User stephan sturm - MathOverflow

Answer by Stephan Sturm for Good books on stochastic partial differential equations?

2013-03-13T01:45:17Z

On the analytical side, I like a lot the book A Concise Course on Stochastic Partial Differential Equations by Prevot and Roeckner. It is a very well written introduction to SPDEs.

Besides this, I know a couple of people who are very fond of Stochastic Equations in Infinite Dimensions by da Prato and Zabczyk.

Answer by Stephan Sturm for Reflected Brownian Motion

2013-03-01T03:12:28Z

As your reflected Brownian motion is nothing else then the absolute value of a Brownian motion you have $$ \{t \in [0,T] : Y_t = 0\} = \{t \in [0,T] : \vert W_t \vert= 0\} = \{t \in [0,T] : W_t = 0\}$$ which is a Lebesgue null set (however, an uncountable one, as remarked in the comments). In point of view of distribution, $Y_t$ has the same law as the difference of the Maximum process of a Brownian motion and the Brownian itself, $\max_{0 \leq s \leq t} W_s - W_t$. For details see, e.g., Karatzas & Shreve, Brownian Motion and Stochastic Calculus, Chapter 3.6.C, p.210ff. More general, your problem is known as Skorohod problem (which should not be confused with the Skorohod embedding problem), and I believe there is a fair amount of queueing theoretic literature about.

Answer by Stephan Sturm for On martingale representation theorem

2013-02-12T05:37:57Z

No, but under some regularity conditions you might represent $a(t)$ in terms of Malliavin calculus by means of the Clark-Ocone formula (see e.g. the Lecture notes by Eulalia Nualart, Section 1.5.3.)

Answer by Stephan Sturm for Representation theorem for continuous uniformly integrable martingales

2012-09-10T21:29:39Z

I think question 1) is reasonably answered by Wolfgang Loehr in his comment. To get a counterexample for your claims in 2), just set $a_u=W_u^2-u$ for your Brownian motion. Ito's formula gives you the martingale representation $$ W_t^2-t = 0+\int_0^t 2W_s dW_s = 0 + \int_0^t \frac{2W_s}{W_s^2-s}(W_s^2-s) dW_s$$ Thus for in your terms you have $v_s(u) = \frac{2W_s}{W_s^2-s}$. Trying to integrate this you get as antiderivative an exponential integral with pole at $\sqrt{s}$, thus $v_s(u)$ is not integrable.

On the positive side you have always (even when you have just a local martingale) that $v_s(u) M_s(u)$ is predictable locally in $L^2$ (cf. Revuz/Yor, Continuous Martingales and Brownian Motion, Theorem V.3.4) and it is square integrable if your $a_u$ is square integrable (Proposition V.3.2). Under some regularity conditions in terms of Malliavin calculus you may calculate $v_s(u) M_s(u)$ even explicitly by means of the Clark-Ocone formula (see e.g. the Lecture notes of Eulalia Nualart, Section 1.5.3.)

Answer by Stephan Sturm for Interesting mathematical documentaries

2012-06-19T19:43:47Z

La lettre scellée du soldat Doblin about Wolfgang Döblin / Vincent Doblin.

Here the synopsis from IMDB: "When France surrendered in 1940 and German soldiers showed up in the Vosgian village of Housseras, an unknown French foot soldier burned his papers and killed himself in a farmer's barn. Four years later he was identified as "soldat Doblin, Vincent". In fact, he was none other than the mathematician Wolfgang Doeblin, son of the famous German novelist Alfred Döblin ("Berlin Alexanderplatz") who was forced to flee Nazi Germany with his family in 1933. A French citizen since October 1936, Wolfgang Doeblin carried on his research into probability theory during his military service and even during the hardships of the "Phoney War" in the winter of 1939-40. In February 1940, four months before his death at the age of 25, he sent his most important manuscripts ("About the Kolmogoroff Equation") as a "sealed envelope" to the Academy of Science in Paris, where they were kept in safe custody for 60 years. Wolfgang Doeblin's short and dramatic life story, almost forgotten, was finally brought into the limelight when the "sealed envelope" was opened in May 2000. Far ahead of their time, his groundbreaking contributions to the theory of random processes place Wolfgang Doeblin among the major innovators of probability, the "mathematics of randomness". Mathematical models for evaluation of chances and risks went on to gain major importance in numerous domains of modern science, in everyday life and especially in contemporary financial mathematics."

