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17h
comment If ${a_i} \to 0$ and $\{ {X_i}\} _{i = 1}^\infty $ is a sequence of iid random variables with zero mean, does ${a_i}{X_i} \to 0$ almost surely?
@Saty, I guess he is mentioning the following version of the Chebyshev inequality $$ \Bbb{P}(|X| > c) \leq \frac{\Bbb{E}\varphi(|X|)}{\varphi(c)}, $$ where $\varphi : [0, \infty) \to [0, \infty)$ is strictly increasing and $c > 0$. Then obviously the Markov's inequality is a special case of this one.
17h
answered Help in finding the integral function.
18h
comment Help in finding the integral function.
Mathematica gives the following partial fraction decomposition, which looks horrible as expected: $$ \scriptsize \frac{1}{(x^3+1)^3} = \frac{8-7 x}{27 (x^2-x+1)^2}+\frac{7-5 x}{27(x^2-x+1)}+\frac{1-2 x}{9 (x^2-x+1)^3}+\frac{5}{27 (x+1)}+\frac{1}{9 (x+1)^2}+\frac{1}{27 (x+1)^3}. $$ So... good luck!
18h
comment Lebesgue Stieltjes measure
I used that term in a very naive sense. If you can show that some property is satisfied for all generating sets and stable under $\sigma$-algebra operations, then this property should hold for all measurable set. In this specific case, Dynkin's $\pi$-$\lambda$ theorem is more suited for our aim. Finally, function-space versions of this theorem are usually called monotone class theorem. To be accurate, what I really wanted to mention was the $\pi$-$\lambda$ theorem.
19h
comment Lebesgue Stieltjes measure
We are not comparing $m$ and $\nu_F$. We are comparing $m$ and the pushforward measure $F_* \nu_F$. But your idea is correct; we can first show that $m((a, b]) = \nu_F(F(x) \in (a, b])$ and then the claim follows from usual monotone class argument.
1d
revised Closed-form of $\int_0^\infty \frac{1}{\left(a+\cosh x\right)^{1/n}} \, dx$ for $a=0,1$
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1d
comment $C(M)=\{A\in M_n(\mathbb{C}) \mid AM=MA\}$ is a subspace of dimension at least $n$.
I guess that we can apply this argument block-wise to the Jordan form of $M$ to reach the conclusion, since each block in the Jordan form is either scalar multiple of identity matrix or something whose minimal polynomial coincides with their characteristic polynomial.
1d
comment $C(M)=\{A\in M_n(\mathbb{C}) \mid AM=MA\}$ is a subspace of dimension at least $n$.
@GAVD, I mean, the minimal polynomial may have degree less than the characteristic polynomial as in the case $$ M = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2 \end{pmatrix}. $$ In this case $M^2 - 3M + 2I = 0$ and $\Bbb{C}[M]$ has dimension 2.
1d
comment $C(M)=\{A\in M_n(\mathbb{C}) \mid AM=MA\}$ is a subspace of dimension at least $n$.
But the subspace $\Bbb{C}[M]$ of $C(M)$ spanned by non-negative powers of $M$ may have dimension lower than $n$; it is in fact equal to the degree of the minimal polynomial, which need not be $n$. How do we guarantee that there are more rooms in $C(M)$?
1d
revised Integral ${\large\int}_0^\infty\frac{dx}{\sqrt[4]{7+\cosh x}}$
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1d
revised Integral ${\large\int}_0^\infty\frac{dx}{\sqrt[4]{7+\cosh x}}$
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1d
answered Integral ${\large\int}_0^\infty\frac{dx}{\sqrt[4]{7+\cosh x}}$
1d
comment Closed-form of $\int_0^\infty \frac{1}{\left(a+\cosh x\right)^{1/n}} \, dx$ for $a=0,1$
Simple and nice trick! (+1)
1d
answered Closed-form of $\int_0^\infty \frac{1}{\left(a+\cosh x\right)^{1/n}} \, dx$ for $a=0,1$
1d
awarded  Guru
1d
comment Why doesn't the “naive” scalar product for $SO(n)$ yield something invariant?
Just for curiosity, but why do you still have kets instead of bras when you take transpose to them?
1d
comment Evaluating $\lim_{n\to\infty} e^{-n} \sum\limits_{k=0}^{n} \frac{n^k}{k!}$
@Did, I agree that the existence of a dominating function is by no means obvious here. I recognized this problem long before, but was lazy enough to leave it unmodified. Now I fixed my proof so that it looks much nicer.
1d
revised Evaluating $\lim_{n\to\infty} e^{-n} \sum\limits_{k=0}^{n} \frac{n^k}{k!}$
Simplified the proof
1d
revised Evaluating $\lim_{n\to\infty} e^{-n} \sum\limits_{k=0}^{n} \frac{n^k}{k!}$
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2d
comment Does $\int_{-1}^1\frac{\arctan x}{\text{arctanh}\,x}\,dx$ have a closed form?
@RandomVariable You are right :)