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I remember in a mathematical probability class, the professor writing down the three basic axioms. The first two are very simple...the third is just additivity
by mturmon 14y ago
I remember in a mathematical probability class, the professor writing down the three basic axioms. The first two are very simple...the third is just additivity:
P(A1 U A2 U ...) = P(A1) + P(A2) + ...
for disjoint Ai's.
He explained the intent, and asked us if we thought the axiom was reasonable. He seemed very disappointed when everyone nodded and smiled that it seemed okay, unobjectionable, even obvious. He re-emphasized that this was an infinite union. He said something to the effect of, you seem too complacent about this one, because the union is allowed to be infinite, and you can play all kinds of games with that.
It was only later that some of us came across the huge literature that alters these axioms (some by weakening the additivity axiom above), to see what more general systems come out.
- btilly 14y agoThe third should be just additivity over countable sets of disjoint Ai's. If you leave out countable, then I can make each Ai be a single real number. Under any continuous probability distribution, they have individual probability 0. Yet all together we want them to have probability 1.
- mturmon 14y agoIndeed. I used the ... notation rather than an abstract index to indicate a countable index set. If you expand out all the details, it's tedious. For instance, each Ai also has to be measurable.
- Someone 14y agoCountable does not make a difference. If you pick a random integer, each has probability zero, but you will pick one of them.
- btilly 14y agoNot a counterexample. There exists no continuous probability distribution over the integers, and therefore there is no way to "pick a random integer, each with probability zero".
- mturmon 14y agoNo, it needs to be countable unions only. And to your specific example: There is no probability measure on the integers that assigns zero probability to each integer.
- Someone 14y agoYep, that was a lapse of mind. I was thinking of divergent double sums vs divergent double integrals.