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A stochastic process is a collection of random variables usually indexed by a totally ordered set.

**2**

votes

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 wi …

answered Feb 1 '13 by Yuri Bakhtin

**0**

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

answered Sep 7 '11 by Yuri Bakhtin

**2**

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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.

answered Oct 23 '11 by Yuri Bakhtin

**2**

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Notice that enumeration of points is not given a priori. The Poisson point configuration is a set of points with no order on them. It also can be viewed as a measure that puts mass 1 at each configura …

answered Dec 19 '13 by Yuri Bakhtin

**1**

vote

Another approach is to try to use Theorem 4.1 from the section on diffusion approximation of the book of Ethier and Kurtz on Markov processes (and their convergence). Of course, this will give weak co …

answered Sep 23 '14 by Yuri Bakhtin

**0**

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

answered Feb 1 '13 by Yuri Bakhtin

**1**

vote

This paper might be relevant: http://arxiv.org/abs/0908.4450

answered Apr 15 '14 by Yuri Bakhtin

**0**

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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.

answered Aug 23 '12 by Yuri Bakhtin

**2**

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The first part of this answer was written when I overlooked the zero mean condition that takes this question to another level of difficulty.
If you want an example where things are computable precise …

answered Dec 11 '13 by Yuri Bakhtin

**0**

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A more useful definition of a Poisson point field is:
Let $\mu$ be a (finite or infinite) measure on a space $E$ with a sigma-algebra $\mathcal{B}$. To each set $A\in\mathcal{B}$ with $\mu(A)<\infty …

answered Sep 16 '13 by Yuri Bakhtin

**3**

votes

This is a specific case of Doob's optional sampling theorem

answered Jul 6 '13 by Yuri Bakhtin

**8**

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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)|d …

answered Feb 1 '13 by Yuri Bakhtin

**3**

votes

A simple counterexample is a process that, starting at zero moves with constant velocity 1 (with probability $1/2$) or with constant velocity -1 (with probability $1/2$).
There are various subtletie …

answered Dec 14 '13 by Yuri Bakhtin

**2**

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You can have weak solutions with quite low regularity. In the compact case, continuity of drift and diffusion is sufficient for existence of a weak solution, and in the non-compact case you will need …

answered Jul 6 '13 by Yuri Bakhtin

**2**

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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.

answered Feb 11 '13 by Yuri Bakhtin