My development of a correspondence between NARS and probability (or,
more specifically, my attempt to embed NARS in probability theory) has
been stalled for a while, so I'm posting what I've come up with so
far. I had planned on doing this sooner, but other events
intervened... some of the following is just repetition (first 4
paragraphs), but the ordering makes the argument clear.
The basic idea is to treat NARS truth values as representations of a
statement's likelihood rather than its probability. The likelihood of
a statement given evidence is the probability of the evidence given
the statement. Unlike probabilities, calculating likelihoods does not
require prior beliefs; the likelihood of a statement is a direct
reflection of the evidence in favor of it. So, I thought likelihoods
were a good match for the experienced-based semantics of NARS.
The second decision was to model inheritance statements with
probability distributions over other inheritance statements;
specifically, A=>B is the conditional probability of A=>X given B=>X
(ie, something like the probability that A will inherit B's intension)
and also the conditional x=>B given X=>A (measuring B's inheritance of
A's extension). This seems to follow from the typical description of
NARS.
Third, I chose to have a single parameter determine this distribution,
ranging from 0 to 1. I simply called it 'par' before, although perhaps
'strength' or something would have been more descriptive...
Given those three assumptions, plus the NARS formula for revision,
there is (I think) only one possible formula relating the NARS
variables 'f' and 'w' to the value of 'par': the probability density
function p(par | w, f) = par^(w*f) * (1-par)^(w*(1-f)). Note: NARS
truth values are more often (I think?) represented by the pair 'f'
'c', where 'c' is computed from 'w' by the formula c=w/(w+k), where k
is a fixed constant. This is of little consequence at this point, and
it was more intuitive to use 'f' and 'w' (at least for me).
Now. What I did not realize before was that the tree constraints, plus
the desire to match the calculation for NARS induction/abduction,
totally determine the formula by which we calculate the probability
distribution that 'par' represents. Thus, the formula I previously
concocted is almost definitely incorrect, but we can obtain the
correct one by deducing it from the constraints.
Here's the math. In NARS, the operation we're interested in is taking
two pools of evidence, one concerning A=>X and the other concerning
B=>X, and combining them to calculate the evidence they lend to A=>B.
So probabilistically, we want to determine the probability of the
evidence for A=>X and B=>X given each possible 'par' value of A=>B.
Call the three parameters parab, parax, and parbx; call the evidence
variables fax, wax, fbx, wbx, fab, wab. We have:
p(fax=v1 & wax=v2 &fbx=v3 & wbx=v4 | parab=v5)
=integrate{p(fax=v1&wax=v2 | parax=v6)*p(fbx=v3&wbx=v4 |
parbx=v7)*p(parax=v6&parbx=v7 | parab=v5)}
The integration is from 0 to 1 over the two new variables that get
introduced when we seperate things out, v6 and v7 (the values of parax
and parbx). The separation works because we assume the two bodies of
evidence are independent (as in NARS) and we know that the evidence is
independent of parab given parax and parbx (because the evidence
variables are independent of everything else when the 'par' they are
associated with is known).
We can expand this somewhat further, because we know the probability
of the 'f' and 'w' variables given their corresponding 'par'
variables:
=integrate{[parax^(fax*wax) *
(1-parax)^((1-fax)*wax)]*[parbx^(fbx*wbx) *
(1-parbx)^((1-fbx)*wbx)]*p(parax=v6&parbx=v7 | parab=v5)}
[note: the way I've written everything, the second expression should
technically be filled with v1, v2, v3..., not fax, wax, fbx...; but it
is far easier to read this way.]
So, the task is to choose the right formula for
p(parax=v6&parbx=v7|parab=v5), so that the integration results in the
distribution dictated by the NARS formulas. Specifically, NARS says
fab=fax, and wab=(fbx*cax*cab + k)/(fbx*cax*cab). Note the
introduction of the 'c' variables, which can be expanded to their 'w'
counterparts as described above. This means that the formula is
dependent on 'k'. Using the formula for the likelihood of 'par' given
'f' and 'w', I get the following formula, representing the desired
result of the integration above:
p(fax=v1&wax=v2&fbx=v3&wbx=v4 | parab=v5)
=
(parab^(fax*((fbx*cax*cab +
k)/(fbx*cax*cab))))*((1-parab)^(fax*(1-(fbx*cax*cab +
k)/(fbx*cax*cab))))
[Again, I'm using the readable names when I technically should be
using the other ones.]
Taking the two together, I need to solve this equation:
integrate{[parax^(fax*wax) * (1-parax)^((1-fax)*wax)]*[parbx^(fbx*wbx)
* (1-parbx)^((1-fbx)*wbx)]*(???)}
=
(parab^(fax*((fbx*cax*cab +
k)/(fbx*cax*cab))))*((1-parab)^(fax*(1-(fbx*cax*cab +
k)/(fbx*cax*cab))))
Obviously, a bit complicated. It's even worse when the 'k' is made explicit :).
The ??? needs to be filled in in a way that makes the equation work.
Additionally, the ??? formula shouldn't contain the variables fax,
wax, fbx, or fwx. There is also the constraint that ??? should
integrate to 1 over the variables parax and parbx, to satisfy
probability theory.
This would give us the formula for p(parax=v6&parbx=v7 | parab=v5). At
that point I could make things easy on myself and give up the
assumption that A=>B represents a conditional (A=>X given B=>X), and
just take A=>B to represent the joint distribution over par values for
A=>X&B=>X. In this case the formula '???' is exactly the distribution
A=>B represents. Alternatively, I could reason as follows:
p(parax=v6&parbx=v7 | parab=v5)
=p(parax=v6|parbx=v7¶b=v5) * p(parax=v6&parbx=v7)/p(parax=v6)
The first factor is then the formula I want for A=>B; but to find it,
I would need to know the prior distributions p(parax=v6&parbx=v7) and
p(parax=v6)....
That's all for now! If anyone can solve that big equation, let me know!!
-------------------------------------------
agi
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