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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.

share|improve this question
    
sorry, is it wrong to pose the same question on both meta-exchange? perhaps I should delete one of them, if it is. I posted it again there due to the lack of interest (and bump) from here, not even a down-vote to criticize this if I posted it in a not well-written form. Thanks. –  Ajat Adriansyah May 16 '12 at 11:36
    
it is generally "bad form". Keeping the question on one stackexchange focuses interest, and keeps all the answers in one place. This way it also makes it easier for future visitors to find the question/answer. In situations like this generally one of the questions will be closed and migrated to the other SE: in this case Stats closed the version there, which now redirects to this question here. –  Willie Wong May 16 '12 at 12:21
    
Ok, Thank you for the explanation and the migration :D –  Ajat Adriansyah May 16 '12 at 16:20
    
This question had ended its bounty period and still has zero answer. What does usually become a common/suggested action that someone should take if something like this happen in stackexchange? Any suggestions? thank you. –  Ajat Adriansyah May 27 '12 at 0:46
    
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). –  Willie Wong May 29 '12 at 9:17

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