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Nov 5, 2022 at 7:01 comment added groceryheist You were right about the covariance structure. I was confused because I thought P(A=Y)=\alpha would imply P(A=1|Y=1) = alpha. we needed a stronger assumption than P(A=Y) = alpha because the false-positive and false-negative rates can be differnet.
Nov 4, 2022 at 4:11 comment added groceryheist So I'm now quite convinced that $P(A=1|Y=1) = \alpha$ and revised the answer I submitted previously to show why.
Oct 22, 2022 at 17:35 history edited Manoel Ribeiro CC BY-SA 4.0
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S Oct 22, 2022 at 16:41 review First answers
Oct 22, 2022 at 18:14
S Oct 22, 2022 at 16:41 history answered Manoel Ribeiro CC BY-SA 4.0