Timeline for Relationship between R2 and correlation coefficient
Current License: CC BY-SA 3.0
7 events
| when toggle format | what | by | license | comment | |
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| May 28 at 21:26 | comment | added | Victor | @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). | |
| Mar 12, 2021 at 23:09 | comment | added | Fortunato | 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. | |
| May 12, 2020 at 18:19 | comment | added | user284839 | 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 | |
| Jul 10, 2018 at 1:17 | comment | added | bespectacled | @Dinesh, can you please explain why Cov(y_hat, e)=0 ? | |
| Dec 22, 2015 at 15:53 | vote | accept | anson9 | ||
| Dec 21, 2015 at 13:07 | comment | added | whuber♦ | +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.. | |
| Dec 21, 2015 at 12:28 | history | answered | Dinesh | CC BY-SA 3.0 |