Answer by Stephan Sturm for Stochastic processes having Markov kernels

2012-05-22T14:27:19Z

No, that both processes have the same one-dimensional marginals is not sufficient. In contrary, when $X$ is an arbitrary elliptic Itô-process, you can always find a Markov process with the same marginals. Cf. I. Gyöngy, Mimicking the One-Dimensional Marginal Distributions of Processes Having an Itô Differential. Probab.Theory Relat.Fields 71(4), 501–516 (1986)

Answer by Stephan Sturm for Properties preserved under passage to augmented filtration

2012-05-19T06:03:13Z

Hi lpdbw,

I think this is a very interesting questions, here at least a partial answer; It depends heavily on the little word "strong" in parentheseis.

Assuming that $(X_t)$ is strong Markov, the answer to your questions seems to be "yes": The completed filtration of an $\mathbb{R}^d$ -valued strong Markov process is already right-continuous (cf. Theorem 2.7.7. of Karatzas/Shreve: Brownian Motion and Stochastic Calculus - note that they use a different terminology for augmented/completed). And just augmenting the filtration does not affect the martingale property (these are just sets of mass zero).

If you look however on Markov processes which does not necessarily has the strong Markov property, the answer seems to be negative as seen in the following counterexample: Let $(W_t)$ be a Brownian motion in its natural filtration and $\xi$ a random variable independent of the Brownian filtration, $\mathbb{P}[\xi =1] = \mathbb{P}[\xi =2] =\frac{1}{2}$ , and define the process $(X_t)$ as $$ X_t = \int_0^t \xi \; \;dW_s. $$

This process is a Markov process and a martingale in it's natural filtration $\mathcal{F}_{t}$ , but just passing to the right continuous filtration $\mathcal{F}_{t+} = \bigcap_{s>t}\mathcal{F}_s$ destroys the Markov property. Note that for $t>0$ we have $\mathcal{F}_t =\mathcal{F}_{t+} = \sigma(W_s; s\leq t) \vee \sigma(\xi)$ , however at $t=0$ they are fundamentally different: $\mathcal{F}_0$ is trivial whereas $\mathcal{F}_{0+} = \sigma(\xi)$ . Adding additional null sets changes again nothing.

Answer by Stephan Sturm for Solving SDE's on subsets of $R^n$.

2012-03-14T19:25:14Z

I think the crucial question is how you (or the authors you refer to) interpret the word "Lipschitz". If $b$ and $\sigma$ are ${\it \mbox{globally}}$ Lipschitz, you may indeed extend them by your favorite method to the whole space. Of course, the solutions may differ, if you use different Lipschitz extensions. However, they agree up to the first exit time of $D$, or equivalently, the processes stopped at $\tau_D$ agree (and I assume that is everything what the authors care about).

If $b$ and $\sigma$ are merely ${\it \mbox{locally}}$ Lipschitz, a Lipschitz continuation may not exist. If you have merely continuous coefficients, you may prove local existence of solutions, but these solutions may only exist locally (up to an explosion time). I think the usual way is to make a one-point compactification of $D$ by adjoining an additional 'cemetery' state $\Delta$ to $D$ and look on the SDE on $D \cup \{\Delta\}$.

A classical reference on this is Chapter III of the book "Stochastic Differential Equations and Diffusion Processes" by Ikeda and Watanabe.

Answer by Stephan Sturm for Any reference on Brownian Motion continuity

2012-02-06T23:22:56Z

The continuity (or more precisely the existence of an almost sure continuous modification) is a direct consequence of your condition 3 by Kolmogorov's continuity criterion(or more generally the Theorem of Kolmogorov-Chentsov). More details can be found e.g in the book of Revuz and Yor. (and, by the way, for standard Brownian motion $\sigma^2$ should be 1).

Concerning your uniform convergence idea you might have a look on Schilder's Theorem (explained e.g. In the book of Dembo and Zeitouni on large deviations).

