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For $i=1,2,\cdots,2^m$, let $v_i$ be dependent random variables. Suppose for $n$ large, the vector $\mathbf{Z}_n=\left(Z_1^{(n)},\cdots,Z_{2^m}^{(n)}\right)$ with $Z_i^{(n)}=\frac{1}{\sqrt{n}}\left(v_i-\frac{n}{2^m}\right)$ has multivariate normal distribution with mean $0$ and $Var\left(Z_i^{(n)}\right)=\frac{1}{2^m}-\frac{1}{2^{2m}}$ and covariance matrix $\Sigma_m$.

Suppose we have computed $\Sigma_m^{1/2}=\frac{1}{2^{m/2}}I_{2^m}-\frac{1}{2^{3m/2}}\mathbf{e}_{2^m}\mathbf{e}_{2^m}^T$. Where $I_{2^m}$ denotes identity matrix of size $2^m \times 2^m$ and $\mathbf{e}_{2^m}$ is a $2^m \times 1$ column vectors with all entries equal $1$. Is it true that for large $n$ we have $$||Q^m\Sigma_m^{1/2}\mathbf{Z}_n||^2 \sim \chi^2\left(2^{m-1}\right)$$

Where $Q=diag\left(\left[\begin{array}{lr}1 &-1\\ -1 &1 \end{array}\right],\cdots,\left[\begin{array}{lr}1 &-1\\ -1 &1 \end{array}\right]\right)$.

[I have this question from a problem related to random numbers test]

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$Q^m$ is computable, since it's blockwise diagonal ($Q^m=2^mQ$, and $Qe_{2^m}=0$, so we have to look at $\frac 1{2^m}\lVert QZ_n\rVert^2$. – Davide Giraudo May 4 '12 at 9:32
@DavideGiraudo I've got $Q^m=2^{m-1}Q$, this is since the minimum polynomial of $Q$ is $t^2-2t$.. then I obtain the identity $\Sigma_m^{1/2}Q\Sigma_m^{1/2}=\frac{Q}{2^m}$, to get $2^{m-2}\lvert\lvert Q Z_n\rvert\rvert^2$. – Ajat Adriansyah May 4 '12 at 9:57
Yes actually it's $2^{m-1}$. I think we have to use the result which states that if $X_1,\ldots,X_p$ are normal independent distributions, with mean $0$ and variance $1$, then $\sum_{j=1}^pX_j^2\sim\chi^2(p)$. – Davide Giraudo May 4 '12 at 9:59

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