in simple linear regression

R-squared is equal to the squared correlation coefficient between the actual y and the predicted y (i.e. 𝑦 hat )

how to prove this relationship?



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):


Proof for the relationship between R2 and correlation coefficient

I hope this answer clears your doubt.

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    $\begingroup$ +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 Dec 21 '15 at 13:07
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    $\begingroup$ @Dinesh, can you please explain why Cov(y_hat, e)=0 ? $\endgroup$ – bespectacled Jul 10 '18 at 1:17
  • $\begingroup$ 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 $\endgroup$ – aman5319 May 12 '20 at 18:19
  • $\begingroup$ 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. $\endgroup$ – Fortunato Mar 12 at 23:09

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