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From the book "Understanding Machine Learning: From Theory to Algorithms", The Realizability Assumption is defined as follows:

There exists $h^{\star}\in \mathcal{H}$ s.t. $L_{(D,f)}(h^{\star})=0$. Note that this assumption implies that with probability 1 over random samples, $S$, where the instances of $S$ are are sampled according to $\mathcal{D}$ and labeled by $f$, we have $L_{S}(h^{\star})=0$. The realizability assumption implies that for every ERM hypothesis we have that $L_{S}(h_S)=0$. However, we are interested in the true risk of $h_S$, $L_{(D,f)}(h_S)$, rather than its empirical risk.

Isn't the true risk $\big{(} L_{(D,f)}(h_S) \big{)}$ already ZERO by the definition of the realizability assumption?, I mean isn't $L_{S}(h_S)=0$ because $L_{(D,f)}(h_S)=0$? and if it isn't, why then the realizability assumption implies that $L_{S}(h_S)=0$?

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  • $\begingroup$ Does $h_S = h^*$ for all $S$? $\endgroup$
    – jbowman
    Commented Feb 21, 2023 at 21:15
  • $\begingroup$ @jbowman Why not? $\endgroup$
    – ammar
    Commented Feb 22, 2023 at 22:35
  • $\begingroup$ @jbowman what are you implying by that? if we assume they're not equal the why would the realizability assumption imply $L_S(h_S)=0$? $\endgroup$
    – ammar
    Commented Feb 23, 2023 at 21:25
  • $\begingroup$ Just because $L_S(h_S)=0$ doesn't mean that $L_{(D,f)}(h_S) = 0$. For the latter to be true, $h_S$ must equal $h^*$, because it's $h^*$ for which $L_{(D,f)}(h) = 0$ ,but $h_S$ is associated with a particular sample, not the population, so there's no reason why that would be so. $\endgroup$
    – jbowman
    Commented Feb 23, 2023 at 21:30
  • $\begingroup$ Does this link help: stats.stackexchange.com/questions/304991/… $\endgroup$
    – jbowman
    Commented Feb 23, 2023 at 21:35

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