Independence and equality of variance in random matrix multiplication and element-wise multiplication

Question

Let $B\in\mathbb{R}^{m\times n}$ be a random matrix, sampled i.i.d. from a uniform distribution $\mathcal{U}$ with bounds $[-a, a]$ and $e\in\mathbb{R}^{m}$ be a vector. We define $e'\in\mathbb{R}^{n}$ with the multiplication $$ e'=eB $$

Under what conditions the vector $e'$ has elements that are independent, have a zero mean, and have equal variance $\sigma^2$?

Let $y\in\mathbb{R}_{\geq 0}^{n}$ be a vector with elements that are positive or zero. If we redefine $e'$ as $$ e'=eB\odot y $$ where $\odot$ is the Hadamard (element-wise) multiplication, what conditions should be satisfied for $e'$ to have elements that are independent, have a zero mean, and have equal variance?

In either case, what would the variance be?

Answer 1

Since all the entries of $B$ are independent, then the entries of $e'$ will be as well.

\begin{align} \mathbb E\left[e'\right] &= \mathbb E\left[Be\right] = \mathbb E\left[B\right]e = \textbf{0} \end{align}

this proves that all entries have a $0$ mean. For the variance you have to compute:

\begin{align} \mathbb V\left[e'_j\right] &= \mathbb V\left[\sum_{k=1}^n B_{j,k} e_k\right]\\ &= \sum_{k=1}^n e_k^2 \mathbb V\left[B_{j,k}\right]\\ &= \frac1{3}a^2 \sum_{k=1}^n e_k^2 = \frac{a^2}3\left\|e\right\|^2 \end{align}

For the last one

\begin{align} \mathbb {V}\left[(Be \odot y)_j\right] &= y_j^2 \mathbb V\left[(Be)_j\right] = \frac{a^2}3y_j^2\left\|e\right\|^2. \end{align}