Timeline for Identification from implicit function
Current License: CC BY-SA 4.0
11 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Feb 2, 2020 at 12:43 | answer | added | Alecos Papadopoulos | timeline score: 2 | |
| Jan 8, 2020 at 14:58 | comment | added | whuber♦ | @Stop This question begins with the precise mathematical statement of the model. Except insofar as additional text might explain the symbols, that overrides all else. | |
| Jan 8, 2020 at 13:27 | comment | added | Jesper for President | @whuber I struggle to make sense of that comment. Is it because rather than using the words "endogenous"/"exogenous" which you perhaps see as to vague you would prefer a more precise mathematical statement? Or does it reflect some other disagreement with respect to approaching the question? | |
| Jan 8, 2020 at 9:26 | vote | accept | bonifaz | ||
| Jan 8, 2020 at 9:03 | comment | added | bonifaz | @Tim, yes, that's what I meant with 'implicitly defined'. | |
| Jan 8, 2020 at 2:26 | comment | added | whuber♦ | @Stop "Endogenous" and "exogenous" do not add anything to the model. If they do, then that needs to be incorporated in the model statement! | |
| Jan 8, 2020 at 0:32 | answer | added | Jesper for President | timeline score: 6 | |
| Jan 7, 2020 at 22:56 | comment | added | Jesper for President | How is that supposed to help when $x$ is assumed exogenous but $y_i$ is assumed endogenous (if that is indeed what @bonifaz means by "obvious endogeneity concerns") | |
| Jan 7, 2020 at 22:07 | comment | added | whuber♦ | Why not write the relation as $x_i=f(y_i,\theta)+\epsilon_i$ where $\epsilon_i=-\varepsilon_i$ and $f(y_i,\theta)=y_i-h(y_i,\theta)$? That's a standard regression setting, presenting no special problems. | |
| Jan 7, 2020 at 21:49 | comment | added | Tim | $y_i$ is generated by a function of $y_i$ itself? Are you sure that this is correct? $h(y_i, \theta) = y_i - x_i - \varepsilon_i $. | |
| Jan 7, 2020 at 21:38 | history | asked | bonifaz | CC BY-SA 4.0 |