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Given a matrix $A = (X_1,X_2,...,X_n)$

How is it that $A^TA$ is the correlation matrix where $\frac{1}{n}(A^TA)_{ij} = Corr(X_i,X_j)$?

I am confused because $\frac{1}{n}(A^TA)_{ij} = \frac{1}{n}\sum_{k=1}^nA_{ki}A_{kj}$

and $\frac{1}{n}\sum_{k=1}^nA_{ki}A_{kj} = E[X_iX_j]$

but how does $Corr(X_i,X_j) = E[X_iX_j]$ ?

I know that $Cov(X_i,X_j) = E[X_iX_j] - \mu_i\mu_j$

but $Corr(X_i,X_j) = \frac{Cov(X_i,X_j)}{(Var(X_i)Var(X_j))^{\frac{1}{2}}}$

Wouldn't this imply the following?

$$Cov(X_i,X_j) + \mu_i\mu_j = \frac{Cov(X_i,X_j)}{(Var(X_i)Var(X_j))^{\frac{1}{2}}}$$

Unless I am just attempting to skip some serious algebra here, I am not sure what I'm missing here...

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1 Answer 1

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This is because \begin{equation} C_{X_i X_j} = \frac{Cov(X_i,X_j)}{(Var(X_i)Var(X_j))^{\frac{1}{2}}} \end{equation} is the correlation coefficient determined by dividing the covariance by the product of the variables standard deviations, while the correlation is \begin{equation} Corr(X_i,X_j)= E[X_i X_j] \end{equation} If $X_i$ and $X_j$ have zero mean, this is the same as the covariance which is defined as \begin{equation} Cov(X_i,X_j)= E[(X_i-\mu_{X_i}) (X_j-\mu_{X_j})] \end{equation} with $\mu_{X_i}$ and$\mu_{X_j}$ the mean of ${X_i}$ and ${X_j}$, respectively.

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  • $\begingroup$ I don't think this answer is correct due to the second definition. Also the question might lack (as stated in your answer) an hypothesis about each coordinate having zero mean, but also having unitary variance. $\endgroup$
    – cabo
    Dec 1, 2023 at 2:55

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