Hey, sorry if it’s mad trivial, but may I ask for a derivation of this? You can start with “P(H) = P(H|E)P(E) + P(H|~E)P(~E)” if that makes it shorter.
(edit):
Never mind, I just did it. I’ll post it for you in case anyone else wonders.
1} P(H) = P(H|E)P(E) + P(H|~E)P(~E) [CEE] 2} P(H)P(E) + P(H)P(~E) = P(H|E)P(E) + P(H|~E)P(~E) [because ab + (1-a)b = b] 3} (P(H) - P(H))P(E) + (P(H) - P(H))P(~E) = (P(H|E) - P(H))P(E) + (P(H|~E) - P(H))P(~E) [subtract P(H) from every value to be weighted] 4} (P(H) - P(H))P(E) + (P(H) - P(H))P(~E) = P(H) - P(H) = 0 [because ab + (1-a)b = b] (conclusion) 5} 0 = (P(H|E) - P(H))P(E) + (P(H|~E) - P(H))P(~E) [by identity syllogism from lines 3 and 4]
Hey, sorry if it’s mad trivial, but may I ask for a derivation of this? You can start with “P(H) = P(H|E)P(E) + P(H|~E)P(~E)” if that makes it shorter.
(edit):
Never mind, I just did it. I’ll post it for you in case anyone else wonders.
1} P(H) = P(H|E)P(E) + P(H|~E)P(~E) [CEE]
2} P(H)P(E) + P(H)P(~E) = P(H|E)P(E) + P(H|~E)P(~E) [because ab + (1-a)b = b]
3} (P(H) - P(H))P(E) + (P(H) - P(H))P(~E) = (P(H|E) - P(H))P(E) + (P(H|~E) - P(H))P(~E) [subtract P(H) from every value to be weighted]
4} (P(H) - P(H))P(E) + (P(H) - P(H))P(~E) = P(H) - P(H) = 0 [because ab + (1-a)b = b]
(conclusion)
5} 0 = (P(H|E) - P(H))P(E) + (P(H|~E) - P(H))P(~E) [by identity syllogism from lines 3 and 4]
P(H) = P(H|E)P(E) + P(H|~E)P(~E)
P(H)*(P(E)+P(~E))=P(H|E)P(E) + P(H|~E)P(~E)
P(H)P(E)+P(H)P(~E)=P(H|E)P(E) + P(H|~E)P(~E)
P(H)P(~E)=(P(H|E)-P(H))*P(E) + P(H|~E)P(~E)
0=(P(H|E)-P(H))*P(E) + (P(H|~E)-P(H))*P(~E)
The trick is that P(E)+P(~E)=1, and so you can multiply the left side by the sum and the right side by 1.