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comment Explanation of “weight function” of inner product in Hilbert space
You don't need the weight function either in finite or infinite dimensions, but you can include it in either case. An inner product creates a geometry on a vector space, and in finite dimensions, you can see adding the weight function as "distorting" that geometry in a linear way (e.g. stretching in some directions more than others). Maybe the same analogy is helpful for infinite dimensions.
Jan
31
comment Functions to represent set operations?!
How would $f(a_i + a_j)$, which is a number, "represent" $A_i \cap A_j$, which is a set? What exactly do you mean by "represent"?
Jan
30
awarded  Revival
Jan
30
answered What does “modulo equivalence relationship” mean?
Jan
30
comment Question about weak convergence of random variables
@Student: I added a counterexample for that. A natural question is whether one could find an example with continuous (or even smooth) densities. I thought about modifying my example but couldn't quite make it work.
Jan
30
revised Question about weak convergence of random variables
another example
Jan
30
answered Question about weak convergence of random variables
Jan
29
answered Random function of random variable
Jan
29
comment Understanding the concept of measurability of random variables
I don't understand what you are asking here. It doesn't seem to correspond to a precise mathematical statement that can be answered yes or no.
Jan
29
comment Why is the drift of an Itō process considered to be a Riemann integral even when it's not even Riemann integrable?
For your second question, sure, replace $X_t$ by $0$ on $N$, or anything else you like. Nobody cares what happens on a null set. All the theorems you want to prove say "a.s." everywhere anyway.
Jan
29
comment Why is the drift of an Itō process considered to be a Riemann integral even when it's not even Riemann integrable?
"But unless $b$ has (surely) continuous paths, it is not Riemann integrable": Why do you say that? There are lots of functions that are Riemann integrable which are not continuous. For instance, every cadlag function has at most countably many discontinuities and hence is Riemann integrable.
Jan
29
revised Application of Doob's inequality
edited tags
Jan
26
comment Are the polynomial functions on $S^1$ dense in $C(S^1,ℂ)$?
@AreaMan: Then Stone-Weierstrass says you do have a dense set.
Jan
25
comment How to picture a first countable space?
At a higher level, "picturing" a first countable space should in some sense be easy, because pretty much every space you can picture is first countable. Spaces that are not first countable tend to be weird, pathological spaces that are hard to visualize (at least from an analyst's perspective).
Jan
22
revised Connection between weak topology in probability and weak* topology in functional analysis
added 679 characters in body
Jan
22
answered Connection between weak topology in probability and weak* topology in functional analysis
Jan
22
comment written set of functions as a union of Borel measurable set
@Tirifilo: I'm not sure I understand - are you asking whether a ball $B_n$ can be written as a countable union of compact sets? It cannot. In any infinite-dimensional Banach space, compact sets are nowhere dense, so by the Baire category theorem, any countable union of compact sets will have empty interior.
Jan
22
answered written set of functions as a union of Borel measurable set
Jan
22
comment Is 1-d Brownian filtration rich enough to admit a 2-d Brownian motion?
Incidentally, I think it might be possible to perform essentially the same construction in a fancier way: consider the stochastic integrals of two integrands with disjoint supports. These should be independent time-changed Brownian motions. Now if you invert the time change, you'll get Brownian motions back, and it should be possible to do this so that the inverse of the time change doesn't "look into the future".
Jan
22
comment Is 1-d Brownian filtration rich enough to admit a 2-d Brownian motion?
As for the convergence of the sum, if I have understood correctly, I think it equals something like $\sum_k (B(2^{-2k}) - B(2^{-(2k+1)}))$. This is a sum of independent random variables whose variances are $2^{-(2k+1)}$, which is a summable sequence. By Kolmogorov's convergence criterion, the sum converges almost surely.