tag:blogger.com,1999:blog-4987609114415205593.post889347141051223196..comments2024-03-28T13:40:26.497+00:00Comments on M-Phi: Aggregating abstaining expertsJeffrey Ketlandhttp://www.blogger.com/profile/01753975411670884721noreply@blogger.comBlogger1125tag:blogger.com,1999:blog-4987609114415205593.post-65325484620187471772017-09-13T00:08:49.371+01:002017-09-13T00:08:49.371+01:00Posted this on facebook then figured it might be m...Posted this on facebook then figured it might be more appropriate to post here.<br /><br />Interesting considerations. I wonder if there is anything useful to be said when there isn't a unique extension of the credences to the whole space but they merely impose a restriction on the remaining values on which they are undefined.<br /><br />For instance, I suspect if Alice assigns a probability to X_1, X_2 and X_3 (which form a partition) and Bob assigns P(X_1)=.99 this highly restricts his assignment to X_2 and X_3 but I suspect for an appropriate credence function for Alice we will find merely aggregating these credences on X_1 will yield a result that couldn't be achieved by aggregating any coherent extension of Bob's credences to the whole space with Alice's. However, since there is no unique extension one needs to do something more complex then merely use the unique extension. Perhaps consider some kind of average of all possible extensions of Bob's credence function with Alice's credence would meet this more demanding criteria.<br /><br />There might be an interesting theorem in here ... I'll have to think about it.TruePathhttps://www.blogger.com/profile/00124043164362758796noreply@blogger.com