tag:blogger.com,1999:blog-10560800.post110749872800573083..comments2024-05-22T21:46:41.318-07:00Comments on Machine Learning, etc: Getting probabilities from studentsYaroslav Bulatovhttp://www.blogger.com/profile/06139256691290554110noreply@blogger.comBlogger17125tag:blogger.com,1999:blog-10560800.post-44158152930025222872021-05-31T11:31:03.280-07:002021-05-31T11:31:03.280-07:00Sohbet
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Lundgrenhttps://www.blogger.com/profile/17611372683875347506noreply@blogger.comtag:blogger.com,1999:blog-10560800.post-40236600567773386892018-09-13T07:22:04.972-07:002018-09-13T07:22:04.972-07:00Graphs are best option to know the result or proba...Graphs are best option to know the result or probabilities as you can see the clear response. Many students believe that this is their own choice to have best plan to study. <a href="https://www.litreview.net/faq-on-how-to-write-my-literature-review/" rel="nofollow">https://www.litreview.net/faq-on-how-to-write-my-literature-review/</a> can help you a lot in many writing cases. <br />martinhttps://www.blogger.com/profile/12223202060259660467noreply@blogger.comtag:blogger.com,1999:blog-10560800.post-6811858586237756952018-09-12T20:42:25.746-07:002018-09-12T20:42:25.746-07:00Nice post. Probability is an important mathematica...Nice post. Probability is an important mathematical function in our high education syllabus. 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If you wants to buy that service so <a href="http://www.thesiswritingservice.com/" rel="nofollow">click here</a> and get remarkable online service. <br /> hayathttps://www.blogger.com/profile/07826528651799350596noreply@blogger.comtag:blogger.com,1999:blog-10560800.post-68995268413473013562018-08-31T11:10:11.187-07:002018-08-31T11:10:11.187-07:00Hemingways and their analysis about the critical s... Hemingways and their analysis about the critical side could have been really effective as they are really saying the right one. <a href="https://www.essayeditor.biz/" rel="nofollow">try this</a> for the students that is very helpful for the writing services.<br />bollingshttps://www.blogger.com/profile/10727115107258018703noreply@blogger.comtag:blogger.com,1999:blog-10560800.post-13497773418028059972010-07-30T17:23:10.309-07:002010-07-30T17:23:10.309-07:00BTW, Shen Yi has written a whole dissertation on i...BTW, Shen Yi has written a whole dissertation on issue of calibrated losses, "LOSS FUNCTIONS FOR BINARY CLASSIFICATION AND CLASS PROBABILITY ESTIMATION", page 16 has this derivation characterizing loss functions that elicit probabilities (ie, "proper scoring rules")Yaroslav Bulatovhttps://www.blogger.com/profile/06139256691290554110noreply@blogger.comtag:blogger.com,1999:blog-10560800.post-86326087621459264232007-08-28T05:40:00.000-07:002007-08-28T05:40:00.000-07:00I know the derivation now. Thanks. What if the pro...I know the derivation now. Thanks. What if the problems are multichoice ones?Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-10560800.post-2114324102677076842007-08-28T01:40:00.000-07:002007-08-28T01:40:00.000-07:00Hi, I cannot figure out why "any utility function ...Hi, I cannot figure out why "any utility function f(x) for which f'(x)/f'(1-x)=(1-x)/x will work. (ie expected utility EU(x) will have a local extremum at point x=p)". <BR/>The link is not available now. So I cannot see your derivation. Could you post it here? Thanks a lot.Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-10560800.post-1107743820708170172005-02-06T18:37:00.000-08:002005-02-06T18:37:00.000-08:00The most interesting answer I got was also the sim...The most interesting answer I got was also the simplest looking:<br /><br />f'(x)/f'(1-x) = (1-x)/x<br /><br />rewrite that as<br /><br />xf'(x) = (1-x)f'(1-x)<br /><br />So basically you can see here is that the necessary condition is that<br />xf'(x) is some m(x) which is invariant under x->1-x<br /><br />So to construct proper f(x) all we need to do is pick some m(x) that's<br />symmetric Yaroslav Bulatovhttps://www.blogger.com/profile/06139256691290554110noreply@blogger.comtag:blogger.com,1999:blog-10560800.post-1107683462056283492005-02-06T01:51:00.000-08:002005-02-06T01:51:00.000-08:00Some smart people in comp.math commented on this, ...Some smart people in comp.math commented on this, and it turns out that you can take any nice function defined on (0,1/2] and extend it to the second half by<br /><br />f(x) = C + int_{1/2}^x (1-t) f'(1-t)/t dt <br />where C is determined in a way to make the two pieces match up.<br /><br />One person suggested that in order for the function to be infinitely differentiable it should look like a/xYaroslav Bulatovhttps://www.blogger.com/profile/06139256691290554110noreply@blogger.comtag:blogger.com,1999:blog-10560800.post-1107568725642544112005-02-04T17:58:00.000-08:002005-02-04T17:58:00.000-08:00So any utility function f(x) for which f'(x)/f'(1-...So any utility function f(x) for which f'(x)/f'(1-x)=(1-x)/x will work. (ie expected utility EU(x) will have a local extremum at point x=p)<br /><br />This sort of looks like a differential equation...is there a good way to characterize the set of solutions?Yaroslav Bulatovhttps://www.blogger.com/profile/06139256691290554110noreply@blogger.comtag:blogger.com,1999:blog-10560800.post-1107545713731827172005-02-04T11:35:00.000-08:002005-02-04T11:35:00.000-08:00Yes, your idea seems to elicit correct probabiliti...Yes, your idea seems to elicit correct probabilities as well. When I said f(x)=x^2 does not work, that was treating f(x) as a way of awarding points. On other hand, you are penalizing, so this equates to f(x)=-(1-x)^2 utility function which *does* achieve expected maximum at x=pYaroslav Bulatovhttps://www.blogger.com/profile/06139256691290554110noreply@blogger.com