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so if given a covariance matrix I can find the eigenvalues and move forward from there... but I seem to have trouble with the step before if I am given a data set and am told to create the covariance matrix. Looking at the notes I see the formula: $$cov(x) = \frac{1}{n-1}\sum(x_i - \bar{x})(y_i -\bar{y}) $$

I'm not too sure what to do with this formula and was hoping you can tell me how.

Data Set:

    X1 | X2
    ---|---
    3  | 7
    2  | 4
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  • $\begingroup$ I saw some sources writing covariance matrix as a product of $X_1 X_2^T/(n-1)$, could anyone help explain why it works? $\endgroup$
    – Logan
    Commented Apr 19, 2017 at 8:12

1 Answer 1

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The variance-covariance matrix has the following structure: $$\begin{bmatrix} var(x) & cov(x,y) \\ cov(x,y) & var(y) \end{bmatrix}$$

where $var(x) = \frac{1}{n-1} \sum (x_i - \bar{x})^2 $ and $cov(x,y) = \frac{1}{n-1} \sum (x_i - \bar{x})(y_i - \bar{y}) $.

for your data,

$\bar{x} = \frac{(3 + 2)}{2} = \frac{5}{2}$

$\bar{y} = \frac{(7 + 4)}{2} = \frac{11}{2}$

$var(x) = (3 - \frac{5}{2})^2 + (2 - \frac{5}{2})^2$

$var(y) = (7 - \frac{11}{2})^2 + (4 - \frac{11}{2})^2$

$cov(x,y) = (3 - \frac{5}{2})(7 - \frac{11}{2}) + (2 - \frac{5}{2})(4 - \frac{11}{2})$

so, all you need to do is calculate these values and put them in the right places in the matrix. Does that make sense?

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  • $\begingroup$ @Brad S. I have a slightly different problem. I also want to obtain a covariance matrix. I need it to use it as input for a generalized $\chi^2$ minimization in order to fit a model when the errors from the data are correlated. In short, I have a dataset $x_i$ of uncorrelated data with $i=1..30$. But then I need to apply the following transformation to the data $y_i = x_{i-1}-2x_i+x_{i+1}$. After that, the errors get correlated and I need to account for that by means of the covariance matrix given by the transformation. Could you please give me a hand or advise on how to get that matrix? $\endgroup$
    – Stefano
    Commented May 20, 2018 at 15:43
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    $\begingroup$ I think variance formula should be 1/N not 1/N-1, am I missing something? $\endgroup$
    – PeerNet
    Commented Oct 13, 2019 at 22:28
  • $\begingroup$ @PeerNet No, when looking at a sample of a population the correct variance is $\frac{1}{N - 1}\sum(x_i - \bar{x})$. You must be confusing this with the formula for the population variance; there we divide by $N$ only. $\endgroup$ Commented Dec 1, 2020 at 1:11

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