Given natural numbers $m, n \geq 2$ and a random vector $\mathbf{r}= (a_1,a_2,\cdots,a_n)\in\mathbb{Z}_2^n$. We define the $m$-circulant of $\mathbf{r}$ by the vector $\overline{\mathbf{r}}_m=(a_1,a_2,\cdots,a_n,a_1,a_2,\cdots,a_{m-1})$ Suppose we divide $\overline{\mathbf{r}}_n$ into $n$ overlapping blocks as follows $\mathbf{b}_1=(a_1,\cdots,a_m),\quad \mathbf{b}_2=(a_2,\cdots,a_{m+1}),\quad \cdots \quad \mathbf{b}_{n}=(a_n,\cdots,a_{m-1})$
For any $\mathbf{u}\in \mathbb{Z}_2^m$ define the random variable
$v_{\mathbf{u}}:=\text{the number of block $\mathbf{b}_j$ such that $\mathbf{b}_j=\mathbf{u}$}$
Therefore there are exactly $2^m$ random variables and each of them depends on $n$, so we give the following index for each of them: $v_1^{(n)},v_2^{(n)},\cdots,v_{2^m}^{(n)}$
We can prove that $v_1^{(n)} \sim b\left(n,\frac{1}{2^m}\right)$ (binomial distribution).
Define the $2^m \times 1$ random variable vector $\mathbf{Z}_n:=(Z_1^{(n)},\cdots,Z_{2^m}^{(n)})^T$, where $Z_i^{(n)}:=\frac{1}{\sqrt{n}}\left(v_i^{(n)}-\frac{n}{2^m}\right)$
Could we prove that for sufficiently large $n$, $\mathbf{Z}_n$ is asymtotically multivariate normal distribution?
Edit[comment truncated]:
My indication for the non-overlapping case:
Define the indicator $X_{\mathbf{u}j}:=1 \quad \text{if $\mathbf{u}=\mathbf{b}_j$ }$ and equals $0$ otherwise. $v_{\mathbf{u}}=\sum_{j=1}^n X_{\mathbf{u}j},$ where all $X_{\mathbf{u}j}$ have bernoulli distribution with $p=\frac{1}{2^m}$. We reindex the $X's$, where $X_{ij}^{(n)}$ is associated to $v_i^{(n)}$ (as $X_{\mathbf{u}j}$ is associated to $v_{\mathbf{u}}=v_i^{(n)}$)
If we define $2^m \times 1$ vectors $\mathbf{X}_j=\left(X_{1j} , \cdots, X_{2^m j}\right)^T$, then by the mean of bernoulli distribution we have $\mu=E[X_j]=\left(\frac{1}{2^m},\cdots,\frac{1}{2^m}\right)^T$ and so
$\mathbf{Z}_n=\frac{1}{\sqrt{n}} \sum_{j=1}^n [\mathbf{X}_j-E(\mathbf{X}_j)] = \sqrt{n} (\overline{X}_n -\mu)$
for non-overlapping case, by the Central Limit Theorem, since each $\mathbf{X}_j$ is independent, then we get the result. But clearly they're not on my case.
thanks.