20111116, 14:45  #56  
Oct 2011
Maryland
2×5×29 Posts 
Quote:


20111116, 15:29  #57 
Mar 2010
2^{6}×3 Posts 
Believe me, it wouldn't be cheating. This is an unsolved problem. If you think that it may help you to read previous results, then do not hesitate. By now, few skillful mathematicians were trying to prove it and partial results were published. There is a danger in this and probably you will experience it. Trying to prove it is time consuming. Also, it is very easy to make a mistake. If at some point you think that you found a simple proof, try rather to find error. This is what my experience says to me. Later I will write in this thread another problem, easier one. This time it will not be open problem (I proved it) and I believe it may be helpful to find a final proof. Last fiddled with by literka on 20111116 at 15:30 
20111116, 17:24  #58  
"Forget I exist"
Jul 2009
Dartmouth NS
2×3×23×61 Posts 
Quote:
let then so the average of is so the average for is although I feel confident this can help I don't quite know how. 

20111116, 17:46  #59  
Apr 2010
5·31 Posts 
Quote:


20111116, 20:31  #61  
"Bob Silverman"
Nov 2003
North of Boston
1D48_{16} Posts 
Quote:
formulation. I have not been able to push through an argument, however. I also looked at an inductive approach. It is easy to prove for dimension 1 or 2. But showing that it is true for n+1 based on its being true for n got rapidly intractable for me. I assume that others have tried induction and failed??? 

20111116, 21:12  #62 
Apr 2010
5·31 Posts 
(Referring to the nomenclature of post #59)
The claim clearly holds in the case that the hyperplane cuts through 2^{[I]n[/I]1} unconnected edges of the hypercube. This case is met if e.g. the hyperplane's normal vector (a_{1},...,a_{n}) has a "dominating" element so close to +/1 (thereby suppressing all other elements in magnitude) that the hyperplane cuts all 2^{[I]n[/I]1} parallel edges of the hypercube running along the corresponding coordinate direction (corresponding to the index of the dominating element). Edit: In fact, cutting through a prolonged edge (overshooting each vertex by one unit length) instead of cutting through the edge itself would suffice. This widens the applicability of the allparalleledges case. Working on that. Last fiddled with by ccorn on 20111116 at 21:47 
20111116, 21:44  #63 
Mar 2010
2^{6}·3 Posts 
Simpler problem (proved)
Let C(n,k) denote binomial coefficient C(n,k)=n!/[k! * (nk)!].
Operator "<=" means "less or equal". Let a_{1}>=a_{2}>=...>=a_{2n}. Let { b_{i} } (1<=i<=2n) be a Bernoulli sequence of independent random variables i.e. they are independent and P(b_{i}=1)=P(b_{i}=1)=0.5 Let a random variable x be defined as the sum x=b_{1}*a_{1}+b_{2}*a_{2}+...+b_{2n}*a_{2n} Then P(x<=a1)>=C(2n,n)/(2^2n) 
20111116, 22:16  #64  
"Forget I exist"
Jul 2009
Dartmouth NS
2×3×23×61 Posts 
Quote:
sum of (n elements)^2 =1 so the average of (elements)^2 = 1/n each element is the square root so I get Last fiddled with by science_man_88 on 20111116 at 22:16 

20111116, 22:22  #65  
Apr 2010
5×31 Posts 
Quote:
Edit: Ah, wait. I overlooked the greater sign in the formula. Edit: No abs() on x? No restriction to a_{2[I]n[/I]} > 0? Last fiddled with by ccorn on 20111116 at 22:41 

20111116, 22:59  #66  
Mar 2010
11000000_{2} Posts 
Quote:
You were right, there is an error in the text. I should be : Let C(n,k) denote binomial coefficient C(n,k)=n!/[k! * (nk)!]. Operator "<=" means "less or equal". Let a_{1}>=a_{2}>=...>=a_{2n}>=0 Let { b_{i} } (1<=i<=2n) be a Bernoulli sequence of independent random variables i.e. they are independent and P(b_{i} = 1) = P(b_{i} = 1) = 0.5 Let a random variable x be defined as x=abs(b_{1}*a_{1}+b_{2}*a_{2}+...+b_{2n}*a_{2n}) where abs(y) means absolute value of y. Then P(x<=a1)>=C(2n,n)/(2^2n) Last fiddled with by literka on 20111116 at 23:09 

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