Why are independence and mean-zero necessary for the symmetrization lemma to hold? I'm going through the proof of the symmetrization lemma (Vershynin), which says
$$ \frac{1}{2} E \left\| \sum_i \epsilon_i X_i \right\| \leq E \left\| \sum_i X_i \right\| \leq 2 E \left\| \sum_i \epsilon_i X_i \right\| $$ where $X_1,\ldots,X_N$ are independent mean-zero random vectors in a normed space. The $\epsilon_i$'s are i.i.d taking on $1$ or $-1$ with equal probability.
However, I don't see why independence or the mean-zero assumption is required anywhere for this to hold because the steps in the proof below don't seem to use either assumption.
====================



Anyone know why this is?
References:
Vershynin, Roman, High-dimensional probability. An introduction with applications in data science (to appear),  ZBL06872475.
 A: Here's why independence is needed. Let's simplify our situation to one-dimensional, and all the $X_i$'s are just Rademacher variables. We'll do something extreme and make everything perfectly correlated. So as $n \to \infty$, the term $E \left\|\sum_i \epsilon_i X_i\right\|$ is the expectation of the absolute value of a Gaussian variable with variance $n$ and is roughly $O(\sqrt{n})$. The middle term on the other hand grows as $O(n)$.
We can show that mean zero is needed in a similar fashion. Consider $X_i = 1$ for all $i$ (I know it's silly but they are independent!) The middle term will now grow as $O(n)$ and the term on the right grows as $O(\sqrt{n})$.
A: Ah..I think I figured this out now. Someone else already answered why mean-zero was necessary due to (6.9).
It's not enough for $\epsilon_i (X_i - X_i')$ and $X_i - X_i'$ to have the same distribution for every $i$ (componentwise). We need it to be true for the vectors $((X_1 - X_1'),\ldots,(X_n-X_n'))$ and $(\epsilon_1 (X_1 - X_1'),\ldots,\epsilon_n (X_n-X_n'))$, which is true if $(X_i)$ is an independent sequence. 
To see why, note that $\epsilon_i (X_i - X_i')$ and $\epsilon_j (X_j - X_j')$ are independent for $i\neq j$ so $X_i - X_i'$ and $X_j - X_j'$ must also be independent, which is the case when $X_i\perp X_j$.
