User yuri bakhtin - MathOverflow

Answer by Yuri Bakhtin for A (strictly) positively correlated E-valued random variables.

2013-02-16T18:35:38Z

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: http://www.stat.cmu.edu/~brian/720/week02/esary-proschan-walkup-1967.pdf This is also tightly related to the so called FKG inequalities.

Answer by Yuri Bakhtin for Derivative of a random process

2013-02-16T18:15:06Z

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.

Answer by Yuri Bakhtin for Uniform law of large numbers for martingale difference

2013-02-11T21:57:08Z

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.

Answer by Yuri Bakhtin for Do there exist (almost surely) $C^{\infty}$-smooth Gaussian random fields?

2013-02-11T21:31:46Z

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.

Answer by Yuri Bakhtin for Exit probability of a Brownian particle.

2013-02-01T15:49:18Z

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.

Also, see http://en.wikipedia.org/wiki/Doob's_martingale_inequality#Application:_Brownian_motion

Answer by Yuri Bakhtin for Limit of a Wiener integral

2013-02-01T19:42:32Z

Here's one way of dealing with it. Integrate by parts to see that the expression under the sup is $$ \Bigl|B(t)-\alpha\int_0^t e^{\alpha(s-t)}B(s)ds\Bigr|\le\alpha\int_0^t e^{\alpha(s-t)}|B(t)-B(s)|ds +e^{-\alpha t}|B(t)|. $$

Now the result follows since $B$ is a.s.-bounded and a.s.-Holder on [0,T].

Answer by Yuri Bakhtin for Is there a good concept of a measurable fibration?

2013-02-01T18:34:57Z

This seems to be similar to the concept of measurable partition. One related result is the Rokhlin theorem, see http://w3.impa.br/~viana/out/rokhlin.pdf This notion is extensively used in ergodic theory, see books by Sinai (and coauthors).

Answer by Yuri Bakhtin for Is there a general process for conditioning a stochastic process above a boundary?

2013-02-01T16:00:10Z

www.math.harvard.edu/~alexb/rm/Doob.pdf

Answer by Yuri Bakhtin for Markov chains: invariant measures and explosion

2012-08-23T18:44:58Z

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.

Answer by Yuri Bakhtin for How to find critical points?

2012-01-07T17:35:09Z

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 $x_1< x_0$). Therefore, it has a fixed point.

Answer by Yuri Bakhtin for A non-degenerate martingale

2011-12-26T01:07:15Z

I have not thought about such questions for a while, but I do not see an immediate mistake in the following reasoning:

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 $P(Y_t=\pm1 )=1/2$.

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 $[-1,1]$.

Is this true, or am I missing something?

Answer by Yuri Bakhtin for Poisson Equation:Why the boundary regularity of the domain is important for the regularity of the solution?

2011-12-11T17:45:47Z

I am not sure if this helps when teaching a basic PDE class, but this is certainly a useful understanding:

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.

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.

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.

Answer by Yuri Bakhtin for Getting $B_t$ from its local times $L^x_t$

2011-11-19T04:06:01Z

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.

Since $B_t$ is uniformly continuous on $[0,T]$, we have $\gamma=\bigcap_n G_n$.

Answer by Yuri Bakhtin for Sum of a Gaussian and an independent second moment constrained random variable

2011-11-19T03:54:01Z

If $X$ has polynomial tails, then so does $Y$.

Answer by Yuri Bakhtin for Is the Hausdorff metric on sub-$\sigma$-fields separable?

2011-11-03T04:51:20Z

I suspect that the following is a dense set:

For each $n\in\mathbb{N}$ take all sub-algebras of the finite sigma-algebra generated by intervals of the form $[i/2^n,(i+1)/2^n)$, $i=0,\ldots,2^n-1$.

Answer by Yuri Bakhtin for Processes approximating a reflected brownian motion.

2011-10-23T14:07:54Z

It looks like it should converge in distribution in Skorokhod space D. Martingale problem approach (see the book by Ethier & Kurtz on Markov processes) should work.

Answer by Yuri Bakhtin for What are two independent, uniformly distributed random variables on the unit interval?

2011-09-30T22:58:18Z

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.

In fact, distributions on almost any reasonable measurable space you can think of are pushforwards of Lebesgue measure on [0,1], see e.g., http://en.wikipedia.org/wiki/Standard_probability_space

Answer by Yuri Bakhtin for exchangeable normal r.v.s

2011-09-07T22:11:53Z

i.i.d. standard normals seem to work, don't they?

Answer by Yuri Bakhtin for Approximation to the ratio of a Gaussian CDF to PDF

2011-09-05T20:31:39Z

Reproducing a lemma from the classic Feller book, first we can write

$$ (1-3x^{-4})\phi(x)<\phi(x)<(1+x^{-2})\phi(x). $$

Integrating this from $x$ to $+\infty$, we obtain

$$(x^{-1}-x^{-3})\phi(x)<1-\Phi(x)< x^{-1}\phi(x),$$

so you easily get an approximation rate $x^{-3}\phi(x)$, too.

Answer by Yuri Bakhtin for A formal definition of Scaling Limits?

2011-09-05T18:52:14Z

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.

