Theorem: If A is evidence of B, then B is also evidence of A.
Proof: To say that A is evidence of B means that P(A|B) > P(A|~B), or in other words that
P(A&B)/P(B) > P(A&~B)/P(~B), which we may write as P(A&B)/P(B) > (P(A)-P(A&B))/(1-P(B)). Algebraic manipulation turns this into P(A&B) > P(A)P(B), which is symmetric in A and B; hence we can undo the manipulations with the roles of A and B reversed to arrive back at P(B|A) > P(B|~A). QED.
Hence, if A implies B, then B also implies A!
Now of course, the strengths of these implications might be vastly different. But that’s a separate matter.
Here, the point is that A implies B with near certainty (where A is “K&S faked burglary” and B is “K&S killed Kercher”); I’m not terribly concerned with how strongly B implies A. I don’t need for B to imply A very strongly to make my point, but Massei and Cristiani would definitely need that in order to enable any charitable reading of their burglary section at all.
Theorem: If A is evidence of B, then B is also evidence of A.
Proof: To say that A is evidence of B means that P(A|B) > P(A|~B), or in other words that P(A&B)/P(B) > P(A&~B)/P(~B), which we may write as P(A&B)/P(B) > (P(A)-P(A&B))/(1-P(B)). Algebraic manipulation turns this into P(A&B) > P(A)P(B), which is symmetric in A and B; hence we can undo the manipulations with the roles of A and B reversed to arrive back at P(B|A) > P(B|~A). QED.
Hence, if A implies B, then B also implies A!
Now of course, the strengths of these implications might be vastly different. But that’s a separate matter.
Here, the point is that A implies B with near certainty (where A is “K&S faked burglary” and B is “K&S killed Kercher”); I’m not terribly concerned with how strongly B implies A. I don’t need for B to imply A very strongly to make my point, but Massei and Cristiani would definitely need that in order to enable any charitable reading of their burglary section at all.