Probability Axioms: Deriving Version 2 from Version 1

AI Thread Summary
The discussion revolves around deriving axiom 4 of Version 2 of probability axioms from Version 1. The user initially struggles with the transition, particularly in expressing intersections using the axioms that primarily address unions. After exploring different approaches, they successfully partition the union of events A and B into distinct components, allowing them to manipulate the terms effectively. The key breakthrough involves recognizing that P(AnB) can be expressed in a way that simplifies the calculation, ultimately leading to the correct formulation. The conversation highlights the interconnectedness of the two versions of probability axioms and the importance of understanding unions and intersections in probability theory.
eddo
Messages
48
Reaction score
0
In my probability class we were given two versions of probability axioms which are equivalent. Let S be the sure event, A and B any arbitrary events, I the impossible event. I will use u to denote union, and n to denote intersection:

Version 1
1. P(S)=1
2. P(A)>=0
3. If AnB=I, than P(AuB)=P(A)+P(B)

Version 2
1. P(S)=1
2. P(A)>=0
3. P(I)=0
4. P(AuB)=P(A)+P(B)-P(AnB)

It is easy to go from version 2 to version 1, and I can see how to show that P(I)=0 using version 1. The problem I'm having is how to derive axiom 4 of version 2 from version 1. I know I will have to consider two events made up of unions and intersections of A and B, to which I can apply axiom 3 of version 1, but I can't quite figure out how to do it. Thanks for any help.
 
Physics news on Phys.org
After some trial and error I got this: partition A u B into A n ~B, A n B, and B n ~A. If you work that out you can get the right expression; one of the terms you'll get is P(A n B), and there is a way to go from P(A n B) to -P(A n B) and the other terms will fall into place.
 
I tried this but wasn't able to work it out. How do you go from P(AnB) to -P(AnB)? Once I get to the second step I'm not sure how to write the two terms other than P(AnB) in any other way, since they involve intersections, and the axioms version 1 don't say anything about intersections. You can rewrite the other two terms (before you make them into 2 separate terms) as (An~B)u(Bn~A)=(AuB)n(~Bu~A), but once again this involves intersections, not unions which doesn't help much.
 
Nevermind I got it. You just have to turn P(AnB) into 2P(AnB)-P(AnB), then group one of each of the positive terms with your other two terms. Axiom 3 of version 1 can than be used in reverse to turn each of these into a single probability, where the events in question simplify, using boolean algebra, to A and B. Thanks for the help Bicycle Tree.
 
I was reading documentation about the soundness and completeness of logic formal systems. Consider the following $$\vdash_S \phi$$ where ##S## is the proof-system making part the formal system and ##\phi## is a wff (well formed formula) of the formal language. Note the blank on left of the turnstile symbol ##\vdash_S##, as far as I can tell it actually represents the empty set. So what does it mean ? I guess it actually means ##\phi## is a theorem of the formal system, i.e. there is a...
Back
Top