Well, it wasn’t actually an equation. That’s why I used the =||= symbol. It was a bientailment. It asserts logical equivalence (in classical logic), and it means something slightly different than an equals symbol. The equation with the plus signs and the logical equivalence shouldn’t be confused.
I calculated the result for about three different sets of probabilities before making the original post. The equation was correct each time. I could have just been mistaken, of course, but even Zack (the commenter above) conceded that the equation is true.
EDIT: Oh, I see now. You have changed all my disjunctions into conjunctions. Why?
Ah, I’m sorry. Do you agree that the equation I quoted above is incorrect, though? I’m going to have to leave now, but the relevant basic equations of probability are P(X v Y) = P(X) + P(Y) - P(X&Y), and P(X&Y) = P(X) * P(X | Y)
Well, it wasn’t actually an equation. That’s why I used the =||= symbol. It was a bientailment. It asserts logical equivalence (in classical logic), and it means something slightly different than an equals symbol. The equation with the plus signs and the logical equivalence shouldn’t be confused.
I’m back and there’s been no response, so I’ll be specific. Starting from
Using p(X v Y) = p(X) + p(Y) - p(XY), we get
.15 = p(A|B) + p(B|B) - p(AB|B) - p(A) - p(B) + p(AB) + p(A|B) + p(~B|B) - p(A~B|B) - p(A) - p(~B) + p(A~B)
= p(A|B) + 1 - p(A) - p(A) - p(B) + p(AB) + p(A|B) - p(A) - p(~B) + p(A~B)
= 2 p(A|B) − 2 p(A) = twice the thing you started from, which is bad.
It looks like you simplified p(AB|B) as p(A), but in fact p(AB|B)=p(ABB)/p(B)=p(AB)/p(B)=p(A|B). (I made a similar mistake earlier.)
I get p(A|B) + p(B|B) - p(AB|B) - p(A) - p(B) + p(AB) + p(A|B) + p(~B|B) - p(A~B|B) - p(A) - p(~B) + p(A~B)
= p(A|B) + 1 - p(A|B) - p(A) - p(B) + p(AB) + p(A|B) + 0 − 0 - p(A) − 1 + p(B) + p(A~B)
= - p(A) - p(B) + p(AB) + p(A|B) - p(A) + p(B) + p(A~B)
= - p(A) + p(AB) + p(A|B) - p(A) + p(A~B)
= p(A|B) −2p(A) +p(AB) + p(A~B)
= p(A|B) −2p(A) + p(A)
= p(A|B) - p(A)
But this is quod erat demonstrandum.
Okay. Yay, a use for the retract button where it’s still good to have the text visible!
Manfred,
I calculated the result for about three different sets of probabilities before making the original post. The equation was correct each time. I could have just been mistaken, of course, but even Zack (the commenter above) conceded that the equation is true.
EDIT: Oh, I see now. You have changed all my disjunctions into conjunctions. Why?
Ah, I’m sorry. Do you agree that the equation I quoted above is incorrect, though? I’m going to have to leave now, but the relevant basic equations of probability are P(X v Y) = P(X) + P(Y) - P(X&Y), and P(X&Y) = P(X) * P(X | Y)