User weakstar - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T04:16:11Z http://mathoverflow.net/feeds/user/2310 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/112650/properties-of-the-euler-discretization-of-a-diffusion Properties of the Euler Discretization of a diffusion weakstar 2012-11-17T05:48:14Z 2012-11-18T18:44:11Z <p>Let $X$ be a continuous 1-d diffusion:</p> <p>$$dX_t = a(X_t)dt + b(X_t)dW_t, X_0 = x.$$ W is a standard Brownian Motion and $a(\cdot)$ and $b(\cdot)$ can have nice regularity properties.</p> <p>Let $Z^n_1,Z^n_2,\ldots$ be i.i.d. standard normal random variables. The Euler discretization $X^n$ with step size $\frac{1}{n}$ is the process given by</p> <p>$$X^n_\frac{k}{n} = X^n_\frac{k-1}{n} + \frac{1}{n}\cdot a(X^n_\frac{k-1}{n}) + \frac{1}{\sqrt{n}}b(X^n_\frac{k-1}{n}) \cdot Z^n_k, X^n_0 = x.$$</p> <p>A (at least to me) reasonable way to embed $X$ and $X^1,X^2,\ldots$ on the same probability space is by taking the $Z_i$ generated by the increments of the Brownian Motion $W$: $Z^n_i = \sqrt{n} \cdot (W_\frac{i}{n} - W_\frac{i-1}{n})$.</p> <p>The reason why I want all process defined on the same space is that I want convergence rates, and I have heard that weak convergence results aren't appropriate for this, although I don't really understand why.</p> <p>It's believable that in some sense $X^n \rightarrow X$. There are several different types of convergence available (expectation over product space, expectation of time supremum). I'm interested in what kind of convergence there is, and specifically in the convergence rates.</p> <p>Motivation: the problem I'm studying is an optimal stopping problem on a diffusion when there is perfect information. When information comes at discrete times, it becomes a weird discretization, which can be approximated by the one above. I'd like to see how fast there is convergence to the perfect information case.</p> http://mathoverflow.net/questions/101057/compactness-of-the-set-of-densities-of-equivalent-martingale-measures/101200#101200 Answer by weakstar for Compactness of the set of densities of equivalent martingale measures weakstar 2012-07-03T02:05:15Z 2012-07-03T11:28:01Z <p>It seems to me that in the statement of the Neyman-Pearson Lemma, equivalence isn't assumed, just absolute continuity. I think in general, it is these sets of absolutely equivalent local martingale measures which are closed.</p> <p>If you think about it, $\mathbf{Z}_P$ will almost never be $L^1$-compact, the reason being that it lacks closedness. Suppose that you had $Z^e$ which comes from an equivalent local martingale measure and $Z^a$, which comes from an absolutely continuous but not equivalent local martingale measure. Then $Z^e > 0$. Therefore, consider the convex combinations $\frac{1}{n}Z^e + \frac{n-1}{n}Z^a$. These are equivalent local martingales measures since $Z^e$ is postive, but they converge to $Z^a$ (in $L^1$), which is not equivalent. So, as soon as you have any nonequivalent local martingale measure, you lose closedness.</p> http://mathoverflow.net/questions/99506/blackbox-theorems/99541#99541 Answer by weakstar for Blackbox Theorems weakstar 2012-06-14T01:12:50Z 2012-06-14T01:12:50Z <p>The existence of Brownian Motion.</p> http://mathoverflow.net/questions/98230/distribution-built-up-from-powers-of-a-log-normal-r-v Distribution built up from powers of a log normal r.v. weakstar 2012-05-28T23:21:20Z 2012-05-28T23:21:20Z <p>Let $W_t$ be a standard brownian motion, and $\lambda$ and $\alpha$ are positive constants. Does the expression $$\int_0^t \lambda e^{\lambda u + \frac{\alpha u X_t}{t} - \frac{\alpha^2 u^2}{2t}} du$$ have a recognizable distribution?</p> <p>This comes from a quickest detection problem when a Brownian motion gains a draft at some unknown time. The expression below is part of the posterior probability that the drift has arrived, based on making a single observation at time $t$.