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Jul 31, 2016 at 16:08 vote accept Erel Segal-Halevi
Jul 31, 2016 at 1:47 history tweeted twitter.com/StackCompSci/status/759566127072706560
Jul 30, 2016 at 22:52 answer added Ariel timeline score: 2
Jul 30, 2016 at 21:38 comment added Erel Segal-Halevi @Ariel I did not understand the last inequality in your comment. Can you please expand? (maybe as an answer)
Jul 20, 2016 at 20:13 comment added Ariel Suppose you're in the realizable model, i.e. you want to learn some $f^*\in \mathcal{H}\subseteq 2^\mathcal{X}$ where $VCdim(\mathcal{H})=d$. Let $T$ be your training set obtained in the halving procedure, then you wish to bound $\Pr\left(h(x)\neq f^*(x) | x\notin T\right)$, where $h$ is the hypothesis which minimizes the empirical error. But $\Pr\left(h(x)\neq f^*(x) | x\notin T\right) = 2\Pr_{x\sim \mathcal{D}}\left(h(x)\neq f^*(x)\right)=2err(h)$, so you can still use the usual bound (with the extra $2$ factor).
Jul 20, 2016 at 13:34 comment added Erel Segal-Halevi @Ariel in the original setting, the training samples and the test samples are independent. In the halving procedure, the two halves are dependent. As an example, suppose there is a total of 1000 samples, one of the samples is X and the other 999 samples are Y. If you treat it as a finite support distribution, there is a chance of 0.001 that X will be in the train set and an independent chance of 0.001 that X will be in the test set, so there is positive chance to see X in both sets. In the halving procedure, there is 0 chance to see X twice.
Jul 19, 2016 at 20:30 comment added Ariel maybe i don't understand your setting, but we have (for finite VC dimension), a bound on the error in terms of the number of samples, which holds for any distribution $\mathcal{D}$. Isn't your case simply taking a distribution with finite support $S\subseteq \mathcal{X}$ (sample space), and asking what is the error when the number of samples is greater than $|S|/2$?
Jul 19, 2016 at 19:37 history asked Erel Segal-Halevi CC BY-SA 3.0