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I am trying to calculate covariance matrix from a 2D data, assumed from coming from a Gaussian Distribution. I am trying to calculate using the equality that $\mathrm{Var}[x] = \mathrm{E}[x^2] - \mathrm{E}[x]^2$, so supposing that D is the data matrix where rows are observations, the MATLAB code is:

[mean(D(:,1).^2) - mean(D(:,1))^2 , mean(D(:,1).*D(:,2)) - mean(D(:,1))*mean(D(:,2)) 
 mean(D(:,1).*D(:,2)) - mean(D(:,1))*mean(D(:,2)) , mean(D(:,2).^2) - mean(D(:,2))^2]

However cov(D) gives me an entirely different covariance matrix. Of course I can use cov() and go on my life but I am using the calculation method above in another different piece of C++ code, so it is nice to learn where I am doing wrong.

I think I am missing a crucial and fundamental information here but could not figure it out. Any help?

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  • $\begingroup$ Exactly how does your formula differ from cov(D)? For instance, is there a constant ratio of the components of cov(D) to the components of your calculation? $\endgroup$
    – whuber
    Jun 29, 2012 at 15:14

2 Answers 2

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For ease of presentation, let us restrict to two random variables, say $X_1$ and $X_2$. The covariance matrix is given by $$\Sigma = \left( \begin{array}{cc} \textrm{Var}(X_1) & \textrm{Cov}(X_1, X_2) \\ \textrm{Cov}(X_2, X_1) & \textrm{Var}(X_2) \end{array}\right).$$ Since $\textrm{Cov}(X_1, X_2) = \textrm{Cov}(X_2, X_1)$, only three entries have to be estimated.

Say you have one sample from each of $X_1$ and $X_2$ available to you: $$\{x_{11}, \ldots, x_{1n}\} \qquad \textrm{and} \qquad \{x_{21}, \ldots, x_{2n}\}.$$ The natural empirical estimates are given by $$\widehat{\textrm{Var}(X_k)} =\widehat{\textrm{Cov}(X_k, X_k)} = \frac{1}{n-1} \sum_{i = 1}^{n} (x_{ki} - \bar{x}_k)^2; \qquad k = 1, 2$$ $$\widehat{\textrm{Cov}(X_1, X_2)} = \frac{1}{n - 1} \sum_{i=1}^n (x_{1i} - \bar{x}_1) (x_{2i} - \bar{x}_2)$$ with $$\bar{x}_k = \frac{1}{n} \sum_{i=1}^n x_{ki}; \qquad k = 1, 2.$$

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    $\begingroup$ Although this is correct, what is the answer to the question? What is the "crucial and fundamental information" that the OP may be overlooking? (I suspect, as I hinted in a comment, that it may be as simple as the use of $1/(n-1)$ instead of $1/n$.) $\endgroup$
    – whuber
    Jun 29, 2012 at 15:50
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I believe @whuber's comment nailed it. Here is a simple way to debug your Matlab code:

n = 1024;
D = randn(n,2);
C = [mean(D(:,1).^2) - mean(D(:,1))^2 , mean(D(:,1).*D(:,2)) - mean(D(:,1))*mean(D(:,2)) ; ...
mean(D(:,1).*D(:,2)) - mean(D(:,1))*mean(D(:,2)) , mean(D(:,2).^2) - mean(D(:,2))^2];  % OP code
C2 = cov(D);
disp(C ./ C2);   % they have the same value! i.e., C = k C2 for scalar k.
disp(n * C - ((n-1) * C2));  % zero + roundoff error.
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