Conditional and Joint

[Baltazar]

you're correct, i wrote nonsense

p(p()) is nonsense as p is not an event but a number.



i need to think more before i post i guess

[doobs]

I guess the problem is that while p( A | B ) is defined as a conditional

probability, I don't know how the EVENT A | B is defined. Hence I do not know what to do with p( (A|B), C ).



Nevertheless this discussion was very helpful.

[Johnny]

The event A|B is defined as "Outcomes occurring in set A, given that the outcome is also in set B", i.e. it's the event that the outcome is in the intersection of A and B, as I wrote below. Just draw a Venn diagram to see it clearly. Then draw a Venn diagram to make sure you see the difference between events A|B and B|A.

[doobs]

Thank Johnny; both your first and second explanations helped a lot.

[functor]

Johnny, you said:

it's the event that the outcome is in the intersection of A and B



Are you saying P(A|B) = P(A \intersect B) ? if so, that is incorrect, as

P(A|B) = P(A \intersect B) / P(B) provided P(B) is non-zero.



There is no 'event' A|B it is not in general an element of your sigma-algebra.



Given P(B) != 0, you can always define a new measure on your probability space by defining Q(A) = P(A \intersect B) / P(B) for any event A, but the sigma algebra itself has not changed.



In my opinion p( (A|B), C) is meaningless, so there is no point in trying to figure out what the author meant.



To quote wikipedia:

http://en.wikipedia.org/wiki/Conditional_event_algebra



"only certain trivial Boolean algebras with very few elements contain, for any given A and B, an event X which satisfies P(X) = P(B|A)"

[Baltazar]

Ok i try to redeem myself



i think the only way it makes sense is if B and C are independant.

then you have C|B= C

so you can say that P((A|B),C)=P((A|B),C|B)=P((A,C)|B)



otherwise if C and B are not independant it is not possible to have both events (A knowing B happened) and (C not knowing anything about B) at the same time so their joint probability does not make sense.

[Johnny]

Are you saying P(A|B) = P(A \intersect B) ? if so, that is incorrect, as

P(A|B) = P(A \intersect B) / P(B) provided P(B) is non-zero.



Functor, I'm afraid you've got this wrong. Your statement would be correct if instead you wrote P(A|B) = P(A union B) / P(B) provided P(B) is non-zero. Do you agree?

[Randumb]

I couldn't find anything like this in any of my text books either.



Is this a question of, "What is the Probability of (A given B) and (C)."  If so, isn't the Probability of (D and C) = P(D)P(C).  And, if D=(A|B) and P(A|B)=P(A 'and' B)/P(B) , Then



P((A|B) 'and' C) = (P(A 'and' B) / P(B)) * P(C)

[functor]

Functor, I'm afraid you've got this wrong. Your statement would be correct if instead you wrote P(A|B) = P(A union B) / P(B) provided P(B) is non-zero. Do you agree?



No, I do not. For if A and B are disjoint then P(A union B) = P(A) + P(B) and so your expression would give a probability greater than one. Intersection reads as 'and', P(A|B) means the probability of A and B happening provided we know B happened, which is why you take the intersection.

[Johnny]

Functor: "Are you saying P(A|B) = P(A \intersect B) ? if so, that is incorrect"



Functor: "... which is why you take the intersection."



You seem to be contradicting yourself.