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komponisto comments on Inherited Improbabilities: Transferring the Burden of Proof - Less Wrong
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Comments (58)
I feel you can demonstrate quite amply that A is not sufficient proof of B, and that A=>B has not been sufficiently proven either.
However, neither of these assertions seems to be your point. You seem to be insisting that you can't prove A, and I see absolutely no evidence of that, unless you take as given the assumption A=>B. I would certainly challenge that assumption.
Am I mistaken in this understanding of your point?
P.S. I feel the evidence suggets Knox is guilty at around a 10% chance, based solely on the evidence in this post. I do not feel a 10% chance of guilt is sufficient. I have not considered any evidence outside this post, as my interest is in the probability math, and not in the actual case itself.
P.P.S. A discussion of the dangers of cognitive biases is, I feel, entirely orthogonal to a discussion on probabilities and mathematics. Given my interest is in the math, not the case, I am going to skip over discussion of such biases.
So you don't agree that if Knox and Sollecito faked the burglary, then they are likely guilty of murder?
There isn't much evidence presented in this post -- hardly any at all. (Plenty of information is linked to, of course...)
Well, then I must say you're on the wrong website!
But if your interest is more in the math than in the case, I'm not sure what you're disagreeing with me about. It's kind of hard to dispute the inequality
isn't it?
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:
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.
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.