Are the entries in matrix/vector product independent This question has been confusing me.  Choose a random square matrix $M$ with $M_{i,j} \in \{1,-1\}$ so that $M_{i,j} = 1$ with probability $1/2$ and $M_{i,j} = -1$ with probability $1/2$. Say all the $M_{i,j}$ are i.i.d. Now select a random vector $v$ so that $v_i = 1$ with probability $1/2$ and $v_i = -1$ with probability $1/2$  and all the $v_i$ are i.i.d. We also have that $v$ and $M$ are independent. Let $y = Mv$.

Are the entries $y_i$ independent of each other?

 A: I saw a similar question on here recently, and this solution is adapted from that one (can't remember the link). Let me know if I've erred.
Write
$$
  y
= Mv
= \sum_{j=1}^n v_j M_j
$$
where $M_j$ is the $j^{th}$ column of the matrix $M$. For a fixed vector $a$ of $1$'s and $-1$'s, of which there are $2^n$, $\sum_{j=1}^n a_j M_j$ is a vector of independent r.v.s taking values in $S := \{ -n,-n+2,\ldots,n-2,n\}$. Note that $S$ has $n+1$ elements. Their distribution is given by
$$
  P\left[\sum_{j=1}^n a_j M_{ij} = n-2k \right]
= \binom{n}{k}\frac{1}{2^n}
$$
for $k = 0,1,...,n$. Thus for any fixed $x \in S^n$ of the form $x = n\vec{1} - 2w$ with 
$w \in \{0,1,...,n\}^n$,
$$
  P\left[\sum_{j=1}^n a_j M_j = x \right] 
= \prod_{i=1}^n \left(\binom{n}{w_i} \frac{1}{2^n} \right)
= \frac{1}{2^{n^2}}\prod_{i=1}^n \binom{n}{w_i}. 
$$ 
Let $A$ be the set of all such $a$. Therefore
\begin{align}
   P[y=x]
&= \sum_{a \in A} P\left[\sum_{j=1}^n a_jM_j = x \mid v=a \right]P[v=a] \\
&= \sum_{a \in A} \frac{1}{2^{n^2}} \left(\prod_{i=1}^n \binom{n}{w_i} \right)
   \frac{1}{2^n} \\
&= \frac{1}{2^{n^2}}\prod_{i=1}^n \binom{n}{w_i}.
\end{align}
(This is a kind of multinomial distribution).
You can use this to determine dependence. For instance, 
\begin{align}
   P[y_k = w_k,y_\ell = w_\ell]
&= \frac{1}{2^{n^2}}\binom{n}{w_k}\binom{n}{w_\ell} \sum_{w}\prod_{i \neq k,\ell}
   \binom{n}{w_i}  \\
&= \frac{1}{2^{n^2}}\binom{n}{w_k}\binom{n}{w_\ell} (2^n)^{n-2}  \\
&= \frac{1}{4}\binom{n}{w_k}\binom{n}{w_\ell} 
\end{align}
and you can verify that this is equal to $P[y_k=w_k]P[y_\ell = w_\ell]$. Therefore the $y_i$'s are independent.
A: Since the matrix rows of $M$ are i.i.d. and the column vector $v$ is i.i.d and $y_i$ is the $i$th row of $M$ times $v$, you won't get any dependence among $y_i$, since there is no dependence among the rows of $M$.
So the answer is yes!
