Timeline for Can the empirical Hessian of an M-estimator be indefinite?
Current License: CC BY-SA 2.5
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Feb 28, 2011 at 4:09 | history | bounty ended | Jyotirmoy Bhattacharya | ||
| Feb 28, 2011 at 4:09 | vote | accept | Jyotirmoy Bhattacharya | ||
| Feb 26, 2011 at 22:16 | history | edited | whuber♦ | CC BY-SA 2.5 |
added 756 characters in body
|
| Feb 26, 2011 at 4:22 | comment | added | Jyotirmoy Bhattacharya | Thanks you both you and @mpiktas for the very nice analysis. I would tend to agree with you that Wooldridge is confusing numerical difficulties with theoretical properties of the estimator. Let's see if there are any other answers. | |
| Feb 25, 2011 at 20:51 | comment | added | whuber♦ | @mpiktas At the boundary the Hessian won't necessarily even be defined. The idea is this: if the Jacobian/Hessian/second derivative matrix is defined at a critical point, then in a neighborhood the function acts like the quadratic form determined by this matrix. If the matrix has positive and negative eigenvalues, the function must increase in some directions and decrease in others: it cannot be a local extremum. This is what concerned @Jyotirmoy about the quotation, which seems to contradict this basic property. | |
| Feb 25, 2011 at 19:19 | comment | added | mpiktas | even at the boundary, hessian will be positive? I'll check out the book, I see that I really lack extensive knowledge in this area. Classical theorems are very simple, so I assumed that there should not be something else very complicated. That maybe one of the reasons why I had so much difficulty answering the question. | |
| Feb 25, 2011 at 18:31 | comment | added | whuber♦ | @mpiktas Yes, I'm sure there exist problems where an interior global minimum has an indefinite Hessian, yet where all parameters are identifiable. But it simply is not possible for the Hessian at a sufficiently smooth interior global minimum to be indefinite. This sort of thing has been proven again and again, such as in Milnor's Topology from a Differentiable Viewpoint. I suspect Wooldridge may have been misled by errant numerical "solutions." (The typos on the quoted page suggest it was written hastily, by the way.) | |
| Feb 25, 2011 at 18:00 | comment | added | mpiktas | +1, nice analysis. I think that is why Wooldridge included the remark. I still think it is possible to think of some example where the hessian will be indefinite. Artificially restricting the parameter space for example. In this example the parameter space is whole plane, that is why the local minimum will give semi-positive hessian. I think the time has come to write a nice email to Wooldridge to get his take on the question:) | |
| Feb 25, 2011 at 17:33 | history | answered | whuber♦ | CC BY-SA 2.5 |