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 $\mathbb{S}^{d1}$, assuming $d$ even, and $\mathbf{v}_1,...,\mathbf{v}_n$ be their halftruncations, 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!
1 Answer
$\newcommand\ep\epsilon\newcommand\v{\mathbf v}$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 $$E\Big\sum_1^n\ep_i\\v_i\_2^2\Big \le\sqrt{E\Big(\sum_1^n\ep_i\\v_i\_2^2\Big)^2} =\sqrt{\sum_1^n E\\v_i\_2^4} = \sqrt{n E\\v_1\_2^4} \le\sqrt n.$$

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