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bio website rbribeiro.wordpress.com
location Brazil
age 26
visits member for 1 year, 9 months
seen 11 hours ago

Eles passarão... Eu passarinho...


Jul
15
comment $f_n, f \in L^1$ and pointwise convergence implies …
Your space has finite measure? Which norm is $| f_n - f|$ ?
Jul
10
answered Text on convergence theorems in probability theory (various modes of convergence)
Jul
10
answered Application of Holder's Inequality
Jun
14
awarded  Enthusiast
Jun
10
comment Number of groups that can be formed
The order you form your group doesn't matter. Then you are counting a selection of 5 groups 5! times. If you don't see this, try to assign names (or numbers) to the players.
Jun
10
answered Measure Theory Question 5
Jun
9
answered Sum of bounded in probability random variables
Jun
2
comment Highest degree vertices in random graph with hidden clique
Wonderful! Now we have a solution! =) Would be possible to reduce the answer? I was wondering, fixed an $l$-set, by Chernoff we have that the probability of find a vertex with degree in $G(l, 1/2)$ greater than $2l/3$ is exponentially small, if you look to all possibilities of $l$-sets this goes to zero if $l > C\sqrt{nlog(n)}$? If yes, this would be enough, because we would prove that the probability of exist a vertex with degree large (> $2l/3$) in some $l$-set goes to zero.
Jun
2
awarded  Informed
Jun
2
comment Highest degree vertices in random graph with hidden clique
You are right, @DPoole. Maybe a combinatorial argument can finish the problem. You have that almost all vertices lie close to $n/2$ and the worst case would be your $k$ vertices already form a clique. So, maybe it would be possible to show that, if $k$ is large enough, the degree of the vertices in $V(G) - k$ couldn't be large... Any ideas? I didn't think about this right now... I will give it a try later...
Jun
1
comment Is expectation countably additive?
Right! You should have the right hypothesis to apply the monotone convergence theorem. In this case, $X$ Positive.
May
30
answered Highest degree vertices in random graph with hidden clique
May
28
comment Integration vs Summation
Usually, when I need to compare the sum with the integral I assume $f$ positive and then I divide the semi-line $[0,\infty] $ in intervals of length 1. If you do this, the sum starting from 1 is the sum of the rectangles' area with height $f(x)$ and width 1. A simple draw can tell you what happens if you start the sum from 1 or zero.
May
21
comment Proving the Principle of Uniform Boundedness for continuous functions on $\mathbb{R}$ - less work than the Banach–Steinhaus theorem?
It just a question of norms. The norm of a operator is the sup of T(x) with x in the unitary ball. Then, if you want to prove that the norm of a operator is bounded, you have to determine this sup over the unitary ball and not over other open set. Got it?
May
21
comment Proving the Principle of Uniform Boundedness for continuous functions on $\mathbb{R}$ - less work than the Banach–Steinhaus theorem?
This principle is a consequence of the Baire Category Theorem and the fact preimage of closed sets through continuos functions is again closed and this is exactly what you showed. When you have operators you do the same, but then you use linearity to translate the open set to the unitary ball and there you calculate the operators norm. This allow you show that the norm of all your operators are bounded... ;)
May
21
answered Characteristic Function Inversion
May
21
answered Chebyshev inequality for $n=1$?
May
19
comment Derivative of average number (density) of clusters in Erdős–Rényi random graph
@antarcticfox exactly! Because of this I said his probability is a lower bound, once you can have a component of k vertices without all the connections.
May
18
comment Derivative of average number (density) of clusters in Erdős–Rényi random graph
@the_candyman, what I didn't understand is that all the vertices must be connected. You can have a k-cluster without connect all his vertices, then the probability you gave is a lower bound to the question... Am I missing something in your argument?
May
18
comment Derivative of average number (density) of clusters in Erdős–Rényi random graph
@antarcticfox, you can't just sum over k? Your variable is just a sum of the_candyman's variables with $k=1,..,n$.