724 reputation
619
bio website
location
age
visits member for 3 years
seen Aug 23 at 2:58

Aug
17
awarded  Yearling
Jul
12
awarded  Good Question
Jul
2
awarded  Curious
May
17
awarded  Nice Question
Apr
4
comment A tricky term-by-term differentiation
"The sum is finite" means that the series converges.
Apr
3
awarded  Notable Question
Mar
26
revised A tricky term-by-term differentiation
edited title
Mar
26
comment A tricky term-by-term differentiation
Presumably $M$ will depend on $j$ though.
Mar
25
revised A tricky term-by-term differentiation
added 4 characters in body
Mar
25
comment A tricky term-by-term differentiation
What is the dominating function for $s_m$? Why is it integrable?
Mar
24
comment A tricky term-by-term differentiation
@AlexBecker: No, my question is how to justify this particular differentiation.
Mar
24
asked A tricky term-by-term differentiation
Dec
9
awarded  Popular Question
Nov
5
awarded  Popular Question
Aug
31
accepted Spherical harmonics give all the irreducible representations of $SO(3)$?
Aug
28
comment Spherical harmonics give all the irreducible representations of $SO(3)$?
It makes sense. Why doesn't this work for $SO(4)$?
Aug
26
awarded  Critic
Aug
26
comment A maximal Hoeffding's inequality?
But the bound in the last line, $\exp(-t^2/n)$, is effectively worse than $n\exp(-t^2)$ (which is immediate from the union identity).
Aug
25
comment A maximal Hoeffding's inequality?
I'm confused. In [2], you need $Y$ to be of mean $0$; furthermore, $v$ is a r.v. instead of a constant by your assumption.
Aug
22
comment A maximal Hoeffding's inequality?
Your identity is wrong. You should instead have $$\mathbb P\Big( \max_{1\le k\le n} \Big | \frac{X_1+\cdots+ X_k}{\sqrt{k}} \Big | \ge t\Big) = \mathbb P\Big( \bigcup_k \Big\{ \Big|\frac{X_1+\cdots+ X_k}{\sqrt{k}} \Big | \ge t \Big\} \Big).$$