6 ms·
Why isn't the answer P = { Inf -> 1, otherwise 0 } ? Axiom 1: P(E) elem N => P(E) >= 0, for all E Trivially satisfied Axiom 2: P(Omega) = 1 Satisfied: Omega
by waps 14y ago
Why isn't the answer P = { Inf -> 1, otherwise 0 } ?
Axiom 1:
P(E) elem N => P(E) >= 0, for all E
Trivially satisfied
Axiom 2:
P(Omega) = 1
Satisfied:
Omega = N
{ Inf } elem N
Axiom 3:
Sigma additivity. Trivially satisfied since it either includes { Inf } or it doesn't, making the outcome 0 or 1.
Where is the problem ?
I think it's pretty clear that this is the only possible solution, because since N is not closed, there is no way to keep a uniform density other than 0.
This does not seem like it's a very useful solution, but it does seem to satisfy the axioms.
- deleted 14y ago[deleted]
- anonymoushn 14y agoInfinity isn't a positive real :(
- waps 14y agoIn most definitions I've seen of it, it is. In fact, it's a natural number.
- jerf 14y agoIn addition to anonymoushn's correct point, I would also observe that even if we do treat infinity as a specific real number, the integration of a function zero at everywhere except one specific point from -inf to +inf is still 0. And further if we are treating it as a real number, you've now violated the uniformity constraint; P(0) != P(inf), therefore it is no longer a uniform distribution, the thing you are putatively building. For an analysis of this and some other interesting things, see the Dirac delta (psuedo-)function: http://en.wikipedia.org/wiki/Dirac_delta_function http://en.wikipedia.org/wiki/Dirac_delta_function