Covariance matrix computation

问题

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?

回答1:

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);
   end
 end

where variance (var) of xi:

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

EDIT
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];

来源：https://stackoverflow.com/questions/3307082/covariance-matrix-computation