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awarded  Nice Answer
Oct
27
answered Conditional Expectation and Interpretation of Projection of $Y$ Onto $X$ vs. $Y$ onto $L^{2}(\sigma(X))$
Oct
26
revised Conditional Expectation and Interpretation of Projection of $Y$ Onto $X$ vs. $Y$ onto $L^{2}(\sigma(X))$
added 46 characters in body
Oct
26
asked Conditional Expectation and Interpretation of Projection of $Y$ Onto $X$ vs. $Y$ onto $L^{2}(\sigma(X))$
Jul
1
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?
added 105 characters in body
Apr
10
revised $dx=\frac {dx}{dt}dt $. Why is this equality true and what does it mean?
added 809 characters in body
Apr
10
revised $dx=\frac {dx}{dt}dt $. Why is this equality true and what does it mean?
added 809 characters in body
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.
added 73 characters in body
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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