# Reconciling two definitions of 'uncorrelatedness'

In this paper, the authors defined uncorrelatedness in the following way:

Let $\mathbf{X}=(X_1,...,X_n)$, and $\mathbf{Y}=(Y_1,...,Y_n)$, where $X_i\sim X$ and $Y_i\sim Y$.

$\mathbf{X},\mathbf{Y}$ are uncorrelated iff $(\mathbf{X}-\bar{X}\mathbf{1})'(\mathbf{Y}-\bar{Y}\mathbf{1})=0$, where $\bar{X}=1/n \sum X_i$ and $\mathbf{1}=(1,...,1)$.

My question is what is the link between this definition of uncorrelatedness and the one usually used (I think), that of (for vectors r.v.) covariance matrix - a matrix where each entry $(i,j)$ is $\text{Corr}(X_i,X_j)$ - being an identity matrix?

Any help would be appreciated.

• Looks like the thing they're computing is a multiple of a sample correlation. If those terms are zero, the off-diagonal elements of the sample correlation matrix will be 0. – Glen_b Aug 15 '14 at 21:21

Note how much more strict is this condition for uncorrelatedness of vectors of numbers, compared to the conditions for statistical uncorrelatedness: the later requires only that the condition $(\mathbf{X}-\bar{X}\mathbf{1})'(\mathbf{Y}-\bar{Y}\mathbf{1})=0$ holds "on average"
• Thanks for your interest in this question Alecos. One doubt: Could you elaborate a bit further on the last part of your answer? How does the statistical definition, $E[(\mathbf{X}-E(\mathbf{X}))(\mathbf{Y}-E(\mathbf{Y}))']= I_n$ relate to $E[(\mathbf{X}-E(\mathbf{X}))'(\mathbf{Y}-E(\mathbf{Y}))]=0$? And what is the link with orthogonality in the econometric sense? I hope you don't mind. (Your profile said you're doing a macro phd, so I have to take advantage. :) ) – An old man in the sea. Aug 15 '14 at 18:21
• I have the impression that you are confusing $E[(\mathbf{X}-E(\mathbf{X}))(\mathbf{Y}-E(\mathbf{Y}))']$ with $E[(\mathbf{X}-E(\mathbf{X}))(\mathbf{X}-E(\mathbf{X}))']$. Moreover this (last) expression gives covariance of the elements of vector $\mathbf X$, not correlation, so it is not equal to the identity matrix -what we want for non-correlation is that it is diagonal. – Alecos Papadopoulos Aug 15 '14 at 18:36