If the variance of two correlated variables is: $$Var(r_1+r_2)=\sigma^2_1+\sigma^2_2+2\textrm{cov}(r_1,r_2)=\sigma^2_1+\sigma^2_2+2\rho\sigma_1\sigma_2$$ where $r_1$ and $r_2$ are vectors, then what is the multivariate representation of this.

So, if $R_1$ and $R_2$ both denote a matrix we get $$Var(R_1+R_2)=\Sigma_1+\Sigma_2+...$$ where $\Sigma_i$ denotes the covariance matrix for $R_i$.

Anyone knows how to fill in the dots?

  • $\begingroup$ You could write each $R_j$ as $\sum_i c_i e_i^T$, where $c_i$ is the $ith$ column of $R_j$ and $e_i$ the standard basis vector. Then apply what you already know about the covariance of vectors, along with the usual bilinearity properties. $\endgroup$ – snarfblaat Jun 27 '16 at 15:22

... if $R_1$ and $R_2$ both denote a matrix we get

$R_1$ and $R_2$ should be vectors.

Then $Var(R_1+R_2)=\Sigma_1+\Sigma_2+2\rho_{12}\sqrt{\Sigma_1} \sqrt{ \Sigma_2}$

where $\Sigma_i$ denotes the variance matrix of $R_1$ and $R_2$ respectively.

  • $\begingroup$ Btw, the square root cannot be correct? I mean what if the covariance matrix has a negative value in it? Taking the square root of it would give a imaginary number, which is not what I am looking for $\endgroup$ – Eren Jun 27 '16 at 21:36
  • $\begingroup$ @Eren $\Sigma_i$ is the $\large{\color{blue}{\texttt{variance}}}$ matrix of $R_1$ and $R_2$ respectively. $\endgroup$ – callculus Jun 27 '16 at 21:45
  • $\begingroup$ So, what is a variance matrix, just variances on the diagonal elements and the off-diagonal elements should be zero? $\endgroup$ – Eren Jun 27 '16 at 21:59
  • $\begingroup$ @Eren Yes, that´s right. $\endgroup$ – callculus Jun 27 '16 at 21:59
  • $\begingroup$ Hmm ok, and if you want to get the covariance matrix of $cov(R_1+R_2)$ where $R_1$ and $R_2$ are correlated? $\endgroup$ – Eren Jun 27 '16 at 22:11

When working with multivariate variances and covariances, it's good to keep this notational advice in mind.

I'll stick with your notation and use $\operatorname{Var}(R)$ to denote the (co)variance matrix of the random vector $R$, i.e. $\operatorname{Var}(R)=\operatorname{cov}(R,R)$.


\begin{align} \operatorname{Var}(R_1+R_2)&=\operatorname{cov}(R_1+R_2,R_1+R_2) \\ &= \operatorname{cov}(R_1,R_1)+ \operatorname{cov}(R_1,R_2)+ \operatorname{cov}(R_2,R_1)+ \operatorname{cov}(R_2,R_2) \\ &= \operatorname{Var}(R_1)+ \operatorname{Var}(R_2)+ 2\operatorname{cov}(R_1,R_2)\;, \end{align}

so the multivariate case is exactly analogous to the univariate case.

  • $\begingroup$ Yes true, but the last part that is $2cov(R_1,R_2)$ how to express that to $2\rho\sigma_1\sigma_2$ in matrix form? Because I need that last part in matrix form :( $\endgroup$ – Eren Jun 27 '16 at 15:35
  • $\begingroup$ @Eren: I don't understand what you mean by "matrix form". $\operatorname{cov}(R_1,R_2)$ is the matrix of covariances between the components of $R_1$ and $R_2$, just like $\operatorname{Var}(R)=\operatorname{cov}(R,R)$ is the matrix of covariances between the components of $R$. They're exactly the same kind of matrices. Since you apparently know how to deal with the latter, I'm not sure what the problem is with the former. Or is the point that you want to express the covariance in terms of the correlation? $\endgroup$ – joriki Jun 27 '16 at 15:41
  • $\begingroup$ Well, bascially my correlation process is governed by a VAR model from which i have the autocorrelation matrix. What I want to do is to apply ρσ1σ2 using the autocorrelation matrix. So, to put it more clearly, I have 2 (co)variance matrices and a correlation matrix and I want to construct a covariance matrix from these three matrices. I just do not know how to do that. $\endgroup$ – Eren Jun 27 '16 at 15:52
  • $\begingroup$ @Eren: When you say you have a correlation matrix, you mean a cross-correlation matrix? Because if you meant what is usually called the correlation matrix, you'd have two, one for each vector. How is your correlation matrix defined? $\endgroup$ – joriki Jun 27 '16 at 15:59
  • $\begingroup$ Okay, uhm is there a chat on this site where I can tell you in detail what I am trying to do? Otherwise I am afraid that it'll just be a long comment discussion. I appreciate the help though and I'll vote up your answer anyway. $\endgroup$ – Eren Jun 27 '16 at 16:11

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