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I'd like to know, if I have the following Normal multivariate structure $$\left( {\begin{array}{*{20}{c}} {{Y_{r_1}}}\\ {{Y_{r_2}}} \end{array}} \right)\sim{\mathcal{N}_{r}}\left( {\left[ {\begin{array}{*{20}{c}} {{\mu_1}}\\ {{\mu_2}} \end{array}} \right],\left[ {\begin{array}{*{20}{c}} {{\Sigma _{11}}}&{{\Sigma _{12}}}\\ {{\Sigma _{21}}}&{{\Sigma _{22}}} \end{array}} \right]} \right),$$ then $Y_{r_1}\mid Y_{r_2} \sim \mathcal{N}_r(\mu, \Sigma)$, where $\mu=\mu_1 + \Sigma_{12}\Sigma^{-1}_{22}(Y_{S_n}-\mu_2)$ and $\Sigma=\Sigma_{11}-\Sigma_{12}\Sigma^{-1}_{22}\Sigma_{21}$.

Another result we have is the correlation one, where $Cor(Y_{r_1},Y_{r_2})=\cfrac{Cov(Y_{r_1},Y_{r_2})}{\sqrt{Var(Y_{r_1})}\sqrt{Var(Y_{r_2})}}$.

Therefore, using this 2 facts, can I use the correlation expression in this following way? (It will help me a lot to prove a result)

$Cor(Y_{r_1},Y_{r_2})= \Sigma_{12}\Sigma_{22}^{-1/2}\Sigma_{11}^{-1/2}$

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    $\begingroup$ Consider the case where $\Sigma_{11}$ is a $k\times k$ matrix and $\Sigma_{22}$ is an $l\times l$ matrix with $k\ne l:$ how are we supposed to make sense of your final formula? $\endgroup$ – whuber Jun 14 at 12:47
  • $\begingroup$ Yes, maybe in case where both have the same dimension, but it doesn't help me anyway. $\endgroup$ – Ga13 Jun 14 at 17:08
  • $\begingroup$ Given that this expression doesn't always make sense, could you tell us specifically what you mean by "the correlation expression"? Note that the formula you initially give is valid only for univariate variables, not multivariate variables. Are you simply asking how to read correlations from a covariance matrix? $\endgroup$ – whuber Jun 14 at 17:47
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    $\begingroup$ I was in doubt whether I could extend the univariate correlation formula in an analogous way to multivariate variables. About your last question, the answer is yes. $\endgroup$ – Ga13 Jun 14 at 18:46
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When there are $n$ variables $X_1, \ldots, X_n,$ their mutual covariances can be assembled into a matrix $\Sigma = (\sigma_{ij})$ where

$$\sigma_{ij} = \operatorname{Cov}(X_i,X_j).$$

Note that the diagonal elements $\sigma_{ii} = \operatorname{Cov}(X_i,X_i) = \operatorname{Var}(X_i)$ are the variances. It will be convenient to extract these elements into a diagonal matrix by setting all off-diagonal entries to zero; sometimes this is written

$$\operatorname{Diag}(\Sigma)_{ij} = \left\{\matrix{\sigma_{ii} & i=j \\ 0 & \text{otherwise.}}\right.$$

Note, too, that a positive square root of this matrix is well-defined and unique because all its entries are positive:

$$\sqrt{\operatorname{Diag}(\Sigma)}_{ij} = \left\{\matrix{\sqrt{\sigma_{ii}} & i=j \\ 0 & \text{otherwise.}}\right.$$

One formula for the corresponding correlation coefficient, as given in the question, is

$$\rho_{ij} = \operatorname{Cor}(X_i,X_j) = \frac{\operatorname{Cov}(X_i,X_j)}{\sqrt{\operatorname{Var}(X_i)\operatorname{Var}(X_j)}} = \frac{\sigma_{ij}}{\sqrt{\sigma_{ii}\,\sigma_{jj}}}\ .$$

These can be assembled into a correlation matrix $P= (\rho_{ij}).$

It follows directly from the definitions of matrix multiplication and inversion that the preceding formula is the result of the following combination of operations:

$$P = \sqrt{\operatorname{Diag}(\Sigma)}^{-1}\ \Sigma\ \sqrt{\operatorname{Diag}(\Sigma)}^{-1}.$$

By focusing on part of the matrices, this formula extends directly to the block matrix setup of the question, where the covariances of the variables in the $k$-vector $Y_{r_1}$ and those in the $m$-vector $Y_{r_2}$ appear in the $k\times m$ matrix $\Sigma_{12},$ $\Sigma_{11}$ is the $k\times k$ covariance matrix of $Y_{r_1},$ and $\Sigma_{22}$ is the $m\times m$ covariance matrix of $Y_{r_2}.$ As before, the correlations between the variables in $Y_{r_1}$ and those in $Y_{r_2}$ can be arranged into a $k\times m$ mutual correlation matrix $P_{12}$ and

$$P_{12} = \sqrt{\operatorname{Diag}(\Sigma_{11})}^{-1}\ \Sigma_{12}\ \sqrt{\operatorname{Diag}(\Sigma_{22})}^{-1}.$$


For example, let $$\Sigma = \pmatrix{1 & -1 & 2 \\ -1 & 4 & -1 \\ 2 & -1 & 9}.$$ Then

$$\operatorname{Diag}(\Sigma) = \pmatrix{1 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & 9},$$

$$\sqrt{\operatorname{Diag}(\Sigma)} = \pmatrix{1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 3},$$

$$\sqrt{\operatorname{Diag}(\Sigma)}^{-1} = \pmatrix{1 & 0 & 0 \\ 0 & \frac{1}{2} & 0 \\ 0 & 0 & \frac{1}{3}},$$

giving the correlation matrix

$$P = \pmatrix{1 & 0 & 0 \\ 0 & \frac{1}{2} & 0 \\ 0 & 0 & \frac{1}{3}}\ \pmatrix{1 & -1 & 2 \\ -1 & 4 & -1 \\ 2 & -1 & 9}\ \pmatrix{1 & 0 & 0 \\ 0 & \frac{1}{2} & 0 \\ 0 & 0 & \frac{1}{3}} = \pmatrix{1 & -\frac{1}{2} & \frac{2}{3} \\ -\frac{1}{2} & 1 & -\frac{1}{6} \\ \frac{2}{3} & -\frac{1}{6} & 1}.$$

Taking $k=1$ and $m=2$ makes $\Sigma_{11} = \pmatrix{1},$ $\Sigma_{22} = \pmatrix{4 & -1 \\ -1 & 9},$ and $\Sigma_{12} = \pmatrix{-1 & 2}.$ Thus the correlations between the first variable ($Y_{r_1}$) and the remaining two variables ($Y_{r_2}$) are given by the $1\times 2$ matrix

$$P_{12} = \pmatrix{\sqrt{1}}^{-1} \pmatrix{-1 & 2}\ \pmatrix{\sqrt{4} & 0 \\ 0 & \sqrt{9}}^{-1} = \pmatrix{-\frac{1}{2} & \frac{2}{3}},$$

which is precisely the upper right hand block of $P,$ as claimed.

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