Consider a Hadamard code $Had:\{0,1\}^n \to \{0,1\}^{2^n}$, where the $\vec{j}$-th coordinate of $Had(\vec{v})$ is $\langle\vec{v},\vec{j}\rangle \bmod(2)$. Here we can think of $\vec{j}$ as a point in the space $\{0,1\}^n$, while $\vec{v}$ is a "linear test" or the evaluation of the linear function $\vec{v}$ at the point $\vec{j}$. Geometrically, $\vec{v}$ divides the space $\{0,1\}^n$ into two partitions - points that evaluate to $1$ and those that evaluate to $0$, and the Hadamard codeword simply lists these evaluations in some fixed order on the points in the space.
It is known, and easy to show, that every (non-zero) vector $\vec{v}$ divides the space $\{0,1\}^n$ into two exactly equal halves. That is, there are precisely $2^{n-1}$ points that evaluate to $1$. This gives us a code of relative distance exactly $1/2$.
Notice that for any coordinate $\vec{j}$ and two distinct non-zero tests $\vec{v}_1$ and $\vec{v}_2$, the evaluations $\langle\vec{v}_1,\vec{j}\rangle \bmod(2)$ and $\langle\vec{v}_2,\vec{j}\rangle \bmod(2)$ are independent in a sense. More formally, fixing two distinct vectors $\vec{v}_1$ and $\vec{v}_2$, and choosing the coordinate $\vec{j}$ uniformly at random, the random variables $\langle\vec{v}_1,\vec{j}\rangle \bmod(2)$ and $\langle\vec{v}_2,\vec{j}\rangle \bmod(2)$ are independent.
Is this independence carried over when we relax the Hadarmard code into an "almost" Hadamard code?
Let me clarify what I mean. Suppose we now have a linear code $C:\{0,1\}^n \to \{0,1\}^{N}$ where the relative weight of every codeword is now $$\frac{1}{2}+\epsilon$$ as opposed to exactly half. Now if we choose a random coordinate $\vec{j}$ (which would still be a vector in $\{0,1\}^n$, though now the support is not the whole of $\{0,1\}^n$ but a subset of size $\log(N)$) and two fixed non-zero and distinct vectors $\vec{v}_1$ and $\vec{v}_2$, would the random variables $\langle\vec{v}_1,\vec{j}\rangle \bmod(2)$ and $\langle\vec{v}_2,\vec{j}\rangle \bmod(2)$ still be close to independent?
Since $C$ is a code, the sum $\langle\vec{v}_1,\vec{j}\rangle+\langle\vec{v}_2,\vec{j}\rangle \bmod(2)$ would correspond to the $\vec{j}$-th coordinate of the codeword corresponding to $\vec{v}_1+\vec{v}_2$, so the sum of these random variables would be close to uniform on $\{0,1\}$. But I am interested in the joint distribution of the two variables.
This whole problem can be rephrased combinatorially, by asking if for any two distinct and non-zero codewords of $C$, if we lay them out side-by-side to get an $(N \times 2)$-matrix over $\{0,1\}$, would the rows have an almost equal number of $00,01,10,11$ entries?
This seems like a natural question in basic coding theory, but I am unaware of it being addressed anywhere. In all places, they work only with the sum of the codewords again being a codeword, but that loses some information.
One example of this kind of construction is the expurgated first-order Reed-Muller code (equivalent to a LFSR code) that is a linear map from $\mathbb F_2^n$ to $\mathbb F_2^{2^n-1}$ which maps $\vec x = (x_1, x_2, \ldots x_n) \in \mathbb F_2^n$ to $$\big(\langle\vec x, \vec 1\rangle, \langle\vec x, \vec 2\rangle, \ldots, \langle\vec x, \vec N\rangle\big)$$ where $N = 2^m-1$, and $\vec m$ denotes the binary representation of the integer $m$. All the nonzero codewords have Hamming weight $2^{n-1}$ (relative weight $\frac{2^{n-1}}{2^n-1} = \frac 12 + \epsilon$), but $\langle \vec x_1, \vec i\rangle>$ and $\langle \vec x_2, \vec i\rangle>$ are not independent random variables.