Central limit theorem for dependent Bernoulli random variables on the edges of a sequence of growing hypercubes? Imagine a growing sequence of hypercube graphs. For each integer n, you have the $n$-hypercube with $2^n$ vertices.
Now, assign $\text{Bernoulli}(1/2)$ i.i.d. random variables to all the vertices.
On each edge, you define a r.v. taking values in $\{0,1\}$ such that it is $1$ if the Bernoulli r.v.s of the two vertices it touches are equal, and it is $0$ otherwise.
So for each given n, you have Bernoulli r.v.s that are correlated on the edges and the correlation decreases with the distance (of the shortest path on the $n$-hypercube).
Note that going from one vertex to another on the hypercube requires that EXACTLY one binary component of the position changes by 1.
For example, with n=3, you can go from (1,1,0) to (1,1,1) because exactly one component (the third one) changes by 1.
This generalizes to all dimensions.
Now, the question is, if you take the sum of the (dependent) Bernoulli r.v.s on the EDGES (not vertices) and normalize it correctly (i.e. you translate by the expectation and divide by the standard deviation), does this normalized sequence converges to a $\text{Normal}(0,1)$ distribution ?
I know there are versions of the CLT for strong mixing sequences, but I didn't find a version that would suit the problem here.
 A: We will work with edge variables taking values $\pm 1$ instead of $0,1$, since applying an affine transformation $x \mapsto 2x-1$ will not affect the result. Thus if $x_u$ are i.i.d. mean zero $\pm 1$ valued variables on the nodes, $x_{uw}=x_u x_w$ are the edge variables. Denote the  sum over all hypercube edges by $Q_n=\sum_{\{uw\}} x_{uw}$. Then  $Q_n=\sum_{i=1}^n S_i(n)$ where $S_i =S_i(n)$ is the sum of the   variables $x_{uw}$ corresponding to edges $\{uw\}$ parallel to the $i$'th axis. With this notation, each $S_i(n) $ has mean $0$ and variance $2^{n-1}$;   the variables $S_i $ are uncorrelated, so $Q_n$ has variance $\sigma_n^2=n2^{n-1}$. Also, observe that
$$E[Q_n^4]=O(n^2 2^{2n}) \quad (*) $$
since there are only $2^{n-2}$ simple cycles of length $4$ in the hypercube, so the main contribution to the fourth moment comes from selecting a pair of edges and taking each of them twice.
Let $k=k(n)$ grow sublinearly e.g., $k  =\lfloor \sqrt n \rfloor$, and define
$R_n=\sum_{i=1}^{n-k} S_i(n)$.
Observe that the ratio Var$(R_n)/\sigma_n^2$ tends to $1$ as $n \to \infty$.
We can write
$$R_n=\sum_{v \in \{0,1\}^k} Q_{n-k}^v$$
where $Q_{n-k}^v$ is the sum along all edges parallel to one of the first $n-k$ directions where the last $k$ coordinates form the vector $v$.
Now by (*),
$$\sum_{v \in \{0,1\}^k} E[(Q_{n-k}^v)^4 ]=2^k O(n^2 2^{2(n-k)})$$
so
$$\sum_{v \in \{0,1\}^k} E[(Q_{n-k}^v/\sigma_n)^4] =O(2^{-k})$$
By the Lyapunov CLT [1] with $\delta=2$, the ratio
$R_n/\sigma_n$ tends in law to $N(0,1)$.
Finally, $Q_n-R_n$ has variance $k2^{n-1}=o(\sigma_n^2)$ so the CLT for $R_n$ implies the CLT  for $Q_n$ with the same normalization.
[1] https://en.wikipedia.org/wiki/Central_limit_theorem#Lyapunov_CLT
A: This is not an answer, but the outline of an idea that might work, but which is too long for a comment.
The edge variables corresponding to the $\ 2^{n-1}\ $ edges parallel to a given axis are Bernoulli$\left(\frac{1}{2}\right)$ i.i.d., because none of them have any vertices in common.  Their sum is therefore Binomial$\left(2^{n-1},\frac{1}{2}\right)$—and hence approximately normal for large $\ n\ $—with mean $\ 2^{n-2}\ $ and variance $\ 2^{n-3}\ $.  If $\ S_i\ $ is the sum of the variables corresponding to the edges parallel to the $\ i^\text{th}\ $ axis, then the total sum is the sum $\ \sum_\limits{i=1}^nS_i\ $ of $\ n\ $ approximately normal random variables.  Although $\ S_1,S_2,\dots,S_n\ $ are not independent, they are uncorrelated, because every term in any one of them is independent of any term in any of the others. Therefore $\ \sum_\limits{i=1}^nS_i\ $ has mean $\ n2^{n-2}\ $ and variance $\ n2^{n-3}\ $, and $\ \sum_\limits{i=1}^n\frac{\big(S_i-2^{n-2}\big)}{2^\frac{n-3}{2}}=\frac{\sum_\limits{i=1}^nS_i-n2^{n-2}}{2^\frac{n-3}{2}}\ $ is the sum of $\ n\ $ uncorrelated, approximately standard normal variates. So $\ \frac{\sum_\limits{i=1}^nS_i-n2^{n-2}}{n2^\frac{n-3}{2}}\ $ should also be approximately standard normal.
