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We are given a collection of models $\mathcal{F}$ from which particular, but arbitrarily chosen models $f$ are sampled and a function $\text{err}(f')$ returns the error in classification of the model $f'$.

Now, we define a model $f^* := \text{argmin}_{ f \in \mathcal{F}} \text{err}(f)$ (the model in $\mathcal{F}$ that produces the minimum error).

Question

Is the following statement correct?

$$ \inf_{f \in \mathcal{F}} \text{ err}(f) = \text{err}(f^*) $$

Additionally, why use infimum instead of minimum?

That's it. Thanks!

Anthony Krivonos
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  • This boils down to understanding the difference between a minimum and an infimum itself. See, e.g., this discussion - https://math.stackexchange.com/questions/342749/what-is-the-difference-between-minimum-and-infimum . – stochasticboy321 Apr 30 '20 at 21:16
  • In the context of your question, the issue becomes that $f^$ is not well defined - it need not always exist. If it does exist, then yes, $\inf \mathrm{err}(f) = \min \mathrm{err}(f) = \mathrm{err}(f^)$. Exercise: cook up a $\mathcal{F}$ and a $\mathrm{err}$ for which no minimiser exists. – stochasticboy321 Apr 30 '20 at 21:18

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