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Topic: Continuous Probability Distributions (Read 568 times) |
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Sir Col
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Continuous Probability Distributions
« on: Nov 16th, 2008, 1:39pm » |
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I have two related queries...
In conventional probability theory for continuous distributions, P(X=a) = 0 for all real values, a. So does the statement, "The sum of all probable outcomes..." have any meaning?
I recall reading a paper many years ago which provided a formal treatment of continuous probabilities and I think that infinitesimals were assigned to particular outcome. However, I cannot find any references now.
Does anyone have any knowledge of this?
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towr
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Re: Continuous Probability Distributions
« Reply #1 on: Nov 16th, 2008, 2:22pm » |
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on Nov 16th, 2008, 1:39pm, Sir Col wrote:| In conventional probability theory for continuous distributions, P(X=a) = 0 for all real values, a. So does the statement, "The sum of all probable outcomes..." have any meaning? |
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I'd say in this case only integrals have any real meaning. P(X=a) is really  aa pdf(X=i) (where pdf is the probability density function.)
So, is the area of under a graph the sum of the areas under each point of a graph?
Can we make something of ab xx f(z) dz dy ? And will it be ab f(x) dy ?
It would seem to imply xx f(z) dz = f(x), which doesn't appear very likely, unless f(x)=0.
hmm.
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Grimbal
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Re: Continuous Probability Distributions
« Reply #2 on: Nov 17th, 2008, 5:21am » |
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I have been taught the approach that in the general case, you should consider P(X in S) where S is a set.
For discrete cases, it comes down to consider P(X=n) for each n.
For continuous cases, you can consider the intervals (-inf,a): P(X < a).
If you call F(x) = P(X < x) you have
P(x <= X < x+dx) = F(x+dx) - F(x) ~ F'(x)·dx = f(x)·dx.
where
f(x) = F'(x) is the density function.
PS:
And to reply to your question, the fact that the sum over all possible outcomes is 1 is not a rule, but a result. In fact, you start with
P(X in A) = 1
where A is the set of all outcomes and you can break it down using
P(X in (A union B)) = P(X in A) + P(X in B)
where A and B are disjoint.
For the discrete case, you can break it down to a sum over all individual outcomes. For the continuous case, you can't.
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| « Last Edit: Nov 17th, 2008, 6:04am by Grimbal » |
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