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 Nov 22 awarded Nice Question May 3 awarded Popular Question Nov 16 awarded Notable Question Oct 26 awarded Popular Question Oct 22 accepted Dimension analysis of an integral Oct 19 comment Dimension analysis of an integral So, if there's no $\alpha$ - say, just $\int_{a}^{b} e^{x}dx$, then I cannot use dimensional analysis, correct? Oct 19 asked Dimension analysis of an integral May 24 awarded Teacher Feb 7 comment Real-world example for utility theory Thanks for the answer and book references. Looks like what I called "evidence vector x" you call "outcomes Z", but in general your definition seems to be the same as mine. And still all my questions are not answered - (1) how to program the utility function, is it a proper function or some kind of a mapper? (2) is it possible (or, is it better) to write an algorithm which would learn the utility function on the go rather than hardcoding it? (3) who defines the probabilities and what's the recommended way to do that, should they be updated over time? Jan 19 revised Real-world example for utility theory edited tags Jan 19 asked Real-world example for utility theory Jan 11 comment How to calculate Vapnik-Chervonenkis dimension Aha that makes sense - thanks! Jan 10 awarded Scholar Jan 10 accepted How to calculate Vapnik-Chervonenkis dimension Jan 10 comment How to calculate Vapnik-Chervonenkis dimension thanks for helping out. I'd try to summarize: [1] if I want to prove VC(H)=n, I may choose any layout for n points but then must prove that all "+/-" combinations can be separated with my classifier (and if that's true for all of them, we say that my set of n points is shattered by the classifier), and also [2] it's enough to find just one such layout of n points to prove VC(H)=n. Does that sound correct? Thanks! Jan 9 comment How to calculate Vapnik-Chervonenkis dimension Well, if VC dimension is about finding the size with at least one set shattered, then imagine 2500 "+" examples located inside a rectangle and 2500 "-" ones located outside. I don't see why we cannot do that, how exactly VC theory prevents us from choosing that one? Jan 5 awarded Supporter Jan 5 comment How to calculate Vapnik-Chervonenkis dimension Hmmm. This "you want to find the highest size with at least one set shattered" is exactly what confused me in the first place. A rectangle hypothesis can shatter 5000 "specially placed" points but somehow we ignore this and say VC(H)=4. Why? Srivatsan's "You choose the points and the adversary selects the labels" rule sounds spot on, because then I won't be able to shatter "much". You seem to be explaining it more rigorously but so far it's been close to what they say in the book, and I feel like I'm not satisfied with their answer... Jan 5 comment How to calculate Vapnik-Chervonenkis dimension So finding a 3 point layout which is not possible to shatter with a line, or finding a 100 point layout which is possible to shatter with a rectangle - they're exceptional cases and don't matter because we consider "all possible" sets for a given N and identify how many of those sets are shattered - is that correct? Also, see Srivatsan's comment with "You choose the points, then the adversary chooses the labeling" - I find it very appealing, seems to solve my problem as well as your explanation. Jan 5 awarded Student