Timeline for Is there a name for the "projection" of a function under argmax?
Current License: CC BY-SA 2.5
10 events
| when toggle format | what | by | license | comment | |
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| Jul 4, 2010 at 7:33 | vote | accept | CommunityBot | ||
| Jul 1, 2010 at 10:21 | comment | added | Michael Greinecker | That still wouldn't work. Take some constant function that has exactly two maximizers. Then you can perturb the whole thing slightly and still have two maximizers. By carefully selecting your maximizers, you can get very pathological functions. | |
| Jul 1, 2010 at 9:52 | comment | added | user6979 | Michael: That's true; on the other hand, "most" continuous functions are not constant. Perhaps I should restrict attention to those f which are not constant on any neighborhood in A×B? | |
| Jul 1, 2010 at 9:11 | comment | added | Michael Greinecker | A lot will depend on your tie-breaking rule. If f is constant, you will have no problem finding nowhere continuous and even nonmeasurable functions g. | |
| Jul 1, 2010 at 8:57 | answer | added | Michael Greinecker | timeline score: 0 | |
| Jul 1, 2010 at 7:39 | comment | added | user6979 | I believe this is standard notation in certain contexts (en.wikipedia.org/wiki/Arg_max , planetmath.org/encyclopedia/ArgMin.html). Thanks to Suresh and Michael Burge for restating my definition very clearly in terms that do not use arg max. | |
| Jul 1, 2010 at 6:31 | comment | added | Michael Burge | I think it's the "argument that maximizes"; that is, fix a value of x and ask which y leads to the largest value of f(x,y). | |
| Jul 1, 2010 at 6:28 | comment | added | Suresh Venkat | arg here means the argument (i.e the y) that achieves the max value. so g(x) is the value of y that maximizes f(x,y) | |
| Jul 1, 2010 at 6:18 | comment | added | Robin Chapman | I'm baffled by the notation: what is "arg" here? Generally that is used for the argument of a complex number, but that can't be the case here. | |
| Jul 1, 2010 at 5:05 | history | asked | user6979 | CC BY-SA 2.5 |