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Feb 5, 2016 at 17:35 comment added Anthony Quas There's a missing parenthesis in the inequality. It should be $\mathbb E\Big( \Big[ \sum \big(1_{A_i}-\mu(A_i)\big)\Big]^2\Big)$. It's positive because square numbers are non-negative.
Feb 5, 2016 at 17:09 comment added Darío G Also, how exactly the last inequality help me to find the two indices satisfying $\mu(A_i\cap A_j)\geq \epsilon^2$? Couldn't it be possible that the indices witnessing the maximum value of $\mu(A_i\cap A_j)$ also depend on N, and the measure of the intersection keep being always strictly less than $\epsilon^2$?
Feb 5, 2016 at 16:03 history answered user83457 CC BY-SA 3.0