So, scaling limits provide approximative descriptions of what your objects look like when "you look at them from a large distance" or "zoom out".

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 $\phi_n(\xi_n)$.

Some more comments:

One point of view is understanding scaling limits as limiting points for renormalization group.
Papers by P.Major from around 1980 on self-semilarity and renormalization are useful in understanding the concept.
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 http://arxiv.org/abs/0909.2283 The construction and scaling used is different from Aldous's continuum trees (and there are strong connections to superprocesses).

Answer by Yuri Bakhtin for Lyapunov Exponent and degree of chaos

2011-08-02T03:30:47Z

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 http://www.scholarpedia.org/article/Pesin_entropy_formula

Quoting this article,

The content of this formula is that the entropy of a measure is given exactly by the total expansion in the system.

Answer by Yuri Bakhtin for Billiard dynamics under gravity

2011-08-01T20:39:15Z

This is not an answer to the specific question asked, but I cannot resist sharing this with you.

There is a beautiful result of Chernov and Dolgopyat, see http://www-users.math.umd.edu/~dmitry/galton11new.pdf , 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., http://mathworld.wolfram.com/GaltonBoard.html).

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.

Answer by Yuri Bakhtin for Counterexamples in PDE

2011-06-25T11:39:46Z

In his classic 1935 work Tikhonov showed that the Cauchy problem for the heat equation with 0 initial data has nonzero solutions. He also identified uniqueness classes, of course.

Answer by Yuri Bakhtin for Shortest Path in Plane

2011-06-05T22:19:55Z

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.

Hence, no algorithm is guaranteed to terminate correctly in finite time.

Answer by Yuri Bakhtin for Is the Laplacian on a manifold the limit of graph Laplacians?

2011-06-05T22:26:40Z

Koltchinskii an Gine study how discrete Laplacians constructed for randomly sampled points on a manifold approximate the true Laplacian:

http://projecteuclid.org/DPubS?service=UI&version=1.0&verb=Display&handle=euclid.lnms/1196284116

Answer by Yuri Bakhtin for Covariance sign

2011-06-03T17:17:11Z

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.

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.)

On the other hand, $Cov(f(X)-f(Y), g(X)-g(Y))=$

$= Cov(f(X),g(X))-Cov(f(X),g(Y))-Cov(f(Y),g(X))+Cov(f(Y),g(Y))=2Cov(f(X),g(X))$

Combining these two lines we get $2Cov(f(X),g(X))>0$.

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.

Answer by Yuri Bakhtin for Levy theorem for signed measures

2011-05-11T04:01:08Z

Thanks to Yemon, I see that my initial counterexample that I posted here was nonsense.

Here is a correct counterexample for the proposed statement with no additional assumptions.

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).

Then the Fourier transform is $F(t)=e^{it(n+1/n)}-e^{itn}=e^{itn}(e^{it/n}-1)\to 0$.

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.

Answer by Yuri Bakhtin for Markov random field with continuous index set

2011-03-13T19:08:37Z

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 $D$.

From Rozanov's book http://www.amazon.com/Markov-Random-Fields-Applications-Mathematics/dp/0387907084/ref=sr_1_1?ie=UTF8&s=books&qid=1300042943&sr=8-1 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).

Answer by Yuri Bakhtin for Has anyone found a way to determine the invariant measure of a one-dimensional jump-diffusion?

2010-12-11T01:46:00Z

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.

I suspect all this can be found in the Ethier&Kurtz book on Markov processes, but I do not have it at hand right now.

Answer by Yuri Bakhtin for Looking for a collection of entry level proofs

2010-10-22T18:51:42Z

The classic "Foundations of Analysis" by Edmund Landau. It pedantically and very carefully derives elementary properties of integers, rationals, etc., from Peano axioms. Or is it too formal?

Comment by Yuri Bakhtin

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.

Comment by Yuri Bakhtin

2013-02-18T04:44:41Z

@Paul: You are welcome.

Comment by Yuri Bakhtin

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.

Comment by Yuri Bakhtin

2012-05-04T00:46:10Z

Isn't $s\nu$ an irreducibility measure ?

Comment by Yuri Bakhtin

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.

Comment by Yuri Bakhtin

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.

Comment by Yuri Bakhtin

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.

Comment by Yuri Bakhtin

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.

Comment by Yuri Bakhtin

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 "as probable as the original one".

Comment by Yuri Bakhtin

2011-11-03T18:28:19Z

I see. I was sure wrong.

Comment by Yuri Bakhtin

2011-10-12T20:49:03Z

This is essentially hexagonal tiling, isn't it?

Comment by Yuri Bakhtin

2011-09-08T22:33:20Z

@Michael Hardy: I see. The words "any" and "could" in your point 4 are misleading. Btw, Halmos in his "How to write mathematics" advises against using "any" in mathematical writing.

Comment by Yuri Bakhtin

2011-08-01T21:24:05Z

@Joseph O'Rourke: yes, thanks for pointing out the misprint. I will correct it.

Comment by Yuri Bakhtin

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.

Comment by Yuri Bakhtin

2011-05-11T22:01:36Z

@Yemon Choi: I think you are right.