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Suppose I have the multivariate linear regression:

$$Y=a_0 + a_1*X_1 + .... + a_k*X_k + \epsilon$$

What is $Cov(Y, \epsilon)$? Testing it empirically with random values of Y and X, I find that the correlation is close to 1. Why is that the case?

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2 Answers 2

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In linear model, $\epsilon$ always results to be orthogonal to de predictors, then $Cov(y, \epsilon) = Var(\epsilon)$. Their correlation depends on the proportion of y which is explained by the model, so $Cor(y, \epsilon)^2 = 1-R^2$.

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You obtain correlation between Y and e of 1 because when using random X and random Y, they are almost orthogonal, so Y is actually almost e. That's why the correlation is 1.

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