# Calculating correlation matrix from covariance matrix - r>1

I have been given a covariance matrix and asked to calculate the correlation matrix, but I get an error when doing this.

The covariance matrix is:

$$\sum = \begin{bmatrix}4&6\\6&1\end{bmatrix}$$

and

$$Corr(X,Y)= \frac{Cov(X,Y)}{\sqrt{V(X).V(Y)}}$$

so here,

$$Corr(X,Y)= \frac{6}{\sqrt{4}}=3!!!$$

But, $$-1 \le Corr(X,Y) \le 1$$

So this is impossible. What am I missing here?

## 1 Answer

Your covariance matrix is not a covariance matrix.

It should be positive (semi) definite so both eigenvalues should be $\ge 0$, but $\det(\Sigma) < 0$ so the eigenvalues have opposite signs.

• So you're saying I've been given the wrong matrix? Oh dear, this is a past exam question! – Pebbles28 Feb 8 '18 at 16:16
• @Pebbles28 I could always be mistaken and am happy to delete my answer with apologies if so. It would be worth double checking your question. – PM. Feb 8 '18 at 16:56
• @Pebbles28 I wonder if you were able to follow this up and double check the matrix against the exam question? – PM. Feb 13 '18 at 17:17
• The matrix in question is the same as the one from the past paper, but the answers are not yet available. I will follow up when they become available (likely in March or April). – Pebbles28 Feb 14 '18 at 18:02
• @Pebbles28 thanks, this is something that would be good to get to the bottom of. – PM. Feb 14 '18 at 18:05