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I tried searching a lot on the net and got the following sources:

  • Source One
  • Source Two
  • The first source seems to be incorrect cause when I calculate it using matlab it comes to be different from what they have given as the answer. As for the second link I cant understand that cause its not completely explaining as to how to calculate. Could anyone please provide me with a sound link or explain how to calculate a co-variance matrix?

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    • $\begingroup$ What covariance matrix? The covariance matrix for the OLS estimator is not the same thing as the covariance matrix for the residuals, for example, if we think of a regression context. You need to be more specific. Do you mean sample (co)variance (whose univariate counterpart is $(n-1)^{-1}\sum_{i=1}^n(x_i-\bar{x})^2$)? $\endgroup$ – hejseb Nov 11 '13 at 20:58
    • $\begingroup$ I got my answer finally. Have posted it below. $\endgroup$ – Sohaib I Nov 12 '13 at 4:23
    • $\begingroup$ Where? ${ }$ ${ }$ ${ }$ ${ }$ $\endgroup$ – draks ... Nov 12 '13 at 6:23
    • $\begingroup$ @draks... Bah! Internet problem. Now done! $\endgroup$ – Sohaib I Nov 12 '13 at 13:38
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    I finally understood the concept behind co-variance. Co-variance is different for population data and sample data.

    Following is the method I followed:

    Let $A$ be a $n \times m$ matrix where $n$ is the number of rows (observations) and $m$ represents the number of columns (variables).
    Let $\mathbf e$ be the $n \times 1$ column vector composed entirely of ones. Then, $$ X= A - \left(\frac{1}{n}\right)\mathbf e\mathbf e^TA $$ Then, denote $$ Y = X^TX. $$

    Next is the step that differs for population data and sample data.

    In case of population data, the covariance matrix $\Sigma$ is given by : $$ \Sigma=\left(\frac{1}{n}\right)Y $$
    and in case of sample data, the covariance matrix $\Sigma$ is given by : $$ \Sigma=\left(\frac{1}{n-1}\right)Y $$ Hope it helps anyone stuck on a similar problem.

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    • $\begingroup$ I've erroneously accepted an edit that was not quite right; now I've tried to edit to adress the confusion that existed prior to this edit. I hope I didn't make things worse. $\endgroup$ – Arnaud D. Apr 25 at 14:27

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