This is a follow up from a previous question: Changing the order of Conditional Expectation.
The reason I asked this question earlier is due to part of a proof that I am trying to understand. In particular, I am trying to understand the second equality. To my understanding, at this line, we should have $\mathbb{E}_g(\| g \|_\infty \mathbb{E}_X \mathbb{E}_\epsilon \| \sum_{i = 1} ^N \epsilon_i X_i \|)$ instead of the the other way around. It seems like the author is claiming this follows from independence, but I am not sure why:
Here is the definition of the subscripts under expectation:
Here is the definition of the $\epsilon$ appeared in the proof and Theorem 6.7.1.:
Here is how I would get from line 2 to line 3. Let $G, X, \epsilon$ be mutually independent random vectors. We can consider $f(\epsilon, X)$ and $h(G)$ for some measurable functions $f, h$. Now, assuming all expectations are finite, we get
\begin{align} E[h(G)E[f(\epsilon, X)|(X,G)]] &\overset{(a)}{=}E[h(G)E[f(\epsilon, X)|X]] \\ &\overset{(b)}{=}E[h(G)]\: E[E[f(\epsilon, X)|X]]\\ &\overset{(c)}{=}E[h(G)] E[f(\epsilon, X)] \end{align} where
Step (a) holds because $E[f(\epsilon, X)|(X,G)]=E[f(\epsilon, X)|X]$ since $G$ is independent of $(\epsilon, X)$;
Step (b) holds because the conditional expectation $E[f(\epsilon, X)|X]$ is a measurable function of $X$, and every measurable function of $X$ is independent of every measurable function of $G$;
Step (c) holds by iterated expectations.
You can use $h(G)=||G||_{\infty}$ and $f(\epsilon, X) = ||\sum_{i=1}^n \epsilon_iX_i||$.
In general I do not like the subscripts on expectations. I know they are usually used to make things simpler, but they can also make things more confusing. In this context, I do not really know what the definitions of those subscripts are (they seem to implicitly be conditional expectations given something). I like using standard iterated expectations like $E[X]=E[E[X|Y]]$.