Your post is entangling three separate issues, and I think that’s making it confusing to discuss (it was certainly confusing to read!)
Mathematics:“P(A) ⇐ P(B) / P(B|A).”
No argument here.
Probability:How does the evidence A impact the probability of conclusion B?
I feel you are using entirely incorrect math for the situation, as stated in my previous posts. Just because the formula is correct, does not mean it is applicable to the problem you are trying to solve.
If A is proven, and A=>B is proven, then B is proven. The prior probability of B cannot negate the proof of A, nor the proof of A=>B, and thus has absolutely no bearing on the situation. Prior probability matters if, and only if, we are discussing p(A) and p(A=>B), at which point we still have new evidence (A, A=>B) that requires us to update to a new new probability of B.
You cannot continue to assert the prior probability of B, despite new evidence that suggests a higher or lower chance of B.
Cognitive Bias:Is the judge properly evaluating p(A) and p(A=>B)?
I feel that there is insufficient information to draw a firm conclusion here. However, based on what you have said, I feel rather strongly that you have misinterpreted his evaluations, because you are assuming that common language and logical language are the same.
If A is proven, and A=>B is proven, then B is proven
Agreed.
The prior probability of B cannot negate the proof of A, nor the proof of A=>B, and thus has absolutely no bearing on the situation
This sentence doesn’t make sense as written. I don’t know what it means for a probability to “negate” a proof, and so I don’t know what you’re trying to say when you assert that this can’t happen.
My best guess is that you’re trying to say that “even if P(A) is small on account of P(B) being small, some finite amount of evidence will still suffice to prove A, and therefore B.” Which is obviously true, and nothing I have written says otherwise.
You cannot continue to assert the prior probability of B, despite new evidence that suggests a higher or lower chance of B.
This sounds like our previous discussion, where you said, and I agreed, that other evidence that Knox and Sollecito killed Kercher could raise the probability of their having faked the burglary. I’ve never disputed this, but have pointed out that this isn’t Massei and Cristiani’s reasoning. They attempted to prove the fake burglary without invoking the other murder evidence.
However, based on what you have said, I feel rather strongly that you have misinterpreted his evaluations, because you are assuming that common language and logical language are the same.
Ahhh, you make so much more sense when you phrase it this way!
“other evidence that Knox and Sollecito killed Kercher could raise the probability of their having faked the burglary”
But my point is, this is backwards. It only works if you assume with near-100% certainty that faking the burglary and being the murderer are correlated. Otherwise “faked the burglary” IS simply evidence that Knox is the murderer.
If we prove that Knox killed Kercher, it proves that any 100% correlation is true. It does NOT prove any less-than-100% correlation. It’s even entirely possible for a correlation to be one-directional (A implies B, but B does not imply A).
Thus, Knox killed Kercher is only proof of a faked burglary if you already assume the correlation is proven and two-directional.
I’m just trying to understand your point a bit better. Hopefully you don’t mind the late reply (I’ve been on vacation for a while)
“In probability, “correlations” are always bidirectional.”
Can’t there be three separate, equally valid points which, if proven, would prove she was the murderer? Even if those three equally valid proofs of her guilt are contradictory? Once we know she is guilty, they can’t all three be true, can they?
I’m not sure how one would accurately express this, given what you’re saying. The probability that A implies Guilt, B implies Guilt, and C implies Guilt can all be 100%, yes? Obviously, the probability that guilt implies all of A+B+C is 0%, since they are contradictory. Therefor, how can it be correct to assume the opposite correlation, that Guilt implies A at 100% certainty?
In general it is not true that P(A|B) = P(B|A). P(A|Guilt) depends on the prior probabilities of A and Guilt, as well as P(Guilt|A). For example, say we have four possible proofs A, B, C, D, and P(Guilt|A or B or C) = 1, and P(Guilt|D) = 0. Our prior is all four are equally likely: P(A) = P(B) = P(C) = P(D) = 0.25. P(Guilt) is then 0.75 = P(Guilt|A)P(A) + P(Guilt|B)P(B)...
P(A|Guilt) isn’t 1. But it’s 33%, which is still higher than the prior %25: that is, Guilt is evidence for A.
By the way I think it might help if you avoid talking in proofs and implication and 100% certainty. In hypothetical examples it’s useful to set things to P(X) = 1, but in the real world evidence is always probabilistic; nothing’s ever 100%.
Ahhh, that helps clear things up. For some reason I’d been understanding you as saying that, given P(Guilt|A) = 1, P(A|Guilt) was also 1. It looks like what you meant was just that Guilt is evidence for, but not necessarily 100% proof of, A. Am I getting that all correct?
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.
Your post is entangling three separate issues, and I think that’s making it confusing to discuss (it was certainly confusing to read!)