Answer by Stephan Sturm for calculate all the equivalent martingale measures

2012-02-05T20:58:42Z

One possible approach is to use the fact that the density process $\left. Z_t =\frac{d\mathcal{Q}}{d\mathcal{P}} \right\vert_{\mathcal{F}_t}$ for every equivalent local martingale measure $\mathcal{Q}$ is a true martingale in the Brownian filtration and you can characterize it via the martingale representation theorem. As a guiding example you may look e.g. on Appendix A of R. Frey's paper "Derivative Asset Analysis in Models with Level Dependent and Stochastic Volatility", CWI Quaterly 10, no 1 (special issue on the Mathematics of Finance) p 1-34 which he links on his webpage: http://www.math.uni-leipzig.de/~frey/vol_survey.ps

Comment by Stephan Sturm

2013-05-06T21:43:59Z

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Comment by Stephan Sturm

2013-04-04T14:33:16Z

@an12 If you use Monte-Carlo techniques, you may use symmetry for variance reduction via antithetic sampling.

Comment by Stephan Sturm

2013-04-04T05:13:44Z

Why calculating the integral and not just trying an acceptance-rejection type approach? - Simulating the $(n+1)$st variable and comparing it with other $n$ normal pdfs? This should be faster then integration. If you want hard bounds, you could try to some Quasi-Monte Carlo based sampling. [But this is maybe a bit trickier, as you have to keep the low discrepancy while transforming uniform distributions in normal ones...]

Comment by Stephan Sturm

2013-04-03T13:29:11Z

Great lecture notes on Donsker's theorem in $\mathbb{R}$. However, I do not see how this could be applied to a generic manifold setting.

Comment by Stephan Sturm

2013-02-28T23:49:57Z

Also your limit will be infinity a.s, as every term in the sequence will be infinity a.s.... If you believe that local time is not the right concept for your question, can you post more background on your problem?

Comment by Stephan Sturm

2013-02-28T23:22:03Z

You will have $G(t) = \infty$ almost surely for $t>0$. But I believe what you are looking for is local time, see en.wikipedia.org/wiki/…

Comment by Stephan Sturm

2013-01-06T21:24:48Z

@ Uwe Franz. I believe you forgot the biscuit monad in your discussion, see en.wikipedia.org/wiki/Leibniz-Keks

Comment by Stephan Sturm

2012-12-10T00:38:15Z

"Aire de Levy" is just French for "Levy area" (the $S_t$ of the question). And the formula follows directly from observing that $S_t$ and $-S_t$ have the same distribution... But I agree with quid and Alexandre that the formulation "show that" implies either that the question is homework or that OP has a way to interact with people that is too rude to be appropriate for this site.

Comment by Stephan Sturm

2012-11-30T01:34:09Z

This is fairly standard and for sure not research level. See en.wikipedia.org/wiki/Ornstein–Uhlenbeck_process and mathoverflow.net/howtoask

Comment by Stephan Sturm

2012-11-29T23:53:03Z

For the second part, it is true that $\tau$ may depend on $n$. But as we choose it for a given $n$ optimally, it follows that for all other $n$ it gives a weaker bound. But as this weaker bound decays exponentially, the optimal Chernoff bound can decay asymptotically only faster than exponentially.

Comment by Stephan Sturm

2012-11-29T23:42:47Z

I did not intend to provide an explicit counterexample, just a line of thought. But if you want to go further, take just the lognormal distribution: It has all finite moments, but no positive exponential one. The Chernoff bound is trivial 1.

Comment by Stephan Sturm

2012-11-29T17:25:03Z

I don't now if I understand your question correctly, but the answer seems to be a straightforward no. Existence of a finite expectation does not imply finite higher moments (and thus for sure not exponential moments). On the other hand, if some exponential moment exists, the decay is clearly exponential, no?

Comment by Stephan Sturm

2012-11-01T21:03:09Z

I think we need a bit more details: reflected where? And are they indepent?it would be great if you could give a little more specifications of your problem.

Comment by Stephan Sturm

2012-09-28T14:07:02Z

Besides missing a factor 2 in the square-root term, you my find this easily by googling "law of the iterated logarithm". One good source is the Brownian Motion book by Moerters and Peres, people.bath.ac.uk/maspm/book.pdf(you find the result on page 119)

Comment by Stephan Sturm

2012-09-10T13:14:41Z

Please clarify what is your filtration (are you assuming indirectly the Brownian?) and for which variable you want uniformity (just t up to u -then Wolfgang Loehr is right) or in both variables (on infinite time horizon)?