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Assume there are given two Gaussian random vectors $\boldsymbol{x}$ and $\boldsymbol{y}$ of equal length $N$ with corresponding means $\boldsymbol{\mu}_x$, $\boldsymbol{\mu}_y$ and covariance matrices $\boldsymbol{C}_{xx}$, $\boldsymbol{C}_{yy}$ as well as their cross covariance matrix $\boldsymbol{C}_{xy} = \boldsymbol{C}_{yx}^T$ which in the case of equal length is a square matrix.

Stacking them together yields a new random vector that is again Gaussian $\boldsymbol{z} = \left( \begin{array}{rr} \boldsymbol{x} \\ \boldsymbol{y} \\ \end{array} \right) $ with mean $\boldsymbol{\mu}_z = \left( \begin{array}{rr} \boldsymbol{\mu}_x \\ \boldsymbol{\mu}_y \\ \end{array} \right)$ and covariance matrix $\boldsymbol{C}_{zz} = \left( \begin{array}{rr} \boldsymbol{C}_{xx} & \boldsymbol{C}_{xy} \\ \boldsymbol{C}_{xy}^T & \boldsymbol{C}_{yy} \\ \end{array} \right)$.

I would now like to express $\boldsymbol{C}_{xy}$ by some sort of generalized correlation coefficients $\rho_1, \cdots, \rho_N$, $\rho_i \in [-1, 1]$ , that describe the correlation between $x_i$ and $y_i$. I.e. I´d be interested in something like $\boldsymbol{C}_{xy} = f(\boldsymbol{C}_{xx}, \boldsymbol{C}_{yy}, \rho_1, \cdots, \rho_N)$. This should be in analogy to the definition of the correlation coefficient for two scalar random variables $\rho_{x,y}$, where it is possible to write

$\boldsymbol{z} = \left( \begin{array}{rr} x \\ y \\ \end{array} \right) $, $\boldsymbol{\mu}_z = \left( \begin{array}{rr} \mu_x \\ \mu_y \\ \end{array} \right)$ and $\boldsymbol{C}_{zz} = \left( \begin{array}{rr} \sigma^2_x & \rho_{x,y} \sigma_x \sigma_y \\ \rho_{x,y} \sigma_x \sigma_y & \sigma^2_y \\ \end{array} \right)$.

However, I am not sure if this is possible, at least I have not found anything relevant yet in this regard.

Thanks for your help and constributions.

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2 Answers 2

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This question belongs more on stats.SE (where many similar questions have been thrashed out in detail) but nonetheless here goes.

Let's take the simplest case of $N=1$. Just because $X$ and $Y$ are Gaussian random variables, it is not necessarily the case that $X$ and $Y$ have a jointly Gaussian distribution. See, for example, this answer on stats.SE for several examples of bivariate nonGaussian distributions whose marginal distributions are nonetheless Gaussian. (Full disclosure: one of the examples is attributed to an answer of mine on stats.SE). But, let's ignore such nitpicks and assume that $X$ and $Y$ do indeed enjoy a bivariate jointly Gaussian distribution and so one parameter $\rho_{X,Y}$ suffices to specify the $2\times 2$ covariance matrix. We note that $N$ equals $1$, and so we think that $N$ parameters will be needed in the general case also.

Generalizing to the case of $N>1$, we have the $N\times N$ cross-covariance matrix $C_{X,Y} = E[(X-\mu_x)^T(Y-\mu_Y)]$ which has $N^2$ entries and the only constraints on the entries are that $|C_{X,Y}[i,j]| \leq \sigma_{X,i}\sigma_{Y,j}$ for all choices of $i$ and $j$ and that the covariance matrix $$\left[\begin{matrix} C_{X,X} & C_{X,Y}\\ C_{Y,X} & C_{Y,Y}\end{matrix}\right]$$ be nonnegative definite. Thus, we have $N^2$ parameters $\rho_{i,j}$ that need to be specified; not $N$ as the OP wants. Put another way, we were misled into thinking that $N$ parameters were needed in general because we were generalizing from the false base case "number of parameters equals dimension" (which is true when $N=1$ but not for $N>1$) instead of the correct base case "number of parameters equals the square of the dimension" (which is true when $N=1$ and also when $N>1$).

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Thank you for this information. I already had a feeling that N coefficients might not suffice. The reason I asked for this was that I wanted to find $\boldsymbol{C}_{\boldsymbol{xy}}$ based on a given degree of fit between $\boldsymbol{x}$ and $\boldsymbol{y}$. Meanwhile, I think I found a way to do it: Based on the relation $$\textrm{Cov}[\boldsymbol{x}-\boldsymbol{y}] = \boldsymbol{C}_{\boldsymbol{xx}} + \boldsymbol{C}_{\boldsymbol{yy}} - 2 \boldsymbol{C}_{\boldsymbol{xy}}$$ I can specify some $\boldsymbol{C}_{\boldsymbol{dd}}$ and then set $\boldsymbol{C}_{\boldsymbol{xy}} = \frac{1}{2}(\boldsymbol{C}_{\boldsymbol{xx}} + \boldsymbol{C}_{\boldsymbol{yy}} - \boldsymbol{C}_{\boldsymbol{dd}})$. Basically $\boldsymbol{C}_{\boldsymbol{dd}}$ now constains the information of the $N^2$ coerrelation coefficients. (still, I need to check that that the resulting covariance matrix $C_{\boldsymbol{zz}}$ is positive semidefinite though).

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