Hmmm. To mess around with equations a bit… what can we say about P(Bunyan | stories about Bunyan) and P(!Bunyan | stories about Bunyan), given P(stories about Bunyan | Bunyan) > P(stories about Bunyan | !Bunyan)?
Let’s genaralise it a bit (and reduce typing). What can we say about P(A|B) and P(!A|B) when P(B|A) > P(B|!A)?
...which means that either (1-P(A)) > P(A) or P(A|B) > P(!A|B), and quite possibly both; and whichever of these two inequalities is false (if either) the ratio between the two sides is closer than the inequality that is true.
To return to the original example; either P(Bunyan | stories about Bunyan) > P(!Bunyan | stories about Bunyan) OR P(!Bunyan) > P(Bunyan).
Also, if P(Bunyan | stories about Bunyan) > P(!Bunyan | stories about Bunyan) is false, then it must be true that P(Bunyan|stories about Bunyan) > P(Bunyan).
Hmmm. To mess around with equations a bit… what can we say about P(Bunyan | stories about Bunyan) and P(!Bunyan | stories about Bunyan), given P(stories about Bunyan | Bunyan) > P(stories about Bunyan | !Bunyan)?
Let’s genaralise it a bit (and reduce typing). What can we say about P(A|B) and P(!A|B) when P(B|A) > P(B|!A)?
Consider Bayes’ Theorem: P(A|B) = [(P(B|A)*P(A)]/P(B). Thus, P(B) = [(P(B|A)*P(A)]/P(A|B)
Therefore, P(!A|B) = [(P(B|!A)*P(!A)]/P(B)
Now, P(!A) = 1-P(A). So:
P(!A|B) = [(P(B|!A)*{1-P(A)}]/P(B)
Solve for P(B):
P(B) = [(P(B|!A)*{1-P(A)}]/P(!A|B)
Since P(B) = [(P(B|A)*P(A)]/P(A|B):
[(P(B|A)*P(A)]/P(A|B) = [(P(B|!A)*{1-P(A)}]/P(!A|B)
Since P(B|A) > P(B|!A)
[(P(B|A)*P(A)]/P(A|B) > [(P(B|!A)*P(A)]/P(A|B)
Therefore:
[(P(B|!A)*{1-P(A)}]/P(!A|B) > [(P(B|!A)*P(A)]/P(A|B)
Since probabilities cannot be negative:
[{1-P(A)}]/P(!A|B) > [P(A)]/P(A|B)
.[1-P(A)]*P(A|B) > [P(A)]*P(!A|B)
...which means that either (1-P(A)) > P(A) or P(A|B) > P(!A|B), and quite possibly both; and whichever of these two inequalities is false (if either) the ratio between the two sides is closer than the inequality that is true.
To return to the original example; either P(Bunyan | stories about Bunyan) > P(!Bunyan | stories about Bunyan) OR P(!Bunyan) > P(Bunyan).
Also, if P(Bunyan | stories about Bunyan) > P(!Bunyan | stories about Bunyan) is false, then it must be true that P(Bunyan|stories about Bunyan) > P(Bunyan).