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bio website mathtm.blogspot.com
location University of California Los Angeles, CA
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visits member for 2 years, 8 months
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I am a recent graduate in applied mathematics at UCLA and currently trying to break into the quantitative investment/trading industry while also continuing to pursue advanced graduate-level mathematics as both a hobby and career necessity.


Dec
21
revised What is the explicit obstruction to almost sure convergence in stochastic integrals?
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Dec
21
comment What is the explicit obstruction to almost sure convergence in stochastic integrals?
Here's another observation that leads me to believe the usual uniform $t/n$ partition is not sufficient to force pointwise convergence. If $E_{n}$ is the "bad" set on which $X_{n}$ is not close to the limit $X$, and if it marches around the sample space, then it will obstruct pointwise convergence if $P(E_{n})\to0$ sufficiently slow such that there is enough mass to always cover a set of points of positive mass. But Borel Cantelli says that if $\sum P(E_{n})<\infty$, then $E_{n}$ loses mass sufficiently fast that it can't do this, resulting in pointwise convergence; but $\sum n^{-1}=\infty$
Dec
21
comment What is the explicit obstruction to almost sure convergence in stochastic integrals?
Since the question is somewhat imprecise, maybe answering this would be easier. Since convergence in probability implies a.s. convergence of a subsequence, if we take a rapidly decreasing sequence of partitions $\Pi_{n}$, we recover pathwise convergence. Is it possible to illustrate two explicit sequences of partitions whereby one results in convergence along both modes and the other just in probability? Furthermore, does the usual $t/n$ partition decrease rapidly enough to achieve a.s. convergence?
Dec
21
awarded  Custodian
Dec
21
revised What is the explicit obstruction to almost sure convergence in stochastic integrals?
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Dec
21
reviewed Reject What is the explicit obstruction to almost sure convergence in stochastic integrals?
Dec
21
asked What is the explicit obstruction to almost sure convergence in stochastic integrals?
Dec
18
awarded  Tenacious
Dec
3
answered Suppose that $f: \mathbb R \to \mathbb R$ is a continuous function and there is a number $p \in [a,b]$ so that $f(p) = q$.
Dec
3
comment Proving uniform convergence of an integral-defined function on compact sets
Ah, then you will have to adjust the limits in the integral accordingly to take into account the domain. But note for instance that if $f$ is supported on $K\subset\mathbb{R}$, then $f_{\epsilon}$ is supported on a subset $K_{\epsilon}\subset K$. The specific range of $f$ is irrelevant, only that it allows for $\int_{\mathbb{R}} f=1$ and that $f\geq0$.
Dec
2
revised Proving uniform convergence of an integral-defined function on compact sets
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Dec
2
comment Proving uniform convergence of an integral-defined function on compact sets
Actually, since uniform convergence is usually harder to demonstrate than pointwise convergence, a good starting point would be to prove pointwise convergence first, even if restricted to compact sets. I elaborated on my answer a bit; hopefully it will get you started.
Dec
2
revised If $\lim_{n \to \infty} f_n=f$ (Almost everywhere) then $\lim_{n \to \infty} f_n=f$ ( in measure on$E$)
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Dec
1
revised $y=ce^{y/x};\quad y'=y^2/(xy-x^2)$
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Dec
1
answered $y=ce^{y/x};\quad y'=y^2/(xy-x^2)$
Dec
1
revised If $\lim_{n \to \infty} f_n=f$ (Almost everywhere) then $\lim_{n \to \infty} f_n=f$ ( in measure on$E$)
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Dec
1
answered Proving uniform convergence of an integral-defined function on compact sets
Dec
1
answered If $\lim_{n \to \infty} f_n=f$ (Almost everywhere) then $\lim_{n \to \infty} f_n=f$ ( in measure on$E$)
Dec
1
comment Proving uniform convergence of an integral-defined function on compact sets
In the statement of your problem, does $f$ have compact support?
Dec
1
comment Orders of growth of typical sequences
Are you using the $\ll$ symbol to mean $P(n)\ll Q(n)$ if there is an absolute constant $C$ such that $P(n)\leq CQ(n)$ or that $P(n)/Q(n)\leq C$ for $n\geq N$ for some $N$ sufficiently large; or is it notation for one quantity being "much smaller" than the other? Also, are the numbers $p_{j},q_{j}$ and $a_{j},b_{j}$ arbitrary pairs with no relation between the indices?