bio | website | |
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location | Rehovot, Israel | |
age | 24 | |
visits | member for | 2 years, 11 months |
seen | 3 hours ago | |
stats | profile views | 794 |
I'm a PhD student in Weizmann Institute, specializing in Probability.
Mar 20 |
asked | Is Wiener's Tauberian theorem true in Wiener space? |
Feb 25 |
answered | Location of maximum of Brownian motion with rough drift |
Feb 16 |
comment |
Location of maximum of Brownian motion with rough drift
I'll try to cook up a readable answer later... |
Feb 16 |
comment |
Location of maximum of Brownian motion with rough drift
... which is not in the $W^{1,2}$ Sobolev space, so $\intop f d \mathsf{E} U$ doesn't converge for some $f \in L^2$. So the distributions of $B$ and $(B + U) / \sqrt 2$ must be singular to each other: one needs the recentering and the other one doesn't. |
Feb 16 |
comment |
Location of maximum of Brownian motion with rough drift
... And this can be done using various invariants of the equivalence class of the distribution of Bessel(3). For instance, the stochastic integrals $\intop f(t) d (U_t - \mathsf{E} U_t)$ converge for all deterministic $f \in L^2$, so also the stochastic integrals $\intop f(t) d ((B_t + U_t) / \sqrt 2 - \mathsf{E} U_t / \sqrt 2)$ converge for such $f$. For the Brownian motion (and any other semimartingale with distribution equivalent to that of a Brownian motion) such integrals would converge without the $\mathsf{E} U_t$ recentering. Note that $\mathsf{E} U_t = \mathsf{const} \cdot t^{1/2}$... |
Feb 16 |
comment |
Location of maximum of Brownian motion with rough drift
The answer to your question is that it's almost surely singular. The basic idea is to view the Brownian motion locally at the point where it achieves its local maximum. By the Williams' decomposition theorem, its distribution in the neighborhood of such a point is, up to absolute continuity, that of a Bessel(3) process. From this point of view your question is equivalent to the following: show that for a Brownian motion $B$ and an independent Bessel(3) process $U$, both starting from $0$, the distribution of $(B + U) / \sqrt{2}$ is singular to that of the Brownian motion... |
Feb 10 |
revised |
Defining functions pointwise vs. almost everywhere (w.r.t. uncountably many mutually singular measures)
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Feb 10 |
revised |
Defining functions pointwise vs. almost everywhere (w.r.t. uncountably many mutually singular measures)
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Feb 10 |
comment |
Homemorphism between $X$ and $\mathcal{P}(\mathcal{P}(X))$
Actually, all infinite-dimensional compact convex sets (in good enough topological vector spaces) are homeomorphic to each other in a nonlinear way, so in particular $\mathcal{P}(X)$ are all homeomorphic to each other for all infinite compact metrizable $X$. Thus $\mathcal{P}(X) \simeq \mathcal{P}(\mathcal{P}(X))$. books.google.co.il/… |
Feb 10 |
revised |
Defining functions pointwise vs. almost everywhere (w.r.t. uncountably many mutually singular measures)
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Feb 10 |
revised |
Defining functions pointwise vs. almost everywhere (w.r.t. uncountably many mutually singular measures)
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Feb 10 |
revised |
Defining functions pointwise vs. almost everywhere (w.r.t. uncountably many mutually singular measures)
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Feb 10 |
revised |
Defining functions pointwise vs. almost everywhere (w.r.t. uncountably many mutually singular measures)
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Feb 10 |
asked | Defining functions pointwise vs. almost everywhere (w.r.t. uncountably many mutually singular measures) |
Jan 29 |
comment |
Is the conditional expectation a contraction in weak $\mathbb L^p$ spaces?
If I'm not mistaken, in the case $X_0 = L^1$, $X_1 = L^\infty$, $\theta = 1 - \frac{1}{p}$, the functional $K(x, t, X_0, X_1)$ defined in that Wikipedia article coincides with $\sup_{\mu[A] = t} \mathsf{E} |x| \mathsf{1}[A]$. Which makes the $\Vert \cdot \Vert_{p,\infty}$ norm exactly equal to the $K(\theta,\infty)$ interpolation norm. |
Jan 28 |
comment |
Is the conditional expectation a contraction in weak $\mathbb L^p$ spaces?
en.wikipedia.org/wiki/Interpolation_space#Real_interpolation |
Jan 5 |
awarded | Self-Learner |
Jan 5 |
awarded | Revival |
Jan 5 |
answered | Is an infinite-dimensional “Lebesgue measure” uniquely determined by a set of positive finite measure? |
Dec 14 |
revised |
Operator arithmetic-harmonic mean inequality with operator-valued weights
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