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6h
revised Is every measurable set a measure-independent limit of open sets
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7h
answered Is every measurable set a measure-independent limit of open sets
Dec
16
comment Independence and Conditional Independence between random variables
@dev_nut: Trivial example: $A$ and $A$ are conditionally independent given $A$, but not independent.
Dec
16
answered Independence of random variables involving Brownian motion
Dec
15
comment Show that the Open Mapping Theorem requires both spaces to be complete
@TylerHilton: Hint: $\frac{1}{n} + \dots + \frac{1}{n}$ ($n$ times) is 1.
Dec
14
awarded  Good Answer
Dec
14
comment Are the polynomial functions on $S^1$ dense in $C(S^1,ℂ)$?
@k.stm: You're right, thanks. I tend to think of Morera as the "if and only if" theorem, but actually Cauchy is the easy direction and Morera is the converse. Of course, the proof itself doesn't actually use either of them.
Dec
14
revised Are the polynomial functions on $S^1$ dense in $C(S^1,ℂ)$?
morera -> cauchy
Dec
14
answered Are the polynomial functions on $S^1$ dense in $C(S^1,ℂ)$?
Dec
13
comment Is $[0,1]^2 \setminus \{(a,b)\}$ connected?
The easiest way might be to show that it is path connected
Dec
13
revised Is $[0,1]^2 \setminus \{(a,b)\}$ connected?
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Dec
12
revised Prove $\sqrt{5}$ is both irrational and algebraic.
format title
Dec
12
comment Brownian motion is almost surely continuous
This is maybe not the best example, since if you define Brownian motion as the canonical process on the space of continuous paths, by definition it is continuous surely. This issue is more relevant if you define Brownian motion on some other probability space (often $\mathbb{R}^{[0,1]}$).
Dec
12
comment Can SLLN be applied to $Y=\frac {1 } {k} \sum _{n=1 } ^ {k} Z _n 2^{-n } $, where $Z _n $ are iid?
@StevenStadnicki: To be extremely pedantic, we haven't excluded the possibility that $Z_n$ takes some other value (say $2^n$), but with probability 0.
Dec
12
comment Can SLLN be applied to $Y=\frac {1 } {k} \sum _{n=1 } ^ {k} Z _n 2^{-n } $, where $Z _n $ are iid?
I assume your sum is meant to be $\frac{1}{n} \sum_{k=1}^n 2^{-k} Z_k$? But in your case $\sum_{n=1}^\infty 2^{-n} Z_n$ already converges almost surely (to a finite limit) so if you multiply by $\frac{1}{n}$ the result will converge almost surely to 0.
Dec
10
revised About the von Neumann decomposition
spelling in title
Dec
10
answered Show that the Open Mapping Theorem requires both spaces to be complete
Dec
10
comment Show almost everywhere convergence for variable with Chi distribution
Since we're dealing with a.s. convergence here rather than i.p., we don't really need the continuous mapping theorem, just the definition of continuity.
Dec
10
comment Difficult general integral definite 0 to 1
@Karl: That gives $\int_{-\infty}^0 u^2 e^{ku}\,du$ if I'm not mistaken. I don't know if that's easier but maybe it makes it more obvious that one should proceed by integration by parts (which could also be done without the substitution as in Martin's answer).
Dec
10
revised Prove that if $f$ is integrable on $[0,1]$, then $\lim_{n→∞}\int_{0}^{1} x^{n}f(x)dx = 0$.
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