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Let $\epsilon_1,...,\epsilon_n$ be i.i.d. random signs, $\mathbf{u}_1,...,\mathbf{u}_n$ i.i.d. uniform random vectors on the unit sphere $\mathbb{S}^{d-1}$, assuming $d$ even, and $\mathbf{v}_1,...,\mathbf{v}_n$ be their half-truncations, that is $\mathbf{v}_i[j] = \mathbf{u}_i[j]$ for all $j \in \left \{1,...,d/2\right\}$, else $\mathbf{v}_i[j] = 0$, where $\mathbf{v}_i[j]$ denotes the $j$-th entry of $\mathbf{v}_i$. We assume that the $\epsilon_i$'s are independent from the $\mathbf{u}_i$'s.
I am looking for nontrivial lower bounds on the following quantity. $$\mathbb{E}\left |\sum_{i = 1}^n \epsilon_i \| \mathbf{v}_i\|^2_2 \right |$$ Any help would be greatly appreciated!

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$\newcommand\ep\epsilon\newcommand\v{\mathbf v}$Conditioning on the $\ep_i$'s and using Jensen's and Szarek's inequalities, we have $$E\Big|\sum_1^n\ep_i\|\v_i\|_2^2\Big| \ge E\Big|\sum_1^n\ep_ic_i\Big|\ge\frac1{\sqrt2}\sqrt{\sum_1^nc_i^2},$$ where $$c_i:=E\|\v_i\|_2^2=1/2,$$ by symmetry. So, $$E\Big|\sum_1^n\ep_i\|\v_i\|_2^2\Big| \ge \frac{\sqrt n}2.$$


This lower bound is of course sharp, up a universal constant factor, because $$E\Big|\sum_1^n\ep_i\|\v_i\|_2^2\Big| \le\sqrt{E\Big(\sum_1^n\ep_i\|\v_i\|_2^2\Big)^2} =\sqrt{\sum_1^n E\|\v_i\|_2^4} = \sqrt{n E\|\v_1\|_2^4} \le\sqrt n.$$

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  • $\begingroup$ Thank you so much for your answer. This is indeed what I was looking for! $\endgroup$
    – TCi
    Commented Aug 6, 2021 at 15:16

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