Take two vectors of normally-distributed random variables

$\mathbf{x} = (x_1, x_2, \ldots x_n)$

$\mathbf{y} = (y_1, y_2, \ldots y_n)$

where the covariance of each pair $(x_i, y_i)$ is known,

$\mathbf{C}_{x_{i}y_{i}} = \pmatrix{\sigma^{2}_{x_i} & \sigma^{2}_{x_{i}y_{i}}\\ \sigma^{2}_{x_{i}y_{i}}& \sigma^{2}_{y_i}}$ for $i = 1, 2, \ldots n$,

and all other covariances are zero. The sum of each vector is

$S_{x} = \Sigma_{i}x_i$

$S_{y} = \Sigma_{i}y_i$

What is the correlation matrix $(C_{S_{x}S_{y}})$ of the sums $S_{x}$ and $S_{y}$ in terms of the $\sigma^{2}_{x_i}$, $\sigma^{2}_{y_i}$ and $\sigma^{2}_{x_{i}y_{i}}$?

  • 1
    $\begingroup$ I presume at the end that one of the "$S_{x_i}$" is intended to be "$S_{y_i}.$" The notation is confusing, though: what does the "$i$" mean in this terminology? Moreover, since the sums appear to be numbers, in what sense would they have a "correlation matrix"? $\endgroup$
    – whuber
    Jan 20, 2019 at 22:07
  • $\begingroup$ I tried to make it clearer. $\endgroup$ Jan 21, 2019 at 15:23

1 Answer 1


I worked it out. The covariance between the two sums is the sum of the covariances of the pairs.

$C_{S_{x}S_{y}} \equiv \mathrm{Cov}(S_x,S_y) = \mathrm{Cov}(x_1 + x_2 + \ldots + x_n, y_1 + y_2 + \ldots + y_n)$

From bilinearity of covariance:

$=\mathrm{Cov}(x_1, y_1) + \mathrm{Cov}(x_2, y_1) + \ldots \mathrm{Cov}(x_n, y_1)$

$+\mathrm{Cov}(x_1, y_2) + \mathrm{Cov}(x_2, y_2) + \ldots \mathrm{Cov}(x_n, y_2)$


$+\mathrm{Cov}(x_1, y_n) + \mathrm{Cov}(x_2, y_n) + \ldots \mathrm{Cov}(x_n, y_n)$

Because only each pair is correlated:

$=\mathrm{Cov}(x_1, y_1) + \mathrm{Cov}(x_2, y_2) + \ldots \mathrm{Cov}(x_n, y_n)$


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