in simple linear regression
R-squared is equal to the squared correlation coefficient between the actual y and the predicted y (i.e. š¯‘¦ hat )
how to prove this relationship?
Thanks!
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1 Answer
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The usual way of interpreting the coefficient of determination R^{2} is to see it as the percentage of the variation of the dependent variable y (Var(y)) can be explained by our model.
For the proof we have to know the following (taken from OLS theory and general statistics):
I hope this answer clears your doubt.
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3$\begingroup$ +1. Welcome to our site, Dinesh. You would likely find it easier to write mathematical expressions (and we would find them easier to read) using the built-in $\TeX$ markup: just enclose them between dollar signs \$. Further help is available.. $\endgroup$– whuber ♦Commented Dec 21, 2015 at 13:07
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1$\begingroup$ @Dinesh, can you please explain why Cov(y_hat, e)=0 ? $\endgroup$ Commented Jul 10, 2018 at 1:17
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$\begingroup$ if I have understood this correctly then e is the error or noise term added to y_hat and since e is a constant, the mean e_mean = e and putting this into covariance formula e-e_mean part will be 0 hence Cov(y_hat,e) = 0 $\endgroup$– user284839Commented May 12, 2020 at 18:19
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2$\begingroup$ Rather than e being a constant, I think the idea is that we assume that the errors are not correlated with the prediction, hence Cov(y_hat, e) = 0. Please correct me if i'm wrong. $\endgroup$ Commented Mar 12, 2021 at 23:09
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1$\begingroup$ @Fortunato It's not a question of constant term or an assumption. In fact, it's an algebraic property of the OLS. Just do $Cov(\hat{y},\hat{e}) = \sum{(\hat{e}_i - \bar{e})(\hat{y}_i - \bar{y})} = \sum{\hat{e}_i(\hat{y}_i - \bar{y})}$ and expand the last part using the fact that $\sum{\hat{e}_i} = 0$ and $\sum{\hat{e}_i x_{ij}} = 0$ by construction (FOC). $\endgroup$– VictorCommented May 28 at 21:26