 2012-08-28T12:14:57Z 2012-08-28T12:14:57Z @Nathanael Berestycki: A measure is invariant if it is preserved by the Markov semigroup (i.e., if you start the process with this initial distribution, then for any positive time the distribution stays the same). This measure is not preserved since the mass is driven to infinity and for any finite subset A of S, the probability that the process is in A goes to zero as time goes to infinity. http://mathoverflow.net/questions/95740/irreducibility-and-minorization-condition Comment by Yuri Bakhtin Yuri Bakhtin 2012-05-04T00:46:10Z 2012-05-04T00:46:10Z Isn't $s\nu$ an irreducibility measure ? http://mathoverflow.net/questions/85137/how-to-find-critical-points/85142#85142 Comment by Yuri Bakhtin Yuri Bakhtin 2012-01-07T18:36:05Z 2012-01-07T18:36:05Z @Mark Sapir: of course, you are right @Bass_1: Because the system is a perturbation of $x'=y, y'=-x$. To be honest, I think now that this question is hardly appropriate here, and it looks like you haven't given it a thought. http://mathoverflow.net/questions/84216/a-non-degenerate-martingale/84290#84290 Comment by Yuri Bakhtin Yuri Bakhtin 2011-12-26T15:08:58Z 2011-12-26T15:08:58Z Yes, I missed the boundedness requirement in the question, sorry. The example given by James is great, I just +1'ed it. http://mathoverflow.net/questions/81265/getting-b-t-from-its-local-times-lx-t/81305#81305 Comment by Yuri Bakhtin Yuri Bakhtin 2011-11-22T16:46:36Z 2011-11-22T16:46:36Z @The Bridge: Sorry, I was answering a wrong question. I do not think there's anything better than the occupation time formula that you already have. http://mathoverflow.net/questions/81265/getting-b-t-from-its-local-times-lx-t/81305#81305 Comment by Yuri Bakhtin Yuri Bakhtin 2011-11-21T21:24:25Z 2011-11-21T21:24:25Z @The Bridge: To be honest, I do not know how to explain my argument more precicely in a short message. It seems obvious that the occupation times do not change if you cut your trajectory in pieces, move them rigidly around (only along time axis) and reglue them again. http://mathoverflow.net/questions/81265/getting-b-t-from-its-local-times-lx-t/81305#81305 Comment by Yuri Bakhtin Yuri Bakhtin 2011-11-21T15:57:45Z 2011-11-21T15:57:45Z @The Bridge: Ah, $t$ is fixed! Then no. Think about a trajectory that passes through a point 3 times before $t$ with 2 excursions between them. If you exchange the order of those two excursions you get a different trajectory that has same local times at terminal point $t$. This trajectory is &quot;as probable as the original one&quot;. http://mathoverflow.net/questions/79869/is-the-hausdorff-metric-on-sub-sigma-fields-separable/79895#79895 Comment by Yuri Bakhtin Yuri Bakhtin 2011-11-03T18:28:19Z 2011-11-03T18:28:19Z I see. I was sure wrong. http://mathoverflow.net/questions/77952/name-this-periodic-tiling Comment by Yuri Bakhtin Yuri Bakhtin 2011-10-12T20:49:03Z 2011-10-12T20:49:03Z This is essentially hexagonal tiling, isn't it? http://mathoverflow.net/questions/74784/exchangeable-normal-r-v-s/74802#74802 Comment by Yuri Bakhtin Yuri Bakhtin 2011-09-08T22:33:20Z 2011-09-08T22:33:20Z @Michael Hardy: I see. The words &quot;any&quot; and &quot;could&quot; in your point 4 are misleading. Btw, Halmos in his &quot;How to write mathematics&quot; advises against using &quot;any&quot; in mathematical writing. http://mathoverflow.net/questions/71803/billiard-dynamics-under-gravity/71827#71827 Comment by Yuri Bakhtin Yuri Bakhtin 2011-08-01T21:24:05Z 2011-08-01T21:24:05Z @Joseph O'Rourke: yes, thanks for pointing out the misprint. I will correct it. http://mathoverflow.net/questions/66990/shortest-path-in-plane/66992#66992 Comment by Yuri Bakhtin Yuri Bakhtin 2011-06-07T16:37:23Z 2011-06-07T16:37:23Z of course, one needs an infinite tesselation for it. Suppose we are given unit squares centered at points of integer lattice. Let the time to cross the square centered at (1,n) be 1/(|n|+1). Let the time to cross the square centerd at (0,n) or (2,n) be 1/100^(n+1). Then to go from (0,0) to (2,0) it is better to go far to cross the ridge at the low values of 1/(|n|+1). The further you go the better off you are. http://mathoverflow.net/questions/64314/levy-theorem-for-signed-measures/64564#64564 Comment by Yuri Bakhtin Yuri Bakhtin 2011-05-11T22:01:36Z 2011-05-11T22:01:36Z @Yemon Choi: I think you are right.