5
$\begingroup$

I'm solving the problem 3.33, from the book "Introduction to Linear Regression Analysis (5th edition)", by Montgomery and I got a doubt.

3.33) Prove that $R^2$ is the square of the correlation between $\pmb{y}$ and $\pmb{\hat{y}}$.

I solved the problem assuming that $Cov(\pmb{\hat{y}}, \pmb{e}) = 0$.

Is this true?

Improve this question
$\endgroup$
2
  • $\begingroup$ Yes, that is correct. Linear regression decomposes y (vector containing all y-values in the sample) into two components- $\hat{y}$ that is a linear combination of the columns of X (each row of X is an observation of predictor variables), and, $\epsilon$ that can't be described as such a linear combination (because it is orthogonal to the plane formed by the columns of X). $\endgroup$
    – StatsML
    Commented Jul 3, 2019 at 12:15
  • $\begingroup$ Thanks for the comment. Do you have any material or a book showing this? $\endgroup$
    – igorkf
    Commented Jul 3, 2019 at 12:29

1 Answer 1

Reset to default
4
$\begingroup$

I found the solution in "Linear Models in Statistics", from Rencher, Exercise 9.1 d):

$Cov[\pmb{e}, \pmb{\hat{y}}] = Cov[(\pmb{I - H})\pmb{y}, \pmb{Hy}] = (\pmb{I - H})(\pmb{\sigma^2I})\pmb{H} = \pmb{\sigma^2}(\pmb{H-H}) = 0$,

where $\pmb{H}$ is the hat matrix.

Improve this answer
$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.