Hey guys, I have a quick question. I am trying to prove that the squared sample correlation between fitted and observed values is equal to $R^2$ (coefficient of determination). I am having a lot of trouble. For the numerator, SEE, I have found that $\mathrm{Cov} (y, \hat{y})$ is equal to $\mathrm{Var} (\hat{y})$. I did this through multiplying out $(\hat{y} - \overline{y})'(y - \overline{y})$ and getting that to equal $\mathrm{var}(\hat{y})$. If anyone has a better way to do this, please let me know. What I am having trouble with is the denominator. I found what it would be in matrix form and that is $y'Qy \cdot \hat{y}'Q\hat{y}$ where $Q$ is the operator that gives us the deviations and from here I don't know how to manipulate the two. So I can get just SST or $y'Qy$. I hope someone out there can help and understands ($'$ means transpose).
2026-04-03 12:58:06.1775221086
Linear Regression Question (Linear Algebra) Help!!
71 Views Asked by Bumbble Comm https://math.techqa.club/user/bumbble-comm/detail AtRelated Questions in LINEAR-ALGEBRA
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