</p> http://mathoverflow.net/questions/78849/processes-approximating-a-reflected-brownian-motion Processes approximating a reflected brownian motion. weakstar 2011-10-22T21:15:32Z 2011-10-23T14:07:54Z <p>Let $W$ be a standard Brownian Motion. Let $\epsilon>0$ be given. Let $X^\epsilon$ be the process which diffuses like $W$ on $(-\epsilon,\infty)$, but when $X^\epsilon$ reaches the level $-\epsilon$, it is immediately brought back to the value zero. It then diffuses again according to $W$ until hitting $-\epsilon$, and then is brought back to zero, and so forth. Let $X^0$ be a reflected Brownian Motion (reflected at zero). Then, as $\epsilon \rightarrow 0$, in what sense does $X^\epsilon \rightarrow X^0$ Are there any references for this? I'm also interested in when $W$ is a diffusion.</p> http://mathoverflow.net/questions/76810/passage-time-distributions-for-poisson-processes Passage Time Distributions for Poisson processes. weakstar 2011-09-29T22:18:53Z 2011-09-29T22:56:09Z <p>Let $(X_t)_{t \geq 0}$ be a standard Poisson process with intensity $\mu$. Let $\tau_b = \inf ( t>0 : X_t= at + b)$, where $a>0$ and $b&lt;0$, and let $\sigma = \inf (t>0 : X_t \geq at)$. Is there any reference for the distributions of $\tau_b$ and $\sigma$, as well as computing $P(\tau_b &lt; \sigma)$?</p> <p>Thanks!</p> http://mathoverflow.net/questions/22255/usefulness-of-frechet-versus-gateaux-differentiability-or-something-in-between Usefulness of Frechet versus Gateaux differentiability or something in between. weakstar 2010-04-22T20:56:08Z 2011-08-12T14:06:38Z <p>If you have a function $V: L \rightarrow \mathbb{R}$, where $L$ is an infinite dimensional topological vector space, there are multiple notions of differentiability. For $x,u \in L$, $V$ is Gateaux differentiable at $x$ in the direction $u$ if the limit $\underset{t \rightarrow 0}{\lim} \frac{V(x + tu) - V(x)}{t}$ exists. Supposing that you are in a Banach space, $V$ is Frechet differentiable if the above limit exists for all $u$ in a ball around $x$, and importantly, with the convergence being uniform over this neighborhood. </p> <p>The question is, what difference does it make for a function to be Frechet differentiable versus Gateaux differentiable, maybe with respect to proving theorems that generalize the finite-dimensional setting, where the two notions of differentiability more or less agree. What kind of pathological behavior can functions exhibit that are merely Gateaux differentiable in every direction? There are also intermediate forms of differentiability between Frechet and Gateaux, defined in terms of uniform convergence of the difference quotients over some preferred family of sets (a bornology). Are there any intermediate kinds of differentiability that are important?</p> http://mathoverflow.net/questions/62347/hardy-spaces-analysis-martingales Hardy spaces: analysis <---> martingales weakstar 2011-04-19T22:50:42Z 2011-04-20T14:21:08Z <p>Let $H^p$ be the Hardy space of analytic functions on the open unit disk $D$: $f \in H^p$ if $f$ is analytic on $D$ and $\sup_{r &lt; 1} \int_0^{2\pi} |f(re^{i\theta})|^p d\theta &lt; \infty$. </p> <p>Consider a filtration generated by a 2-d (complex) Brownian Motion $B$. The martingale hardy space $\mathcal{H}^p$ defined on some time interval $[0,T]$, say, is the set of martingales $M$ such that $M^* = \sup_{t \in [0,T]} |M_t| \in L^p$. This definition is mostly interesting for $p=1$, as for $p>1$, $\mathcal{H}^p$ can be associated with a regular $L^p$ space of martingales.</p> <p>If $B$ starts at zero, let $\tau$ be the hitting time of the boundary of $D$. Then a connection between these two spaces is the following: for $f$ analytic on the unit disk, $f(B_{t \wedge \tau}) \in \mathcal{H}^p$ if and only if $f \in H^p$, and this mapping is continuous.