numpy covariance matrix

Question

Suppose I have two vectors of length 25, and I want to compute their covariance matrix. I try doing this with numpy.cov, but always end up with a 2x2 matrix.

Answer 1

Try this:

import numpy as np
x=np.random.normal(size=25)
y=np.random.normal(size=25)
z = np.vstack((x, y))
c = np.cov(z.T)

Answer 2

As pointed out above, you only have two vectors so you'll only get a 2x2 cov matrix.

IIRC the 2 main diagonal terms will be sum( (x-mean(x))**2) / (n-1) and similarly for y.

The 2 off-diagonal terms will be sum( (x-mean(x))(y-mean(y)) ) / (n-1). n=25 in this case.

Answer 3

What you got (2 by 2) is more useful than 25*25. Covariance of X and Y is an off-diagonal entry in the symmetric cov_matrix.

If you insist on (25 by 25) which I think useless, then why don't you write out the definition?

x=np.random.normal(size=25).reshape(25,1) # to make it 2d array.
y=np.random.normal(size=25).reshape(25,1)

cov =  np.matmul(x-np.mean(x), (y-np.mean(y)).T) / len(x)

Answer 4

according the document, you should expect variable vector in column:

If we examine N-dimensional samples, X = [x1, x2, ..., xn]^T

though later it says each row is a variable

Each row of m represents a variable.

so you need input your matrix as transpose

x=np.random.normal(size=25)
y=np.random.normal(size=25)
X = np.array([x,y])
np.cov(X.T)

and according to wikipedia: https://en.wikipedia.org/wiki/Covariance_matrix

X is column vector variable
X = [X1,X2, ..., Xn]^T
COV = E[X * X^T] - μx * μx^T   // μx = E[X]

you can implement it yourself:

# X each row is variable
X = X - X.mean(axis=0)
h,w = X.shape
COV = X.T @ X / (h-1)