3 ms·
In the standard measure theory formalization of probability, that outcome set does have measure exactly 0. Perhaps we could use a variation on in which somehow
by drdeca 7y ago
In the standard measure theory formalization of probability, that outcome set does have measure exactly 0.
Perhaps we could use a variation on in which somehow supports some infinitesimal values, but if one wishes to keep both countable additivity and have the distribution be uniform in the sense of translation invariant, then we run into difficulties.