</p> <p>This allows you to associate $H^p$ to a subspace of $\mathcal{H}^p$. For studying $\mathcal{H}^p$, it would be useful to have a more complete representation of part of $\mathcal{H}^p$ in terms of functions evaluated on $B$. Specifically, for martingales that run on the whole time interval. Can this be obtained by using another hardy space, such as the Hardy space $h^p$ on $\mathbb{R}^2$? Can anything else be said relating hardy spaces of martingales and hardy spaces of functions?</p> http://mathoverflow.net/questions/57184/filtrations-generated-by-cadlag-martingales Filtrations generated by cadlag martingales. weakstar 2011-03-02T23:42:31Z 2011-03-03T02:05:50Z <p>Let $(\Omega,P,\mathcal{F})$ be a probability space with filtration $\mathbb{F} = (\mathcal{F}_t), t \in [0,T]$, where $T$ can be finite or infinite. Let $M$ be a cadlag (local) martingale with respect to $\mathbb{F}$, and let $\mathbb{F}^M$ be the filtration generated by $M$ and then completed with respect to $P$. </p> <p>Question: Is $\mathbb{F}^M$ a right-continuous filtration?</p> <p>Some facts:</p> <ol> <li>If $X$ is a strong markov process, then the completion of $\mathbb{F}^X$ is right-continuous. This is in Karatzas and Shreve.</li> <li>A sort of converse: If $M$ is a local martingale in a right continuous and complete filtration, it has a right continuous modification.</li> </ol> <p>One possible idea: A continuous local martingale can be expressed as a time-changed Brownian Motion, which is strong markov.</p> http://mathoverflow.net/questions/55885/why-semigroups-could-be-important/55996#55996 Answer by weakstar for Why semigroups could be important? weakstar 2011-02-19T17:33:13Z 2011-02-19T17:33:13Z <p>Given a group $G$, the Block Monoid $B(G)$ consists of sequences of elements in $G$ that sum to zero. So for example, an element of $B(\mathbb{Z})$ is $(-2,-3,1,1,3)$. The monoid operation is concatenation, and the empty block is the identity element. </p> <p>Given a Dedekind domain, one can take its ideal class group, and consider the block monoid over that group. Note that in the obvious way, elements of the block monoid can be irreducible or not. One can study irreducible factorization in the Dedekind domain by studying irreducible factorization in the block monoid.</p> http://mathoverflow.net/questions/45581/is-there-an-extension-of-the-arzela-ascoli-theorem-to-spaces-of-discontinuous-fun Is there an extension of the Arzela-Ascoli theorem to spaces of discontinuous functions? weakstar 2010-11-10T18:23:47Z 2010-11-11T02:42:50Z <p>The Arzela-Ascoli function basically says that a set of real-valued continuous functions on a compact domain is precompact under the uniform norm if and only if the family is pointwise bounded and equicontinuous.</p> <p>Are there any analogs of this kind of result for spaces of noncontinuous functions? The specific set I have in mind is the c&agrave;dl&agrave;g functions, which are right continuous and have left limits. Essentially, I want to know if there is any relationship between compactness and equicontinuity. Here, equicontinuity would obviously have to be relaxed to account for jumps, and compactness in the uniform topology could be relaxed to compactness in, say, the <a href="http://en.wikipedia.org/wiki/C%C3%A0dl%C3%A0g" rel="nofollow">Skorokhod topology</a>, or something weaker than uniform.</p> http://mathoverflow.net/questions/29707/path-continuity-for-closed-martingales Path continuity for (closed) martingales? weakstar 2010-06-27T13:17:43Z 2010-09-29T14:14:29Z <p>Take a time interval $[0,T]$, and a filtered probability space $(\Omega,P,\mathcal{F},\mathcal{F}_t)$. If $X \in L^1(\mathcal{F}_T)$, then $M_t = E [X \ | \ \mathcal{F}_t]$ is a martingale. If I want the martingale $M$ to have continuous or right continuous paths, is there a condition I can impose on the filtration to ensure this?