This document fixes some mathematical errors in the previous, and gives intuitive derivations of NARS belief revision, induction, abduction, and deduction from probabilistic assumptions. The derivations are "intuitive" because the math turned out to be more complicated then I have time to do at the moment; I need to integrate over a couple of density functions if I want to exact formulas. Instead, I argue that the boundary conditions are correct for abduction and induction. For deduction, I not some specific differences in boundary conditions.
[Pei: The last sentence has typo.]

The basic definition: 

A=>B represents the probability distribution of A=>C or C=>B given B=>C or C=>A. 
[Pei: A=>B is a statement, not a distribution. Maybe you mean p(A=>B)? But that will be a number, while a “distribution” usually refers to a function.]

More formally, we can say the distribution is the average of p(A=>C | B=>C) and p(C=>B | C=>A), each with randomly chosen C. 
[Pei: The idea is meaningful.]

The probabilistic assumptions: 

1. For all A=>b and C, p(A=>C | B=>C) = p(C=>B | C=>A). This assumption is needed to justify ignoring which direction we are going, but it is not critical. If we drop it, we're just ignoring some information when we do our inferences (because we could be keeping track of evidence for those two distributions separately to get more exact estimates of them). 
[Pei: NARS does not assume this. Actually, in NAL-6, “extensional inheritance” and “intensional inheritance” can be clearly distinguished from each other, by using variable terms. In NAL-1, extensional evidence and intensional evidence are simply merged.]

2. All of these distributions are density functions of the form p(z) = y*(x^z * (1-x)^(1-z)) + (1-y)*(prior), where x is a parameter defining the distribution A=>B, y is the value of that parameter for the given (B=>C or C=>A), and z is the value of that parameter for the distribution being conditioned (A=>C or C=>B). The parameter ranges between 0 and 1, but it is not a probability; it is a way of describing a distribution. Let's call the parameter 'par', just to give it a name. 

3. The prior is even over values of par. 
[Pei: I need to think more about what the above assumptions means to NARS.]

The system is supposed to work with likelihoods rather than probabilities, reflecting the experienced-based semantics of NARS. (The even prior means this doesn't matter *too* much, but it will be more intuitive to explain it in terms of likelihoods I think.) 
[Pei: I still don’t see why likelihoods are more experience-based than probabilities. Conditioning probability and experience-based measurement are not the same, because what can be expressed as condition is quite limited. See http://nars.wang.googlepages.com/wang.bayesianism.pdf]

The likelihood of a statement given experience is the probability of experience given that statement. Since the probabilities are density functions over par, the likelihoods will also be density functions over par (except they aren't required to sum to 1). The nature of the likelihoods happens to lend itself to an easy representation: we can record the likelihood as a pair <f, w>, with f ranging from 0 to 1 and w from 0 to infinity. The following calculation defines the likelihood of par based on f and w: 

par^(w*f) * (1-par)^(w*(1-f)) 
[Pei: I need to think about what it means.]

This definition happens to coincide with the likelihood of heads-to-tails ratios given coin-flipping data. ('par' would be the probability of heads, 'f' the experimental frequency, and 'w' the number of trials.) That analogy makes it very tempting to call 'f' the frequency and 'w' the weight of evidence, as in NARS. But, keep in mind that there are no real coin flips going on. Confusion results from forgetting this, the end result of which is to think of 'par' as a probability. It is not. 
[Pei: This is fine.]

