bio | website | |
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location | Rehovot, Israel | |
age | 25 | |
visits | member for | 3 years, 4 months |
seen | Aug 23 at 1:19 | |
stats | profile views | 865 |
I'm a PhD student in Weizmann Institute, specializing in Probability.
Jun
15 |
comment |
Generalization of Krull dimension for commutative rings
Cardinals can be constructed as a sub-class of ordinals, so they are well-ordered. |
Jun
14 |
comment |
Generalization of Krull dimension for commutative rings
In any case, there is always a smallest cardinality $\alpha$, such that there are no chains of length $\ge \alpha$, which can serve as an alternative notion of dimension. |
Jun
14 |
revised |
Generalization of Krull dimension for commutative rings
added 115 characters in body |
Jun
14 |
answered | Generalization of Krull dimension for commutative rings |
Jun
12 |
comment |
$BMO$-property via a John-Nirenberg type estimate?
I meant any of the equivalent norms in the Orlicz space with a function that grows exponentially - e.g. $\Phi(x) := \cosh x - 1$. For the definition of Orlicz space see en.wikipedia.org/wiki/Birnbaum%E2%80%93Orlicz_space. |
Jun
11 |
revised |
$BMO$-property via a John-Nirenberg type estimate?
added 10 characters in body |
Jun
11 |
revised |
$BMO$-property via a John-Nirenberg type estimate?
added 10 characters in body |
Jun
11 |
answered | $BMO$-property via a John-Nirenberg type estimate? |
Jun
4 |
awarded | Nice Answer |
Jun
4 |
answered | which norms can be realized as operator norms? |
Jun
1 |
awarded | Revival |
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 |
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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 |
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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 $\mathsf{E} U_t = \mathsf{const} \cdot t^{1/2}$... |