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Sum and product rule of measures 
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#1
Feb2212, 11:56 AM

P: 982

I'm wondering what requirements must exist for a measure, m, to have the following properties:
m(Sum(P_i))= Sum(m(P_i)) the sum rule m(Prod(P_i))=Prod(m(P_i)) the product rule Where P_i are underlying sets or single propositions. Thank you. 


#2
Feb2212, 12:34 PM

Mentor
P: 18,293

So P_{i} are sets??
Then how did you define Sum P_{i} and Prod P_{i}?? 


#3
Feb2212, 01:02 PM

P: 982

For example, as I understand measure theory, in order that m(P1 union P2)=m(P1)+m(P2), there must be the restriction that P1 intersect P2 = empty set. In other words, P1 and P2 must be disjoint. In all the books I've looked at, this seems to be given as an axiom that's not proven. Yet, I wonder if there is a similar or perhaps dual requirement for Prod? Also, my goal is to be able to somehow put a measure on the space of propositions so that disjunction and conjunction get translated to addition and multiplication of measures on propositions or on sets of propositions. Can one get from the more traditional treatments of measures on sets to getting measures on propositions by letting the number of elements in a set go to 1 element? Then propositions could be labeled synonymously with its set. Would this turn unions and intersections into disjunction and conjunction? Any help would be very much appreciated. 


#4
Feb2212, 03:55 PM

Sci Advisor
P: 6,070

Sum and product rule of measures
For product, you need to look at probability theory, where the concept of independence comes in. P(A∩B∩C) = P(A)P(B)P(C) if A,B,C are independent.



#5
Feb2212, 08:55 PM

P: 65

... I'm not sure that is quite applicable.
Anyway, that's how a measure is defined. And then we prove theorems about things that satisfy that property (hopefully). 


#6
Feb2312, 10:38 AM

P: 982

But if this can be made to be consistent with adding probabilities, then I wonder if this multiplication can be made more general for other kinds of measures. Or is there something in the normalization procedure that gives us this multiplication of probabilities for independent events. 


#7
Feb2312, 03:52 PM

Sci Advisor
P: 6,070

I don't know of anything other than the obvious. m(A∪B) + m(A∩B) = m(A) + m(B).



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