tag:blogger.com,1999:blog-3891434218564545511.post7506166562378884638..comments2024-03-28T19:56:42.305-05:00Comments on Alexander Pruss's Blog: Domination and probabilistic consistencyAlexander R Prusshttp://www.blogger.com/profile/05989277655934827117noreply@blogger.comBlogger7125tag:blogger.com,1999:blog-3891434218564545511.post-16350048046549276932020-02-19T09:08:24.321-06:002020-02-19T09:08:24.321-06:00Ian:
I've written up a proof on Saturday. It...Ian: <br /><br />I've written up a proof on Saturday. It works for an infinite spaces, but remember that I'm only interested in *simple* utility functions, i.e., ones that have a finite number of values. So no St Petersburg.Alexander R Prusshttps://www.blogger.com/profile/05989277655934827117noreply@blogger.comtag:blogger.com,1999:blog-3891434218564545511.post-17903327596043190512020-02-18T22:32:37.012-06:002020-02-18T22:32:37.012-06:00Does ‘finitely-additive’ in the post imply that Ω ...Does ‘finitely-additive’ in the post imply that Ω may be infinite?<br /><br />I have a sketch proof that works for finite Ω. (Like you, I have not actually written out all the details.) But for infinite Ω, St Petersburg-like possibilities make things tricky.IanShttps://www.blogger.com/profile/00111583711680190175noreply@blogger.comtag:blogger.com,1999:blog-3891434218564545511.post-54463414795499570752020-02-17T20:53:26.029-06:002020-02-17T20:53:26.029-06:00Oops : ( wasnt paying attention.
--benOops : ( wasnt paying attention.<br /><br />--benThe Forgetful Apologisthttps://www.blogger.com/profile/16991040857248725326noreply@blogger.comtag:blogger.com,1999:blog-3891434218564545511.post-3580356081428006422020-02-17T20:51:27.650-06:002020-02-17T20:51:27.650-06:00This comment has been removed by the author.The Forgetful Apologisthttps://www.blogger.com/profile/16991040857248725326noreply@blogger.comtag:blogger.com,1999:blog-3891434218564545511.post-43598236492583453412020-02-17T20:48:47.823-06:002020-02-17T20:48:47.823-06:00This comment has been removed by the author.The Forgetful Apologisthttps://www.blogger.com/profile/16991040857248725326noreply@blogger.comtag:blogger.com,1999:blog-3891434218564545511.post-16029299532738887352020-02-17T15:52:24.573-06:002020-02-17T15:52:24.573-06:00Ben:
Your ≼ isn't a total preorder. For insta...Ben:<br /><br />Your ≼ isn't a total preorder. For instance, define f(x)=x and g(x)=1-x. Then neither f≼g nor g≼f is true.Alexander R Prusshttps://www.blogger.com/profile/05989277655934827117noreply@blogger.comtag:blogger.com,1999:blog-3891434218564545511.post-25104985224056591212020-02-17T14:39:37.508-06:002020-02-17T14:39:37.508-06:00Take \Sigma=\{0,1\} and define f ≼ g iff f \leq g,...Take \Sigma=\{0,1\} and define f ≼ g iff f \leq g, so that Ef=min f.<br /><br />Then 1=P(\{0,1\})=P\{0\}+P\{1\}=E1_{\{0\}}+E1_{{\1\}}=0, a contradiction.Ben Wallishttps://www.blogger.com/profile/00131358613835119782noreply@blogger.com