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.
In 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
Re:Zerocoin: making Bitcoin anonymous
by anonymoushn on Friday April 12, 2013 @12:41AM
(#5536670)
Infinity isn't a positive real :(
In fact everything that we refer to as "normal" distributions in the real world technically aren't, as the finite nature of the universe means the probability of the extremes is simply zero (give or take being totally wrong about the nature of the universe in which case all bets are off anyhow) rather than very, very small, and in many cases there's a sharp cutoff at 0, or some other arbitrary boundary, which a true normal distribution doesn't have. But it's often still the best mathematical approximation, with negligible error. (... until it isn't.... caveat emptor.)