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Topic: Probability 0. (Read 360 times) |
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James Lu
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Is it possible to prove that the probability of something is zero?
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Grimbal
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Re: Probability 0.
« Reply #1 on: Feb 9th, 2005, 9:20am » |
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If it were not possible, then the probabiliy of finding such a proof would be zero. Wouldn't it? 
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towr
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Re: Probability 0.
« Reply #2 on: Feb 9th, 2005, 9:35am » |
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yes, it's 'just' mathematics after all..
f.i. call PN(1) = 1/N the probability of selecting 1 from a uniform distribution over integers 1..N
then lim N[to][infty] PN(1) = 0
So the probability of selecting 1 from a uniform distribution over the positive integers is 0. (But interestingly enough, it is not impossible)
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| « Last Edit: Feb 9th, 2005, 9:45am by towr » |
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Icarus
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Re: Probability 0.
« Reply #3 on: Feb 9th, 2005, 3:50pm » |
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Actually, that proof is incomplete, as you have not shown that PN [to] P[subinfty] as N [to] [infty]. And in fact, P[subinfty], a uniform distribution over all positive integers, does not even exist.
However, to prove that P(A) = 0, all you need to do is prove that P(A) < [epsilon] for all [epsilon] > 0. This is usually fairly easy to do, given a particular set A and probability distribution P.
For example, uniform probability over the unit interval [0,1] is defined by P(A) = [int]A1 dx, and is definable for any A for which the integral exists. In particular P([a,b]) = b-a, for a, b [in] [0,1]. And if A [subseteq] B (and both A and B have definable probability), then it is clear that P(A) [le] P(B).
In particular, for [epsilon] > 0, P(1/2) := P({1/2}) [le] P([(1-[epsilon])/2, (1+[epsilon])/2]) = [epsilon]. Since this holds for all [epsilon]>0, P(1/2) = 0.
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