NARS revision

If we have calculated <f1,w1> for a statement based on some data, and <f2,w2> based on another independent data collection, then the situation is mathematically analogous to pooling together evidence from two different sessions of flipping the same coin. The likelihood of different weights of the coin can be calculated simply by multiplying the likelihoods that result for each dataset individually:

( par^(w1*f1) * (1-par)^(w1*(1-f1)) ) * ( par^(w2*f2) * (1-par)^(w2*(1-f2)) ) 

= par^(w1*f1+w2*f2) * (1-par)^(w1*(1-f1)+w2*(1-f2)) 

= par^(w1*f1+w2*f2) * (1-par)^((w1-w1*f1)+(w2-w2*f2))

= par^(w1*f1+w2*f2) * (1-par)^(w1+w2 - (w1*f1+w2*f2))

= par^((w1*f1+w2*f2)/(w1+w2) * (w1+w2)) * (1-par)^((1 - (w1*f1+w2*f2)/(w1+w2)) * (w1+w2))

That distribution can be represented by the pair <(w1*f1+w2*f2)/(w1+w2) , w1+w2>, matching NARS.
[Pei: No problem. As far as weight of evidence is additive, this is the only choice.]

NARS induction/abduction

If we have evidence <f1,w1> for A=>C and <f2,w2> for B=>C, and the two rely on independent data, then finding the likelihood of A=>B is analogous to finding the likelihood of a coin's bias when we only have limited evidence suggesting the result of one coin flip, and furthermore only have limited evidence suggesting that that coin was really the one we were interested in the first place. A desperate situation!

A=>B is our coin, A=>C is the coin flip, and the par-value of B=>C is the probability that A=>B was created by our coin. 
[Pei: I’m lost here. You mean “B=>C was created …”?]
The directness of this analogy can be seen by re-examining the distribution formula:

"All of these distributions are density functions of the form p(z) = y*(x^z * (1-x)^(1-z)) + (1-y)*(prior), where x is a parameter defining the distribution A=>B, y is the value of that parameter for the given (B=>C or C=>A), and z is the value of that parameter for the distribution being conditioned (A=>C or C=>B)."

With y=1, the formula is identical to the formula for the probability of heads or tails (z=1 or z=0) for a coin where p(heads)=x. If we take intermediate values of z, it is as if part of a coin could decide to be heads and part tails-- or if you prefer, as if a large number of coins were flipped, but the resulting fraction was only given the evidential weight of one coin. This is like the ideal situation, in which we know par[ac] perfectly. With limited evidence, the weight will turn out to be less then 1. This corresponds to the NARS calculation, in which the weight climbs to 1 as evidence for A=>C accumulates. (However, the exact math may not correspond.)

As y goes from 1 to 0, the distribution gets mixed with the prior distribution; it is as if somebody flipped our coin with P=y, but just flipped some random coin otherwise. This also corresponds to the NARS calculation, in which the evidence is considered relevant to A=>B in proportion to the 'f' value for C=>B. (Again, the exact math may not line up.)
[Pei: I need to think more about what is going on here.]

Of course, the case for concluding A=>B from C=>A and C=>B is essentially the same.

NARS deduction

Deduction is easy if we use the even prior to convert likelihoods to probabilities. If we have the par-values of A=>B and B=>C, then by definition A=>B gives us p(A=>C | B=>C). But we know B=>C, so we're done. Of course, we don't have A=>B and B=>C; we have Bayesian distributions on them. So, we expand A=>B to a distribution over possible distributions; we multiply this carefully by the distribution of possible B=>Cs; and then we integrate over the extra variables to collapse it to a distribution on A=>C. If we like, we can multiply by 1 and call it a likelihood again.

Now, does this have the same boundary conditions as the NARS equivalent? Not quite. The boundary conditions on the weight will definitely differ: in NARS, the weight is 0 if the weight of evidence for either A=>B or B=>C is zero. But probabilistically, we can use the prior in the lack of evidence to draw some conclusion about A=>C so long as we have some information about one of the two. The weight of evidence in NARS is also taken to be zero if both frequencies are zero; this does hold in the probabilistic version. What about the boundary conditions for frequency? NARS takes the conclusion frequency to be high if both evidence frequencies are high; this is obviously correct. If one evidence frequency is high and the other is low, NARS takes the resulting frequency to be low. I am not sure what the behavior is for the probabilistic version. Again, if both frequencies are low, NARS and probability agree that there is little evidence for the conclusion one way or the other.
[Pei: Need further consideration here.]

--Abram Demski