Proof related to the expected value of cardinality

baiyang11

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Jul 9, 2013
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Consider \(\displaystyle N\) random variables \(\displaystyle X_{n}\) each following a Bernoulli distribution \(\displaystyle B(r_{n})\) with \(\displaystyle 1 \geq r_{1} \geq r_{2} \geq ... \geq r_{N} \geq 0\). If we make following assumptions of sets \(\displaystyle A\) and \(\displaystyle B\):

(1) \(\displaystyle A \subset I \) and \(\displaystyle B \subset I\) with \(\displaystyle I=\{1,2,3,...,N\}\)

(2) \(\displaystyle |A \cap I_{1}| \geq |B \cap I_{1}|\) with \(\displaystyle I_{1}=\{1,2,3,...,n\}, n<N\)

(3) \(\displaystyle |A|=|B|=n\)

Do we have \(\displaystyle \mathbb{E}(\Sigma_{a\in A} X_{a}) \geq\mathbb{E}(\Sigma_{b\in B} X_{b})\)?

To avoid confusion, \(\displaystyle \mathbb{E}\) means expected value.


Thanks!
 
So now you are going to insist not only \(\displaystyle 1 \geq r_{1} \geq r_{2} \geq ... \geq r_{N} \geq 0\) but that the first n elements of I comprise A? That indeed is loss of generality in the (re)ordered set of r's.
 
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