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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.

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    It means either no-one knows the answer or no-one cares enough to write it up. Unfortunately there is no way to find out which, and even if there were, there's no way to guarantee an answer be posted. If you have time you can try waiting a while and open another bounty later (hoping more users stop by).2012-05-29

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