On Philosophy

May 12, 2007

A Calculus For Knowledge

Filed under: Epistemology,Logic — Peter @ 12:00 am

Just as we have a calculus for probability that allows us to deduce the probability of a statement given other known probabilities we would like to have a knowledge calculus as well, one that tells us how confident we can be in any statement given our confidence about certain other statements. It would be nice if the probability calculus was the same as the knowledge calculus, since we have a handle on the probability calculus, but unfortunately that doesn’t seem possible, as I will argue. But let us assume for the moment that they are the same. Certainly the probability calculus has some features that we would want to see in a knowledge calculus. For example, the probability of a statement and its negation are complements of each other (P(S) = 1 – P(~S)). And this is easily understood in terms of knowledge, a probability of .5 means that we know nothing about a statement, a probability of 1 means that we are completely sure of it, and a probability of 0 means that we are sure of its opposite.

However, assigning meanings to the results is only one part of the showing that the probability calculus is an adequate knowledge calculus. We also want to ensure that certainty carries over in an acceptable way from premises to deductions. Let’s consider a case in which the correct answer is intuitively obvious. Suppose that we are certain that A implies B, or in other words, that P(A→B) = 1. Given this we would assume that, knowing no other facts, we could be of certain of B as we were of A. Or in other words that it would follow from this that P(A) = P(B), barring any other deductions that strengthen the case for B.

Thus we derive:
P(A→B) = P(~A∨B), because since A→B is equivalent to ~A∨B we can be as certain of one as we are of the other.
P(~A∨B) = P(~A) + P(B) – P(~A∧B), by the definition of the probability of disjunctions
P(~A) + P(B) – P(~A∧B) = P(~A) + P(A∧B) + P(~A∧B) – P(~A∧B) = P(~A) + P(A∧B), because P(B) = P(A∧B) + P(~A∧B)
P(~A) + P(A∧B) = P(~A) + P(B)*P(A|B), by the definition of the probability of conjunctions.
Thus we have P(A→B) = P(~A) + P(B)*P(A|B).
Assume that P(A) = b, and thus P(~A) = 1 – b. Now we can solve for P(B), and we get:
P(B) = b / P(A|B).
Unfortunately we don’t know what the value of P(A|B) is. But since neither P(B) nor P(A|B) can’t be greater than 1 we know that b ≤ P(A|B) ≤ 1.
If P(A|B) = 1 then the probability calculus does indeed give the right result. So let’s consider an arbitrary situation in which P(A→B) = 1, as illustrated below:

In the above situation there are n outcomes in which both A and B are false, o outcomes where A is false and B is true, and n outcomes where both A and B are true. As you can see P(A|B) is not necessarily equal to one. In fact it is only equal to 1 when o is equal to 0.

Now this isn’t completely unacceptable, because it is never the case that it will tell us that the probability B is less certain than that of A, which would be very strange. But it still isn’t the kind of calculus we want. We want our calculus to be able to give us a value for K(B) given only K(A→B) and K(A). And we certainly wouldn’t expect the value of K(B) to become less as K(A≡B) increases, but that is what the probability calculus would lead us to believe (we would expect it to be completely unaffected). And because of this we can conclude that any acceptable knowledge calculus cannot strictly track the probability calculus.

And simply reflecting on this example reveals an even deeper problem. Let us assume that we are curious as to how sure we are of the statement that A→B as a matter of natural law. Now the probability calculus says that if in exactly 75% of the cases we know of A→B holds then the probability of A→B is .75. But we were investigating the probability that the statement A→B is a natural law. And any known cases in which A→B does not hold should make us completely certain that it is not a natural law. But note that this does not make us completely sure that ~(A→B) is a natural law; perhaps in the majority of cases A→B does hold. Rather it makes us completely sure of the claim that “A→B is not always true”. This seems to indicate that any satisfactory knowledge calculus will be some extension to first order or modal logic, and not of sentential logic as the probability calculus is. But that is a possibility I will leave for another day.

Advertisements

6 Comments

  1. Peter,
    Have you made a calculus like this? Or are you just working on it? I’m curious what uses you think it would be best for. Also, I have come to believe that set theory (and so logic) is not strong enough for representing knowledge — that only a form of arithmetic will do. What are your insights on that?

    Bruce

    Comment by Bruce — May 12, 2007 @ 12:46 pm

  2. I assume you know that every arithmatic statment can be translated into a set theoretic statement, which set theory, plus a few additional axioms will prove if and only if it is an arithmatical truth. Or, in other words that arithmatic isn’t more powerful than set theory.

    Comment by Peter — May 12, 2007 @ 1:13 pm

  3. Yes, of course I know that. And Zermelo–Fraenkel (ZFC) set theory can be described by Logic plus “a few additional axioms.” It’s the “few additional axioms” that I believe (reasons not given here) are necessary.

    As an analogy, think of logic as assembly-language, set theory as the C language, and a knowledge-database as a software package. If you say that any knowledge-database-software can be see as equivalent to the output from a C compiler plus a little code, yes, that is true. But we still need that code.

    My point is that statements in logic are different in form from statements in logic+ZFC, and that statements in arithmetic are different in form from ZFC+number theory, and that the lower-level forms are not, by themselves, sufficient. But if you have to use number theory, why use a number-theory/logic hybrid? Why not just use number-theory?

    Actually, I didn’t have a point. I was just asking about your ideas on a knowledge calculus; whether you have a new idea you are going to share or are you just writing an article expressing already known ideas.

    Comment by Bruce — May 12, 2007 @ 2:10 pm

  4. I’m still working on the knowledge calculus itself, I have to sit down and work through some applications of Bernoulli’s theorem to get the right formulas, and that looks like it is going to have to wait until I have more free time. Just as a general rule of thumb: everything I write here is my original work unless I specifically say that I am summarizing or responding to the ideas of someone else.

    Comment by Peter — May 12, 2007 @ 2:19 pm

  5. Hmmm. How is Bernoulli’s Theorem relevant?

    Comment by Bruce — May 12, 2007 @ 3:54 pm

  6. Because in the more general case we will be asking how surely we are of a law such as “A->B holds 25% of the time”, with the universal quantifier being a degenerate case where we want to know how sure we are that A->B holds 100% of the time. The certainty will then logically be a function of how many samples you have collected and how many standard deviations they are from where you expect them to be.

    Comment by Peter — May 12, 2007 @ 4:12 pm


RSS feed for comments on this post.

Blog at WordPress.com.

%d bloggers like this: