# Confused about The Realizability Assumption

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 $$L_{S}(h_S)=0$$?

• Does $h_S = h^*$ for all $S$? Commented Feb 21, 2023 at 21:15
• @jbowman Why not? Commented Feb 22, 2023 at 22:35
• @jbowman what are you implying by that? if we assume they're not equal the why would the realizability assumption imply $L_S(h_S)=0$? Commented Feb 23, 2023 at 21:25
• 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. Commented Feb 23, 2023 at 21:30
• Does this link help: stats.stackexchange.com/questions/304991/… Commented Feb 23, 2023 at 21:35