What notions of independence exist for these random variables?

Question

I think I've gotten a bit confused about the notions of independence that the following random variables satisfy. Any help would be greatly appreciated.

Let $x_1, ..., x_k \in \mathbb{Z}_q^n$, each distinct (not random, I want this result to hold in general for any choice of these $x$'s). Let $A_j \in \mathbb{Z}_q^{m \times n}, m < n, b_j \in \mathbb{Z}_q^m$ both uniformly distributed, and independently. Then define:

\begin{equation} y_{i,j} = A_jx_i + b_j \end{equation}

What can I say about the independence of the $y_{i,j}$, i.e., pairwise independence, etc,. I think for each $j$ $y_{i,j}$ are independently and uniformly distributed over $\mathbb{Z}_q^m$, (since they must differ in at least one coordinate), but I am generally confused.

Thanks

Answer 1

I presume $\mathbb Z_q$ denotes a finite field. Are the $b_j$ mutually independent, and independent of the $A_j$ and $x_i$? Then for each $i$ the $y_{i,j}$ are independent and uniform on $\mathbb Z_q^m$, because this is true conditional on the $A_i x_j$.

EDIT: If $Z_q$ is the integers mod $q$ where $q$ is not prime, the $y_{i,j}$ for different $i$ but the same $j$ may not be pairwise independent. Thus consider $y_{1,j} - y_{2,j} = A_j (x_1 - x_2)$. If $x_1 - x_2$ has all entries divisible by some nontrivial divisor $p$ of $q$, then $y_{1,j} - y_{2,j}$ also has all its entries divisible by $p$.

Answer 2

If you don't have a finite field, you can obtain distributions in ideals of $\mathbb{Z}_q$ instead of uniform distributions.

Let $q=27,$ say, and let all $x_i \in 3 \mathbb{Z}_{27}^n,$ then if the distributions of the $b_i$ are also supported on $3 \mathbb{Z}_{27}^n,$ you won't get a uniform distribution since all components of the $y_{i,j}$ you obtain will be in $3 \mathbb{Z}_{27}^m.$