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Aug 14, 2022 at 14:01 comment added newandlost I still don't understand how to get from 2.11 to 2.12, $E_X$ is not a constant, if it the expectation value of everything the follows to the right of that symbol if I understand it correctly. I mean expression 2.11 is $\int dx p(x) \int dy (y-f(x))^2 p(y|x)$ , so how come not to considere the integration over $x$?
Aug 14, 2022 at 13:53 comment added newandlost I still don't fully understand it: Assuming $E_{Y|X} ([Y-c]^2|X=x) = \int dy (y^2+c^2-2yc)p(y|x)$, we can get the minimum by taking the derivative, setting to zero and solving c yields: $c = \frac{E_{Y|X} (Y|X=x) }{\int dy\; p(y|x)}$ -> ??? Where is my mistake?
Jul 2, 2017 at 11:31 comment added David Epstein @usεr11852 Not that good any more (too old)
Jul 2, 2017 at 11:29 history edited David Epstein CC BY-SA 3.0
improved notation at one point
Jul 2, 2017 at 0:26 comment added usεr11852 (+1) Totally unrelated but I have to ask based on your location: Are you "really, really good" in Geometry?
Jul 1, 2017 at 20:31 history answered David Epstein CC BY-SA 3.0