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# Search Results

Results tagged with Search options answers only user 38566
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A stochastic process is a collection of random variables usually indexed by a totally ordered set.

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
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
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
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
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 $A = \{X\,:\,\limsup_{n \to \infty} {1\over n}\s … answered Apr 23 '18 by Martin Hairer The answer is no, pretty much whatever your definition of Feller is. Take for$Q_t$the heat semigroup on$\mathbb{R}$,$X = U = (0,1)$, and for example$f(x,y) = cos(y/x)\exp(-y^2)$. If you assume th … answered Jun 8 '16 by Martin Hairer Yes. First, note that$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 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 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 As Bjørn already mentioned, the CLT shows that the limit is indeed normal with variance$3 T$. A more interesting case is that of $$I = \lim_{\Delta \to 0} \sum_{i=0}^n f(W_{i\Delta},B_{i\Delta})\De … answered Aug 1 by Martin Hairer Here is a counterexample with continuous paths in dimension 2. Let B be a standard Brownian motion and set$$ x_t = {B_t^3\over 1+B_t^4}\;,\quad y_t = {B_t\over 1+B_t^4}\;. $$It clearly has mean ze … answered Dec 11 '13 by Martin Hairer Let's call your state space \mathcal{X}. Then, you want to be able to find a function V \colon \mathcal{X} \to \mathbf{R}_+ and a small set C such that$$ PV \le V - 1 + K \mathbf{1}_C\;,$$whe … answered Mar 23 '18 by Martin Hairer 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 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 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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