Comment author:Nick_Hay2
13 August 2007 09:55:16PM
16 points
[-]

Perhaps this formulation is nice:

0 = (P(H|E)-P(H))*P(E) + (P(H|~E)-P(H))*P(~E)

The expected change in probability is zero (for if you expected change you would have already changed).

Since P(E) and P(~E) are both positive, to maintain balance if P(H|E)-P(H) < 0 then P(H|~E)-P(H) > 0. If P(E) is large then P(~E) is small, so (P(H|~E)-P(H)) must be large to counteract (P(H|E)-P(H)) and maintain balance.

Comment author:potato
07 May 2012 11:42:33PM
*
5 points
[-]

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]

## Comments (77)

OldPerhaps this formulation is nice:

0 = (P(H|E)-P(H))*P(E) + (P(H|~E)-P(H))*P(~E)

The expected change in probability is zero (for if you expected change you would have already changed).

Since P(E) and P(~E) are both positive, to maintain balance if P(H|E)-P(H) < 0 then P(H|~E)-P(H) > 0. If P(E) is large then P(~E) is small, so (P(H|~E)-P(H)) must be large to counteract (P(H|E)-P(H)) and maintain balance.

*5 points [-]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]

*3 points [-]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.