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Given the following minimization problem: $$ \text{argmin}_{\theta} \mathbb{E}_{(x,y)}[L(f_{\theta}(y),x)] $$ which can be rewritten as: $$ \text{argmin}_{\theta} \mathbb{E}_x \mathbb{E}_{(y|x)}[L(f_{\theta}(y),x)] \,. $$ Can I say that the minimum of the loss function L, being $\text{argmin}_{\theta} \mathbb{E}_{(y|x)}[L(f_{\theta}(y),x)]$ is preserved and thus $$ \text{argmin}_{\theta} \mathbb{E}_{(x,y)}[L(f_{\theta}(y),x)] = \text{argmin}_{\theta} \mathbb{E}_x ( \text{argmin}_{\theta} \mathbb{E}_{(y|x)}[L(f_{\theta}(y),x)] ) $$ ?

Thank you.

Basically I am trying to get the proof behind what is stated is section 2 of this article: https://arxiv.org/pdf/1803.04189.pdf

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  • $\begingroup$ What do you actually mean by $\arg\min_\theta \mathbb{E}_x \arg\min_{\theta} (...)$? Once you choose a minimizer inside the expectation, what do you mean by averaging the minimizers and how can you choose a minimizer again? $\endgroup$
    – Steve
    Jun 9 at 9:33
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    $\begingroup$ Are you asking whether the solution $\theta$ to your original problem also has the property that for all $x$, $\theta$ solves the minimization problem conditioned on $x$? $\endgroup$
    – usul
    Jun 9 at 15:57
  • $\begingroup$ I am basically asking if $\text{argmin}_{\theta} \mathbb{E}_{(y|x)}[L(f_{\theta}(y),x)]$ is still a solution of $\text{argmin}_{\theta} \mathbb{E}_{(x,y)}[L(f_{\theta}(y),x)]$, with x and y as pair of input data $\endgroup$
    – emasasso
    Jun 10 at 13:11
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Some issues. First, $\arg\min_\theta \mathbb E[L(f_\theta(y), x)|x]$ is a function of $x$, so it makes no sense with the outside $\arg\min$ at the RHS of your relationship. Second, even if we remove the outside $\arg\min$, we still can't interchange minimum with expectation without further assumptions.

However, it can be true (in some sense) if we are dealing with non-parametrized $f$. In that case, I would rather consider conditioning on $y$ instead of $x$, and suppose that $f^*(y) = \arg\min_f \mathbb E[L(f(y), x)|y]$, then it follows directly that, for any measurable $f$: $$\mathbb E[L(f(y), x)] = \mathbb E[\mathbb E[L(f(y), x)|y]] \geq \mathbb E[\mathbb E[L(f^*(y), x)|y]] = \mathbb E[L(f^*(y), x)]$$ which concludes: $$\min_{f}\mathbb E[L(f(y), x)] = \mathbb E[\min_f \mathbb E[L(f(y), x)|y]]$$ and $$\arg\min_f \mathbb E[L(f(y), x)|y]] \subset \arg\min_{f}\mathbb E[L(f(y), x)]$$

Note that this is closely related to Bayes estimator in decision theory. In our case, we can think of $x$ as a parameter with some distribution $\Lambda$, and $f(y)$ as an arbitrary estimator. $f_\Lambda(\cdot) = \arg\min_{f}\mathbb E[L(f(y), x)]$ is called Bayes estimator, and $\arg\min_f \mathbb E[L(f(y), x)|y]]$ is a way to find it (See Theorem 1.1 in Chapter 4 and related background in the very famous Theory of Point Estimation).

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To be properly formulated, all occurrences of argmin should be replaced by min.

Then the hypothesized relation is false.

The RHS of the hypothesized relationship is obtained from $ \text{min}_{\theta} \mathbb{E}_x \mathbb{E}_{(y|x)}[L(f_{\theta}(y),x)] \,. $ by moving (part of the) minimization to between the expectations. I.e., interchanging minnmization with expectation (integration), which is not a valid operation.

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