Timeline for Can the empirical Hessian of an M-estimator be indefinite?
Current License: CC BY-SA 2.5
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| Feb 27, 2011 at 20:06 | history | edited | vqv | CC BY-SA 2.5 |
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| Feb 26, 2011 at 13:52 | comment | added | whuber♦ | +1 Good points in the update, especially the last paragraph. When the Hessian is available--as is implicitly assumed throughout this discussion--one would automatically use its positive-definiteness as one of the criteria for testing any critical point and therefore this issue simply could not arise. This leads me to believe the Wooldridge quotation must concern the Hessian at a putative global minimum, not at a mere critical point. | |
| Feb 26, 2011 at 5:45 | history | edited | vqv | CC BY-SA 2.5 |
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| Feb 26, 2011 at 5:37 | history | edited | vqv | CC BY-SA 2.5 |
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| Feb 26, 2011 at 5:32 | history | edited | vqv | CC BY-SA 2.5 |
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| Feb 25, 2011 at 8:44 | comment | added | probabilityislogic | could you adapt your answer so that it matches the notation of the question? To what is $x^2-y^2$ referring? Where does this get inserted into the equations given in the question? | |
| Feb 24, 2011 at 15:42 | history | edited | vqv | CC BY-SA 2.5 |
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| Feb 24, 2011 at 15:33 | history | answered | vqv | CC BY-SA 2.5 |