</p> <p>A standard result says that if the filtration is right-continuous, meaning that $\cap_{s>t} \mathcal{F}_s = \mathcal{F}_t$, then there exists a modification of $M$ with right continuous paths (in fact right continuous with left limits). However, I want to say something about the original process, and not a modification.</p> http://mathoverflow.net/questions/28147/why-do-we-care-about-lp-spaces-besides-p-1-p-2-and-p-infinity/28209#28209 Answer by weakstar for Why do we care about L^p spaces besides p = 1, p = 2, and p = infinity? weakstar 2010-06-15T03:47:35Z 2010-06-15T03:47:35Z <p>$L^0$, which for a finite measure is just the set of all measurable functions, is important in probability. When you're modeling some physical phenomenom, there is often no canonical choice of probability measure, so you sometimes work with a class of equivalent probabilities, i.e. ones that have the same null sets. The only two $L^p$ spaces that are invariant under changing to an equivalent probability are $L^\infty$ and $L^0$. The former space is often too small for modeling purposes, and so you are forced to work with $L^0$. It's topologized by convergence in measure/probability, and it's pretty horribly non-convex.</p> http://mathoverflow.net/questions/24402/quantitative-questions-about-the-size-of-a-finite-epsilon-net Quantitative questions about the size of a finite epsilon net weakstar 2010-05-12T14:57:44Z 2010-05-12T18:01:40Z <p>Let $X$ be a metric space, and let $U \subset X$ be any set. A finite set $N = N(\epsilon) \subset U$ is called a finite $\epsilon$-net of $U$ if every point of $U$ is at most a distance of $\epsilon$ from some point of $N$.</p> <p>It is easy to show that if $U$ is compact, then for any $\epsilon>0$, a finite $\epsilon$-net exists. I am interested in the behavior of the function $|N(\epsilon)|$ as $\epsilon$ goes to zero. If there are answers in this generality, great. I am mostly interested in the particular case where $U$ is also convex, and where $X$ is also an infinite dimensional topological vector space, but any answers are of course welcome.</p> <p>Edit: I'm also interested in weakening the notion of $\epsilon$-net, so instead of requiring every point of $U$ to be close to a point in $N$, we could require every point of $U$ to be close to a point in the convex hull of $N$. The motivation for this comes from looking at objects which are "convexly compact"; this means (in a metric space) that given any sequence $(f_n)$ in $U$, there exist $g_j \in \text{conv}(f_j,f_{j+1},\ldots)$ such that $g_j \rightarrow g$ in $U$.</p> http://mathoverflow.net/questions/115090/total-variation-distance-between-two-solutions-of-sde Comment by weakstar weakstar 2012-12-02T03:57:35Z 2012-12-02T03:57:35Z Those random variables have distribution functions which are discontinuous. http://mathoverflow.net/questions/115090/total-variation-distance-between-two-solutions-of-sde Comment by weakstar weakstar 2012-12-01T19:38:34Z 2012-12-01T19:38:34Z Yeah but your construction of $\mu$ and $\nu$ is a weak one, in that it says nothing about dependence on particular values of $\omega \in \Omega$. Using Grownwall's inequality you can get a convergence rate for (strong) $L^1$ convergence, $E \left|X_t^1 - X_t^2 \right|$ tends to zero in some sense. This implies total variation convergence. http://mathoverflow.net/questions/115090/total-variation-distance-between-two-solutions-of-sde Comment by weakstar weakstar 2012-12-01T17:55:32Z 2012-12-01T17:55:32Z Isn't a bound on $|X_t^1 - X_t^2|$ considerably stronger than a bound in total variation distance? One is strong and the other is weak? http://mathoverflow.net/questions/112650/properties-of-the-euler-discretization-of-a-diffusion/112777#112777 Comment by weakstar weakstar 2012-11-19T02:05:13Z 2012-11-19T02:05:13Z