S.O.S. Mathematics CyberBoard

Your Resource for mathematics help on the web!
It is currently Tue, 23 Sep 2014 03:18:48 UTC

All times are UTC [ DST ]




Post new topic Reply to topic  [ 5 posts ] 
Author Message
 Post subject: Probability (1)
PostPosted: Tue, 16 Aug 2011 02:21:08 UTC 
Online
Moderator
User avatar

Joined: Wed, 30 Mar 2005 04:25:14 UTC
Posts: 14052
Location: Austin, TX
A friend of mine on IRC recommended the following problem:

Let K_n be the size of the largest family of independent events in a finite probability space with n points (the uniform probability measure is assumed). What is the value of K_n?

For n=0 the answer is 1 trivially as all we have is \{\}. For n=1, the answer is 2, \{\},\{1\} and similarly for n=2 we have the maximal family \{\},\{1\},\{1,2\}.

Edit: Correction as per outermeasure's observation.

_________________
(\ /)
(O.o)
(> <)
This is Bunny. Copy Bunny into your signature to help him on his way to world domination


Top
 Profile  
 
 Post subject: Re: Probability (1)
PostPosted: Tue, 16 Aug 2011 02:31:41 UTC 
Offline
Member of the 'S.O.S. Math' Hall of Fame

Joined: Sun, 24 Jul 2005 20:12:39 UTC
Posts: 4216
Location: Ottawa Ontario
Shadow wrote:
A friend of mine ......

You got a friend :?:

_________________
I'm just an imagination of your figment...


Top
 Profile  
 
 Post subject: Re: Probability (1)
PostPosted: Tue, 16 Aug 2011 06:43:30 UTC 
Offline
Moderator
User avatar

Joined: Mon, 29 Dec 2008 17:49:32 UTC
Posts: 6882
Location: On this day Taiwan becomes another Tiananmen under Dictator Ma.
Shadow wrote:
A friend of mine on IRC recommended the following problem:

Let K_n be the size of the largest family of independent events in a finite probability space with n points. What is the value of K_n?

For n=0 the answer is 1 trivially as all we have is \{\}. For n=1, the answer is 2, \{\},\{1\} and similarly for n=2 we have the maximal family \{\},\{1,2\}.


Independent according to which probability measure? (My guess is the uniform distribution on n points?) Why isn't \{1\} included in your n=2?

_________________
\begin{aligned}
Spin(1)&=O(1)=\mathbb{Z}/2&\quad&\text{and}\\
Spin(2)&=U(1)=SO(2)&&\text{are obvious}\\
Spin(3)&=Sp(1)=SU(2)&&\text{by }q\mapsto(\mathop{\mathrm{Im}}\mathbb{H}\ni p\mapsto qp\bar{q})\\
Spin(4)&=Sp(1)\times Sp(1)&&\text{by }(q_1,q_2)\mapsto(\mathbb{H}\ni p\mapsto q_1p\bar{q_2})\\
Spin(5)&=Sp(2)&&\text{by }\mathbb{HP}^1\cong S^4_{round}\hookrightarrow\mathbb{R}^5\\
Spin(6)&=SU(4)&&\text{by the irrep }\Lambda_+\mathbb{C}^4
\end{aligned}


Top
 Profile  
 
 Post subject: Re: Probability (1)
PostPosted: Tue, 16 Aug 2011 06:46:49 UTC 
Online
Moderator
User avatar

Joined: Wed, 30 Mar 2005 04:25:14 UTC
Posts: 14052
Location: Austin, TX
outermeasure wrote:
Shadow wrote:
A friend of mine on IRC recommended the following problem:

Let K_n be the size of the largest family of independent events in a finite probability space with n points. What is the value of K_n?

For n=0 the answer is 1 trivially as all we have is \{\}. For n=1, the answer is 2, \{\},\{1\} and similarly for n=2 we have the maximal family \{\},\{1,2\}.


Independent according to which probability measure? (My guess is the uniform distribution on n points?) Why isn't \{1\} included in your n=2?


Yes uniform is the right way to go. And you're right, {1) should be there. My friend is usually very good with probability, so I didn't really think about checking his examples. I'll fix the original post.

_________________
(\ /)
(O.o)
(> <)
This is Bunny. Copy Bunny into your signature to help him on his way to world domination


Top
 Profile  
 
 Post subject: Re: Probability (1)
PostPosted: Tue, 16 Aug 2011 07:21:08 UTC 
Offline
Moderator
User avatar

Joined: Mon, 29 Dec 2008 17:49:32 UTC
Posts: 6882
Location: On this day Taiwan becomes another Tiananmen under Dictator Ma.
Spoiler:
prime divisor counting function (with multiplicity)

Am I on the right track here? Hmm... looks like I am.
Spoiler:
Consider those nontrivial proper subsets --- there are at most \Omega(n) of them:
Claim: For each prime p such that n=p^em, (m,p)=1, we can have at most e subsets that have size p^{f}r, with 1\leq r<m and f<e.
Proof: If there are more than e of them, the product of probabilities would be \dfrac{1}{p^{e+b}}\dfrac{s}{t} with b>0, s and t both coprime to p, contradicting the probability must be of the form \dfrac{k}{n}\quad(k\in\{0,\dots,n\}).
Now throw in the empty set and the whole space, giving a total of \Omega(n)+2 independent events, for all n>1.

_________________
\begin{aligned}
Spin(1)&=O(1)=\mathbb{Z}/2&\quad&\text{and}\\
Spin(2)&=U(1)=SO(2)&&\text{are obvious}\\
Spin(3)&=Sp(1)=SU(2)&&\text{by }q\mapsto(\mathop{\mathrm{Im}}\mathbb{H}\ni p\mapsto qp\bar{q})\\
Spin(4)&=Sp(1)\times Sp(1)&&\text{by }(q_1,q_2)\mapsto(\mathbb{H}\ni p\mapsto q_1p\bar{q_2})\\
Spin(5)&=Sp(2)&&\text{by }\mathbb{HP}^1\cong S^4_{round}\hookrightarrow\mathbb{R}^5\\
Spin(6)&=SU(4)&&\text{by the irrep }\Lambda_+\mathbb{C}^4
\end{aligned}


Top
 Profile  
 
Display posts from previous:  Sort by  
Post new topic Reply to topic  [ 5 posts ] 

All times are UTC [ DST ]


Who is online

Users browsing this forum: No registered users


You cannot post new topics in this forum
You cannot reply to topics in this forum
You cannot edit your posts in this forum
You cannot delete your posts in this forum

Search for:
Jump to:  
Contact Us | S.O.S. Mathematics Homepage
Privacy Statement | Search the "old" CyberBoard

users online during the last hour
Powered by phpBB © 2001, 2005-2011 phpBB Group.
Copyright © 1999-2013 MathMedics, LLC. All rights reserved.
Math Medics, LLC. - P.O. Box 12395 - El Paso TX 79913 - USA