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location Rehovot, Israel
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I'm a PhD student in Weizmann Institute, specializing in Probability.


Apr
24
comment Quadratic variation and predictable quadratic variation for martingales
On second thought, since you define $\langle M \rangle$ in such a way that it doesn't always exist, it's not even clear what your statement means...
Apr
24
comment Quadratic variation and predictable quadratic variation for martingales
Besides, for continuous martingales $[M]$ and the properly defined $\langle M \rangle$ (i.e. $M^2 - \langle M \rangle$ is a local martingale) are equal.
Apr
24
comment Quadratic variation and predictable quadratic variation for martingales
There is absolutely no reason for $\mathsf{E} \left[ (M_t - M_s)^2 \, \middle| \, \mathcal{F}_s \right]$ to be almost surely finite, so your statement is certainly not true without additional integrability assumptions. That being said, I believe your statement is indeed true for martingales that are bounded in the $L^2$ norm.
Apr
8
awarded  Yearling
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)
added 82 characters in body
Feb
10
revised Defining functions pointwise vs. almost everywhere (w.r.t. uncountably many mutually singular measures)
added 124 characters in body
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)
added 2 characters in body
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