Timeline for Comparing Regression Coefficients from a "log-log" to an Alternative De-meaning Procedure
Current License: CC BY-SA 4.0
6 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jan 14, 2020 at 12:45 | comment | added | Jesper for President | I do not think so. | |
| Jan 14, 2020 at 12:02 | comment | added | km5041 | This is very helpful. And I get very similar estimates as well but they are not exact. Is there a way to “undo” the Taylor approximation in order to write $\alpha$ as a function of $\beta$? | |
| Jan 14, 2020 at 11:29 | answer | added | Serge Kashlik | timeline score: 0 | |
| Jan 14, 2020 at 10:49 | comment | added | Jesper for President | IF you include a constant the OLS estimates of the coefficients of $(x/\bar x)$ and $\log(x)$ should not change that much. | |
| Jan 14, 2020 at 10:46 | comment | added | Jesper for President | You probably need to provide either code from which the simulation can be read or a detailed description of the simulation. Personally I get very similar estimates. If you make a first order Taylor approximation of the log function around $\bar x$ then you get $\log(x) \approx \log(\bar x) + (x-\bar x)/\bar x$ this identity can be used to go from model 1 to model 2. $\log(y) = \lambda + \log(x)\alpha + u$ becomes $\log(y) = \lambda + [\log(\bar x) + (x-\bar x)/\bar x]\alpha + u$ which becomes $\log(y) = (\lambda + \log(\bar x)\alpha - \alpha) + (x/\bar x)\alpha + u$. | |
| Jan 14, 2020 at 1:07 | history | asked | km5041 | CC BY-SA 4.0 |