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Consider a data generating process $$Y=f(X)+\varepsilon$$ where $\varepsilon$ is independent of $x$ with $\mathbb E(\varepsilon)=0$ and $\text{Var}(\varepsilon)=\sigma^2_\varepsilon$. According to Hastie et al. "The Elements of Statistical Learning" (2nd edition, 2009) Section 7.3 p. 223, we can derive an expression for the expected prediction error of a regression fit $\hat f(X)$ at an input point $X=x_0$, using squared-error loss:

\begin{align} \text{Err}(x_0) &=\mathbb E[(Y-\hat f(x_0))^2|X=x_0]\\ &=(\mathbb E[\hat f(x_0)−f(x_0)])^2+\mathbb E[(\hat f(x_0)−\mathbb E[\hat f(x_0)])^2]+\sigma^2_\varepsilon\\ &=\text{Bias}^2\ \ \ \quad\quad\quad\quad\quad\;\;+\text{Variance } \quad\quad\quad\quad\quad\quad+ \text{ Irreducible Error} \end{align}

(where I use the notation $\text{Bias}^2$ instead of $\text{Bias}$).

Question: What are the expectations taken over? What is held fixed and what is random?

The question arose in the comments of the thread "Why is there a bias variance tradeoff? A counterexample".

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  • $\begingroup$ $x_0$ and $f(x_0)$ are assumed fixed. $\hat{f}(.)$, however, depends on the training data, which is taken as random. Finally, there is randomness in $Y|X=x_0$, which is assumed independent of $\hat{f}(.)$ (conditional on $X=x_0$). $\endgroup$ – Tim Mak Jun 9 '20 at 10:01
  • $\begingroup$ @TimMak, sounds reasonable. So what are the expectations taken over? The random sampling of the training set jointly with the randomness in $\varepsilon$, i.e. a double integral? I have added the qualification that $\varepsilon$ is independent of $x$, but should that be the case? (I guess it should.) $\endgroup$ – Richard Hardy Jun 9 '20 at 10:16
  • $\begingroup$ Actually, I read the book in more detail and found that in their presentation $X$ is actually assumed fixed. Section 7.3 refers back I think, to Section 2.9, where they said "For simplicity here we assume that the values of $x_i$ in the sample are fixed in advance (nonrandom)". However, in other presentations, it is common to have $X$ assumed random also. Either way, $\hat{f}(.)$ depends on both $X$ and $Y$, and hence is random even when $X$ is held fixed. $\endgroup$ – Tim Mak Jun 10 '20 at 2:04
  • $\begingroup$ @TimMak, since you are knowledgeable about the matter, consider writing up an answer. $\endgroup$ – Richard Hardy Jun 10 '20 at 5:52
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$X$ is assumed fixed; see Section 2.9, p. 37:

For simplicity here we assume that the values of $x_i$ in the sample are fixed in advance (nonrandom).

Then the only source of random variation here is $\varepsilon$. Hence, the expectations are taken w.r.t. to the distribution of $\varepsilon$.

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