Timeline for MLE exists and is unique for iid series
Current License: CC BY-SA 4.0
11 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Mar 23, 2021 at 14:52 | vote | accept | statwoman | ||
| Mar 23, 2021 at 9:01 | history | edited | Ben | CC BY-SA 4.0 |
deleted 464 characters in body
|
| Mar 23, 2021 at 9:01 | comment | added | Ben | Indeed. But the condition you have specified in your question is $a>1$. I think you can just compute the derivatives from scratch in this case. | |
| Mar 23, 2021 at 6:57 | comment | added | Yves | It is quite well-known that if $g(\boldsymbol{\theta})$ is a non-negative convex function of $\boldsymbol{\theta}$ then $g(\boldsymbol{\theta})^a$ is convex for $a \geq 1$. Yet, this does not give a strict convexity, and does not provides us with derivatives fot the optimisation. | |
| Mar 23, 2021 at 5:06 | comment | added | Ben | @statwoman: Sorry about that --- I misread the initial density function. I have edited the answer to correct the analysis. | |
| Mar 23, 2021 at 5:05 | history | edited | Ben | CC BY-SA 4.0 |
added 2393 characters in body
|
| Mar 23, 2021 at 1:55 | history | edited | Ben | CC BY-SA 4.0 |
deleted 3181 characters in body
|
| Mar 23, 2021 at 1:51 | comment | added | statwoman | You wouldn't have the log in the first equation, since it is $exp(−|x−\theta|^a)$. | |
| Mar 22, 2021 at 22:54 | history | edited | Ben | CC BY-SA 4.0 |
added 881 characters in body
|
| Mar 22, 2021 at 20:23 | vote | accept | statwoman | ||
| Mar 23, 2021 at 3:11 | |||||
| Mar 22, 2021 at 9:00 | history | answered | Ben | CC BY-SA 4.0 |