Timeline for Why can't we use $R^2$ for different dependent variables?
Current License: CC BY-SA 3.0
16 events
| when toggle format | what | by | license | comment | |
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| May 9, 2018 at 18:05 | vote | accept | An old man in the sea. | ||
| May 9, 2018 at 18:04 | answer | added | An old man in the sea. | timeline score: 0 | |
| Jul 5, 2017 at 10:49 | history | bumped | CommunityBot | This question has answers that may be good or bad; the system has marked it active so that they can be reviewed. | |
| Jun 3, 2017 at 18:52 | history | bumped | CommunityBot | This question has answers that may be good or bad; the system has marked it active so that they can be reviewed. | |
| May 14, 2017 at 8:08 | comment | added | An old man in the sea. | @AlecosPapadopoulos What's your opinion on this? Assume we're in the case of only a different dependent variable. | |
| May 14, 2017 at 4:01 | comment | added | chan1142 | It depends on how you define "better suited". $R^2_y$ and $R^2_{\log(y)}$ are two different things. One tells you how well $y$ is explained by the model, and the other how well $\log(y)$ is explained by the model. If my goal is to explain $y$ best, I would choose a model which explains $y$ best, not $\log(y)$ best. $R^2_{\log(y)}$ does not give you information on $R^2_y$, vice versa. | |
| May 5, 2017 at 16:48 | comment | added | An old man in the sea. | @Henry by the way, what's your take on this question? can we use R^2 or not? | |
| May 4, 2017 at 20:01 | comment | added | An old man in the sea. | @Henry It seems that you're right... I searched a bit. It's variance stabilizing, not reduction. Thanks for the comment. ;) | |
| May 4, 2017 at 18:38 | answer | added | Dave Harris | timeline score: 1 | |
| May 4, 2017 at 8:21 | comment | added | Henry | Suppose your $x$ data is $1000, 2000, 3000$ and your $y$ data is $4000, 6000, 8000$. Taking logarithms will reduce $R^2$ | |
| May 4, 2017 at 8:11 | comment | added | An old man in the sea. | @Henry at first thought, I would tend to disagree if the values are big enough(maybe 10^3 or more?). The log transformation is known precisely by its variance reduction properties. | |
| May 4, 2017 at 8:08 | comment | added | An old man in the sea. | @AlecosPapadopoulos I'm interested in both cases, while keeping the same number of regressors. | |
| May 3, 2017 at 22:29 | answer | added | Andrew M | timeline score: 0 | |
| May 3, 2017 at 21:32 | comment | added | Henry | Unless you have information not shown here, there is not particular reason to suppose $\log$ will improve the $R^2$. It might, or it might have the opposite effect | |
| May 3, 2017 at 21:29 | comment | added | Alecos Papadopoulos | You are only transforming the dependent variable and keep the regressors the same? Or the question is more general "Can we use $R^2$ as a model selection criterion"? | |
| May 3, 2017 at 20:35 | history | asked | An old man in the sea. | CC BY-SA 3.0 |