Timeline for Finding Initial Parameter Values for Gompertz Model with Intercept
Current License: CC BY-SA 4.0
14 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Nov 25, 2018 at 2:11 | vote | accept | Remy | ||
| Nov 24, 2018 at 19:22 | answer | added | dave fournier | timeline score: 1 | |
| Nov 22, 2018 at 20:21 | comment | added | dave fournier | I found the post if it is not clear I can say some more. stats.stackexchange.com/questions/62995/… | |
| Nov 22, 2018 at 20:10 | comment | added | dave fournier | well you don't HAVE to impose the constraint, but then you are using a completely different family of curves. the two families are separated by the singular solution where $\beta_3=0$. To solve your problem you use the fact that the model is linear in y(x(1)) and y(x(n)) and use those values as independent parameters instead of $\alpha$ and $\beta$. I wrote all this up to discuss fitting a 4-5 parameter logistic but it seems to have disappeared | |
| Nov 22, 2018 at 19:57 | comment | added | Remy | My textbook doesn't show those constraints but I see online that you are correct. I would be interested in seeing your approach with those constraints in mind. | |
| Nov 22, 2018 at 17:21 | comment | added | dave fournier | Anyway there is a solution satisfying the usual positivity constraints. It is 1403.91 11799.4 9.83262 0.00340595 with a sum of squares of 854256. which is higher than the solution which violates the constraints. After doing all this work I would appreciate knowing whether the constraints should be satisfied or not. | |
| Nov 22, 2018 at 5:34 | comment | added | dave fournier | this certainly fits the data but thought that for a Gompertz curve $\beta_2$ and $\beta_3$ are $\gt 0$ I guess it depends on what you are trying to do. | |
| Nov 18, 2018 at 6:00 | history | tweeted | twitter.com/StackStats/status/1064035603942178816 | ||
| Nov 18, 2018 at 5:35 | comment | added | JimB | I think that the resulting values of $\hat{\alpha}$ and $\hat{\beta}$ don't look as expected is because many of us would initially expect that $\hat{\alpha}$ would be the value of the intercept. But the intercept is $\alpha+\beta_1 \exp{(-\beta_2)}$. The point is that you've found the correct maximum likelihood estimates. | |
| Nov 18, 2018 at 3:42 | history | edited | Remy | CC BY-SA 4.0 | added 918 characters in body |
| Nov 18, 2018 at 0:44 | comment | added | Remy | Graphing the data, it appears that the pattern is logarithmic, rather than S-shaped. So, I don't think it's a problem if it doesn't fit the data well. I'm only looking to run this specific model though. | |
| Nov 18, 2018 at 0:34 | comment | added | James Phillips | I used the Differential Evolution genetic algorithm to determine initial parameter estimates and an equation search found several sigmoidal equations the fit the posted data well, but I could not find a good fit to the equation you posted. If it might be of use I can post the top few results of the equation search. If you can use Python I can post source code for a graphical fitter for these equations. | |
| Nov 17, 2018 at 22:32 | history | edited | Remy | CC BY-SA 4.0 | added 109 characters in body |
| Nov 17, 2018 at 22:12 | history | asked | Remy | CC BY-SA 4.0 |