Why does $\mathbb{E}[\varepsilon^{-2}\|\Sigma (x - y)\|_2^2] = \|x-y\|_2^2$? I was recently reading this cstheory answer and am confused by the following.
Let $x$ and $y$ both be $n$-dimensional vectors with elements from $\{0,1\}$. Let us assume $0 \leq \varepsilon \leq 1$. If $\Sigma$ is a $\varepsilon^{-2} \times n$ matrix of iid unbiased $\pm 1$ random variables then is the following actually true?

$\mathbb{E}[\varepsilon^{-2}\|\Sigma (x - y)\|_2^2] = \|x-y\|_2^2$

I believe we are meant to assume $\varepsilon^{-2}$ is an integer.

I believe there was a typo in the linked answer and it should be

$\mathbb{E}[\varepsilon^{2}\|\Sigma (x - y)\|_2^2] = \|x-y\|_2^2$

Is there a straightforward proof?
 A: I am inclined to believe in user felipa's comment that the $\epsilon^{-2}$ on the LHS should be $\epsilon^2$, else the statement is not correct. Without loss of generality, take $m = \epsilon^{-2}$ and $z = x - y$. Then we attempt to prove the statement
\begin{align*}
\frac{1}{m}\mathbb{E}\|\Sigma z\|_2^2 = \|z\|_2^2
\end{align*}
If we denote the entries of $\Sigma$ as $\sigma_{ij}$ for $1 \le i \le m$ and $1 \le j \le n$, each $\overset{\text{i.i.d}}{\sim} \{-1, 1\}$ with probability 1/2 (Rademacher distribution), then
\begin{align*}
(\Sigma z)_{i} = \sum_{j=1}^{n}\sigma_{ij}z_j
\end{align*}
hence the LHS is
\begin{align*}
\frac{1}{m}\mathbb{E}\sum_{i=1}^{m}\left(\sum_{j=1}^{n}\sigma_{ij}z_j\right)^2 = \frac{1}{m}\sum_{i=1}^{m}\sum_{j=1}^{n}\sum_{j'=1}^{n}\mathbb{E}[\sigma_{ij}\sigma_{ij'}]z_jz_{j'}
\end{align*}
for all $j \neq j'$, independence gives us $\mathbb{E}[\sigma_{ij}\sigma_{ij'}] = \mathbb{E}[\sigma_{ij}]\mathbb{E}[\sigma_{ij'}] = 0$. When $j = j'$, we have $\sigma_{ij}\sigma_{ij'} = 1$ with probability 1. Hence,
\begin{align*}
\frac{1}{m}\mathbb{E}\sum_{i=1}^{m}\left(\sum_{j=1}^{n}\sigma_{ij}z_i\right)^2 = \frac{1}{m}\sum_{i=1}^{m}\sum_{j=1}^{n}z_j^2 = \|z\|_2^2
\end{align*}
and we are done.
