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 Post subject: I need a decider
PostPosted: Wed, 7 Mar 2012 06:14:44 UTC 
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I'm having a debate with my roommate.

In short my roommate stated in a statistics homework assignment that chebychev's theorem supported a claim that a particular distribution was a skewed distribution. I do not think it's possible to derive any conclusion about a distribution's graphical shape using chebychev's. Would anyone care to speculate? Thanks

MF

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let:
a=b
ab=b^2
ab-a^2=b^2-a^2
a(b-a)=(b+a)(b-a)
a=b+a
a=2a
1=2


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 Post subject: Re: I need a decider
PostPosted: Wed, 7 Mar 2012 09:44:57 UTC 
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mgfortes wrote:
I'm having a debate with my roommate.

In short my roommate stated in a statistics homework assignment that chebychev's theorem supported a claim that a particular distribution was a skewed distribution. I do not think it's possible to derive any conclusion about a distribution's graphical shape using chebychev's. Would anyone care to speculate? Thanks

MF


Chebyshev's inequality (one- or two-sided) doesn't tell you anything about skewness. And it is a pretty crude bound.

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\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}


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 Post subject: Re: I need a decider
PostPosted: Wed, 7 Mar 2012 14:48:53 UTC 
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The usual form of Chebychev's Inequality only requires two moments of the distribution, so it is valid even if a distribution does not have higher moments. And yes, it's a crude bound because it has to work for any distribution with a finite variance.

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