Mathematics: “P(A) ⇐ P(B) / P(B|A).”
No argument here.
Probability: How does the evidence A impact the probability of conclusion B?
I feel you are using entirely incorrect math for the situation, as stated in my previous posts. Just because the formula is correct, does not mean it is applicable to the problem you are trying to solve.
If A is proven, and A=>B is proven, then B is proven. The prior probability of B cannot negate the proof of A, nor the proof of A=>B, and thus has absolutely no bearing on the situation. Prior probability matters if, and only if, we are discussing p(A) and p(A=>B), at which point we still have new evidence (A, A=>B) that requires us to update to a new new probability of B.
You cannot continue to assert the prior probability of B, despite new evidence that suggests a higher or lower chance of B.
Cognitive Bias: Is the judge properly evaluating p(A) and p(A=>B)?
I feel that there is insufficient information to draw a firm conclusion here. However, based on what you have said, I feel rather strongly that you have misinterpreted his evaluations, because you are assuming that common language and logical language are the same.
Agreed.
This sentence doesn’t make sense as written. I don’t know what it means for a probability to “negate” a proof, and so I don’t know what you’re trying to say when you assert that this can’t happen.
My best guess is that you’re trying to say that “even if P(A) is small on account of P(B) being small, some finite amount of evidence will still suffice to prove A, and therefore B.” Which is obviously true, and nothing I have written says otherwise.
This sounds like our previous discussion, where you said, and I agreed, that other evidence that Knox and Sollecito killed Kercher could raise the probability of their having faked the burglary. I’ve never disputed this, but have pointed out that this isn’t Massei and Cristiani’s reasoning. They attempted to prove the fake burglary without invoking the other murder evidence.
You’ll have to be more specific here.
Ahhh, you make so much more sense when you phrase it this way!
“other evidence that Knox and Sollecito killed Kercher could raise the probability of their having faked the burglary”
But my point is, this is backwards. It only works if you assume with near-100% certainty that faking the burglary and being the murderer are correlated. Otherwise “faked the burglary” IS simply evidence that Knox is the murderer.
If we prove that Knox killed Kercher, it proves that any 100% correlation is true. It does NOT prove any less-than-100% correlation. It’s even entirely possible for a correlation to be one-directional (A implies B, but B does not imply A).
Thus, Knox killed Kercher is only proof of a faked burglary if you already assume the correlation is proven and two-directional.
In probability, “correlations” are always bidirectional. Bayes theorem:
P(A|B =\frac{P(B|A)P(A)}{P(B)})
If P(B|A) > P(B), then P(A|B) > P(A). By the same factor even:
The analogy to biconditionality in deductive logic would be P(A|B)= P(B|A) which obviously isn’t always true.
I’m just trying to understand your point a bit better. Hopefully you don’t mind the late reply (I’ve been on vacation for a while)
“In probability, “correlations” are always bidirectional.”
Can’t there be three separate, equally valid points which, if proven, would prove she was the murderer? Even if those three equally valid proofs of her guilt are contradictory? Once we know she is guilty, they can’t all three be true, can they?
I’m not sure how one would accurately express this, given what you’re saying. The probability that A implies Guilt, B implies Guilt, and C implies Guilt can all be 100%, yes? Obviously, the probability that guilt implies all of A+B+C is 0%, since they are contradictory. Therefor, how can it be correct to assume the opposite correlation, that Guilt implies A at 100% certainty?
It isn’t!
In general it is not true that P(A|B) = P(B|A). P(A|Guilt) depends on the prior probabilities of A and Guilt, as well as P(Guilt|A). For example, say we have four possible proofs A, B, C, D, and P(Guilt|A or B or C) = 1, and P(Guilt|D) = 0. Our prior is all four are equally likely: P(A) = P(B) = P(C) = P(D) = 0.25. P(Guilt) is then 0.75 = P(Guilt|A)P(A) + P(Guilt|B)P(B)...
Given this, we have:
P(A|Guilt) isn’t 1. But it’s 33%, which is still higher than the prior %25: that is, Guilt is evidence for A.
By the way I think it might help if you avoid talking in proofs and implication and 100% certainty. In hypothetical examples it’s useful to set things to P(X) = 1, but in the real world evidence is always probabilistic; nothing’s ever 100%.
Ahhh, that helps clear things up. For some reason I’d been understanding you as saying that, given P(Guilt|A) = 1, P(A|Guilt) was also 1. It looks like what you meant was just that Guilt is evidence for, but not necessarily 100% proof of, A. Am I getting that all correct?
Yes.
P(Guilt|A) = P(A|Guilt) only when P(A) = P(Guilt). In which case it would be 100% proof. But that is a rare situation.
Nitpick: the two conditionals also be equal if A and Guilt were mutually exclusive. (in that case, of course, the two conditionals would be both zero)
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.