Covariance matrix computation

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asked Jul 22, 2010 by singularity

Input : random vector X=xi, i=1..n.
vector of means for X=meanxi, i=1..n
Output : covariance matrix Sigma (n*n).
Computation :
1) find all cov(xi,xj)= 1/n * (xi-meanxi) * (xj-meanxj), i,j=1..n
2) Sigma(i,j)=cov(xi,xj), symmetric matrix.
Is this algorithm correct and has no side-effects?

1 Answer

0 votes
answered Jul 22, 2010 by gacek

Each xi should be a vector (random variable) with it's own variance and mean.

Covariance matrix is symmetric, so you just need to compute one half of it (and copy the rest) and has variance of xi at main diagonal.

 S = ...// your symmetric matrix n*n
 for(int i=0; i<n;i++)
   S(i,i) = var(xi);
   for(j = i+1; j<n; j++)
     S(i,j) = cov(xi, xj);
     S(j,i) = S(i,j);

where variance (var) of xi:

v = 0;
for(int i = 0; i<xi.Count; i++)
  v += (xi(i) - mean(xi))^2;
v = v / xi.Count;

and covariance (cov)

cov(xi, xj) = r(xi,xj) * sqrt(var(xi)) * sqrt(var(xj))

where r(xi, xj) is Pearson product-moment correlation coefficient

or, since cov(X, Y) = E(X*Y) - E(X)*E(Y)

cov(xi, xj) = mean(xi.*xj) - mean(xi)*mean(xj);

where .* is Matlab-like element-wise multiplication.
So if x = [x1, x2], y = [y1, y2] then z = x.*y = [x1*y1, x2*y2];

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