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14h
comment is intersection of a countable collection of dense, open subsets of a complete metric space also dense in X?
Yes - en.wikipedia.org/wiki/Baire_category_theorem
May
9
awarded  Yearling
Apr
26
awarded  Popular Question
Apr
13
awarded  Popular Question
Apr
12
answered Show that a Cauchy sequence has a fast-Cauchy subsequence
Apr
10
comment $dx=\frac {dx}{dt}dt $. Why is this equality true and what does it mean?
No precise meaning, and none is needed really. The point of my answer (the second part anyway) is to view the differentials in the expression as quantities $\delta x$ and $\delta t$ - when they are finite, you get the same form of the expression plus a small error; as you send the quantities closer to $0$, the error disappears, but so do the quantities - you fix this by dividing and examining the ratio (which in turn leads right back to the first interpretation). Thinking of $dx(\cdot)$ as a linear functional of $\delta t$ is likely beyond the scope of the OP's knowledge.
Apr
10
comment $dx=\frac {dx}{dt}dt $. Why is this equality true and what does it mean?
Correct, so you didn't read it apparently.....
Apr
10
comment $dx=\frac {dx}{dt}dt $. Why is this equality true and what does it mean?
I don't follow you - did you read the entire response?
Apr
10
revised $dx=\frac {dx}{dt}dt $. Why is this equality true and what does it mean?
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Apr
10
revised $dx=\frac {dx}{dt}dt $. Why is this equality true and what does it mean?
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Apr
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revised $dx=\frac {dx}{dt}dt $. Why is this equality true and what does it mean?
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Apr
10
answered $dx=\frac {dx}{dt}dt $. Why is this equality true and what does it mean?
Mar
14
revised Let $M$ be a bounded subset of the space $C_{[a,b]}$. Prove that the set of all functions $F(x)=\int^{x}_{a}f(t)dt$ with $f\in{M}$ compact.
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Mar
14
comment Let $M$ be a bounded subset of the space $C_{[a,b]}$. Prove that the set of all functions $F(x)=\int^{x}_{a}f(t)dt$ with $f\in{M}$ compact.
I saw "$M$ compact" in the title and made that assumption, which makes the proof $\int M$ is compact trivial because of the uniform convergence norm. But I see now the assumption is that $M$ is merely bounded.
Jan
30
awarded  Popular Question
Jan
17
revised Deriving the definition of stochastic integrals with respect to Ito processes from first principles
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Jan
17
comment Deriving the definition of stochastic integrals with respect to Ito processes from first principles
@TheBridge - Regarding your second comment, the convergence is true for each $t\geq0$ in $L^{p}(\Omega)$ ($1\leq p\leq2$) no matter how you construct the sequence of partitions $\Pi_{n}$ as long as $||\Pi_{n}||\to0$. As is well known, you can construct a particular sequence $\Pi_{n}$ so that for a fixed $\omega$, $\text{QV}^{2}(W)(\omega)=\alpha$ for any $\alpha>0$ including $+\infty.$ However, you can avoid all of this by demanding that $||\Pi_{n}||\to0$ such that $\sum_{n=1}^{\infty}||\Pi||_{n}<\infty$; then by Borel-Cantelli the QV process does converge pathwise $\omega$-a.s. to $t$.
Jan
17
comment Deriving the definition of stochastic integrals with respect to Ito processes from first principles
@TheBridge - Regarding your first comment, please provide your definition if you disagree with mine; however, I am using the definition that is regularly encountered in mathematical finance texts. Also, there really isn't any need to be pedantic here with the conditions of the coefficient processes; they are more or less implied and in any case you can just assume the ones regularly imposed: adadpability, $L^{2}(\Omega\times[0,t])$ integrable, and $\omega$-a.s. continuous as a function of $t$.
Jan
15
revised Deriving the definition of stochastic integrals with respect to Ito processes from first principles
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Jan
15
revised Deriving the definition of stochastic integrals with respect to Ito processes from first principles
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