Also, do you know any general references about convergence properties for optimal stopping problems? For example, the above kind of uniform convergence is something I need, but by itself does not imply convergence of optimal stopping, since i.e. hitting times aren't continuous with respect to the uniform norm. http://mathoverflow.net/questions/112650/properties-of-the-euler-discretization-of-a-diffusion/112777#112777 Comment by weakstar weakstar 2012-11-19T00:51:26Z 2012-11-19T00:51:26Z Thanks, this is the convergence that I was most interested in. Do you have a reference for this fact, and is the convergence rate $O(t)$? http://mathoverflow.net/questions/112650/properties-of-the-euler-discretization-of-a-diffusion/112693#112693 Comment by weakstar weakstar 2012-11-17T22:59:54Z 2012-11-17T22:59:54Z Thanks for your answer. It seems like strong convergence doesn't have much to say about convergence of optimal stopping problems, unless you can leverage some additional information, like martingality, to use a maximal inequality. Kloeden/Platen also don't have anything to say about this. Do you know of anything in this direction? http://mathoverflow.net/questions/110585/modification-of-doob-inequality/110586#110586 Comment by weakstar weakstar 2012-10-24T21:44:35Z 2012-10-24T21:44:35Z Maybe you can get a positive answer by considering symmetric processes. http://mathoverflow.net/questions/98917/harmonic-functions-and-itos-formula-with-respect-to-the-generator-of-a-reflectin Comment by weakstar weakstar 2012-06-06T00:25:45Z 2012-06-06T00:25:45Z i think the operator is the same, but its defined on the functions that have zero one-sided derivative on the boundary. http://mathoverflow.net/questions/78849/processes-approximating-a-reflected-brownian-motion Comment by weakstar weakstar 2011-10-23T14:28:32Z 2011-10-23T14:28:32Z One thought: the reflected BM is a standard BM plus a local time term at zero. Using the interpretation of local time in terms of downcrossings should possibly do the trick. http://mathoverflow.net/questions/78849/processes-approximating-a-reflected-brownian-motion/78896#78896 Comment by weakstar weakstar 2011-10-23T14:28:05Z 2011-10-23T14:28:05Z Thanks for the reference, will take a look. http://mathoverflow.net/questions/76810/passage-time-distributions-for-poisson-processes Comment by weakstar weakstar 2011-09-29T22:56:27Z 2011-09-29T22:56:27Z Sorry, it's fixed. http://mathoverflow.net/questions/62246/what-does-progressively-measurable-actually-entail/62311#62311 Comment by weakstar weakstar 2011-04-19T20:26:24Z 2011-04-19T20:26:24Z In particular, for $\theta_t(\omega)$ progressively measurable, $\theta_T \in \mathcal{F}_T$ when $T$ is an arbitrary stopping time. If $\theta$ is just adapted, this will only hold true when $T$ is countably valued. I think an ok analogy is progressively measurable is to adapted as strong markov is to markov. http://mathoverflow.net/questions/57184/filtrations-generated-by-cadlag-martingales/57187#57187 Comment by weakstar weakstar 2011-03-03T01:45:19Z 2011-03-03T01:45:19Z Right, I meant your second suggestion, for brownian motion starting away from zero. Just seeing if any affirmative answer can be salvaged. http://mathoverflow.net/questions/57184/filtrations-generated-by-cadlag-martingales Comment by weakstar weakstar 2011-03-03T01:16:11Z 2011-03-03T01:16:11Z For a brownian motion, those kinds of sets have probability zero. I'm not sure if that's the only kind of obstruction, but since martingales have the same paths as brownian motion, I had hoped you might be able to say something. http://mathoverflow.net/questions/57184/filtrations-generated-by-cadlag-martingales/57187#57187 Comment by weakstar weakstar 2011-03-03T01:07:03Z 2011-03-03T01:07:03Z Thanks. Maybe i'm being dumb, but can this counterexample be extended to a nonzero starting position?