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I know the covariance matrix of five variables $(a,b,c,x,y)$:

$C = \left(\begin{array}{ccccc} \sigma^2_x&0&0&0&0\\ 0&\sigma^2_y&0&0&0\\ 0&0&\sigma^2_a&c_{a,b}&c_{a,c}\\ 0&0&c_{a,b}&\sigma^2_b&c_{b,c}\\ 0&0&c_{a,c}&c_{b,c}&\sigma^2_c\\ \end{array}\right)$

How can I compute the value of $\text{cov}(ax+2b+c,ay+3b+c)$?

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  • $\begingroup$ Do the variables follow a multivariate Gaussian distribution? $\endgroup$ – Alessandro Mar 19 '18 at 17:17
  • $\begingroup$ Yes they do, sorry I forgot to mention it. $\endgroup$ – Mathieu Mar 23 '18 at 8:25
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Given that the variables follow a multivariate Gaussian distribution (as stated in the comments), then you can compute the covariance. In the next steps I will define $\mu_x,\dots,\mu_c$ as the expected values of $x,\dots,c$ and I will use the same properties mentioned here: Beginner's question: covariance of products of independent variables

$cov(ax+2b+c,ay+3b+c)=cov(ax,ay)+cov(ax,3b)+cov(ax,c)+cov(2b,ay)+cov(2b,3b)+cov(2b,c)+cov(c,ay)+cov(c,3b)+cov(c,c)$

where

$ cov(ax,ay) = \sigma^2_a \mu_x \mu_y \quad \text{(see link above)}\\ cov(ax,3b) = 3cov(ax,b) = 3 [E(axb)-E(ax)E(b)]=3[E(x)E(ab)-E(a)E(x)E(b)]=3[E(x)(cov(a,b)+E(a)E(b))-E(a)E(x)E(b)]=3c_{a,b}\mu_x\\ cov(ax,c) = c_{a,c}\mu_x\\ cov(2b,ay) = 2c_{a,b}\mu_y\\ cov(2b,3b) = 6\sigma^2_b\\ cov(2b,c) = 2c_{b,c}\\ cov(c,ay) = c_{a,c}\mu_y\\ cov(c,3b) = 3c_{b,c}\\ cov(c,c) = \sigma^2_c $

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From the symbols in the covariance matrix, I presume the variables are given there in the order (x,y,a,b,c). You cannot compute the covariance unless you have information on Cov(ax,ay), for example, as this is effectively a fourth moment in the mentioned original variables. Knowing that covariances (second moments) are zero, as shown by the block structure of the matrix, is not enough unless, say, normality is assumed.

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