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I've been looking all over the interwebs and have been having trouble finding good uses of a covariance matrix to find the correlation coefficient.

For example, given a problem like:

Suppose that $x = \begin{bmatrix} x_1 \\ x_2 \end{bmatrix}$ is a vector with a mean of $0$ and covariance $P = \begin{bmatrix}4.5 & -1 \\ -1 & 2 \end{bmatrix}$. What is the correlation coefficient of $x_1$ and $x_2$?

I know that to find the correlation coefficient of $x_1$ and $x_2$, it is: $$P_{x_1 x_2} = \frac{cov(x_1, x_2)}{\sigma_1 \sigma_2}$$

Furthermore, I believe the $\sigma$ can be derived from the diagonals of the covariance matrix, but I'm not sure how to find $cov(x_1, x_2)$. How can one derive a single covariance from the matrix?

Thanks for your help in advanced!

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There's no derivation needed. All you have is to read the covariances appropriately from their corresponding position in the covariance matrix. The covariance matrix is read as follows \begin{equation} P = \begin{bmatrix} var(X_1) & cov(X_1,X_2) \\\\ cov(X_1,X_2) & var(X_2) \\ \end{bmatrix} \end{equation} where $\sigma_1^2 = var(X_1)$ and $\sigma_2^2 = var(X_2)$. So, yes, as you say, the $\sigma^2_k$'s are at the diagonals and the covariances are at the off-diagonals. Therefore \begin{equation} P_{x_1x_2} = \frac{cov(x_1, x_2)}{\sigma_1 \sigma_2}= \frac{-1}{\sqrt{4.5}\sqrt{2}} \end{equation}

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  • $\begingroup$ Awesome, answer, thank you so much! Is there ever a possibility where the opposite $cov(X_1,X_2)$ diagonals could be different? $\endgroup$ – Billy Thorton Sep 17 '18 at 3:06
  • $\begingroup$ No you always have $cov(X_1,X_2) = cov(X_2,X_1)$. $\endgroup$ – Ahmad Bazzi Sep 17 '18 at 3:06
  • $\begingroup$ btw, if you found the answer helpful, you could upvote and mark it as correct so that other people searching the same topic could know that it was helpful. $\endgroup$ – Ahmad Bazzi Sep 17 '18 at 3:07

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