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

      5
      votes
      The answer is no to both questions, here are counterexamples. Take for $X$ a sequence of i.i.d. Bernoulli ($\pm 1$) random variables and take for $Z$ an independent sequence. To construct a counterexa …
      answered Mar 30 '14 by Martin Hairer
      9
      votes
      This is true, but the limiting process is driven by a Brownian motion $B$ which is different from $W$. To prove this, use first the Dambis-Dubins-Schwarz representation of a continuous martingale (see …
      answered Sep 19 '14 by Martin Hairer
      4
      votes
      In the case of a diffusion, (1) is implied for example by having bounded coefficients. This follows immediately from applying BDG to $X_t-x$ and doesn't require (2) which is much harder to get. Note …
      answered Sep 5 '17 by Martin Hairer
      7
      votes
      Here is a not so trivial example: take for $P_t$ the transition semigroup for the SDE $$ dx = -x^3\,dt + dW\;, $$ on $\mathbf{R}$. This is not Feller in your sense because solutions come in from infin …
      answered Apr 20 '15 by Martin Hairer
      8
      votes
      No, this is not enough. Take $Y_n$ i.i.d. Gaussians, $r$ an independent Bernoulli taking values in $\{-1,1\}$, and set $X_n = r + Y_n$. Then the events $A = \{X\,:\,\limsup_{n \to \infty} {1\over n}\s …
      answered Apr 23 '18 by Martin Hairer
      1
      vote
      The answer is no, pretty much whatever your definition of Feller is. Take for $Q_t$ the heat semigroup on $\mathbb{R}$, $X = U = (0,1)$, and for example $f(x,y) = cos(y/x)\exp(-y^2)$. If you assume th …
      answered Jun 8 '16 by Martin Hairer
      2
      votes
      Yes. First, note that $P_t$ is strongly continuous on $C(\bar D)$ so that, for every $f \in C(\bar D)$, $P_t f \to f$ in $C(\bar D)$ as $t \to 0$. For every $t > 0$, $P_t f$ belongs to every Sobolev s …
      answered Oct 3 '13 by Martin Hairer
      3
      votes
      Take the solution to $$ dx = \tanh^3(x)\,dt + \sigma x \, dW\;. $$ For $\sigma > 0$, solutions diverge if you start anywhere except at $0$. For any $\sigma \neq 0$ on the other hand, solutions converg …
      answered Feb 11 by Martin Hairer
      3
      votes
      In general, it can self-interesect, as can be seen in the example of the simple rotation $$ f(x,y) = (-y,x)\;,\qquad \mathcal{R} f(x,y) = (x,y)\;. $$ In this case, the deterministic solution turns aro …
      answered Feb 10 '14 by Martin Hairer
      4
      votes
      As Bjørn already mentioned, the CLT shows that the limit is indeed normal with variance $3 T$. A more interesting case is that of $$ I = \lim_{\Delta \to 0} \sum_{i=0}^n f(W_{i\Delta},B_{i\Delta})\De …
      answered Aug 1 by Martin Hairer
      2
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      Here is a counterexample with continuous paths in dimension 2. Let $B$ be a standard Brownian motion and set $$ x_t = {B_t^3\over 1+B_t^4}\;,\quad y_t = {B_t\over 1+B_t^4}\;. $$ It clearly has mean ze …
      answered Dec 11 '13 by Martin Hairer
      4
      votes
      Let's call your state space $\mathcal{X}$. Then, you want to be able to find a function $V \colon \mathcal{X} \to \mathbf{R}_+$ and a small set $C$ such that $$ PV \le V - 1 + K \mathbf{1}_C\;, $$ whe …
      answered Mar 23 '18 by Martin Hairer
      3
      votes
      This question has recently been addressed in detail in the article "Averaging along irregular curves and regularisation of ODEs" by Catellier and Gubinelli. They give a purely analytical condition on …
      answered Mar 20 '15 by Martin Hairer
      5
      votes
      Yes, RPT allows you to define a notion of stochastic integral against fBm for a class of integrands that is larger than what Young's theory allows. Assuming that you're really interested in solving SD …
      answered Jul 6 '18 by Martin Hairer
      2
      votes
      Here is the long version of my comment. First, let us show that if $X$ is any $1D$ strong Markov process with $X_0 = 0$ which is symmetric under sign inversion (i.e the Markov semigroup commutes with …
      answered Dec 15 '13 by Martin Hairer

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