On Fri, Oct 10, 2008 at 4:24 PM, Ben Goertzel <[EMAIL PROTECTED]> wrote:
>> 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)).
>
> Why is this the only possible formula?
Let's see... let's call the function we're looking for L(f,w). To
satisfy NARS revision it must have the property L(f1,w1)*L(f2,w2)=L{
(w1*f1+w2*f2)/(w1+w2) , w1+w2 }. Taking f1=f2 and w1=w2, we have:
L(f,w)^2=L{ (2*w*f)/(2*w) , 2*w}
L(f,w)^2=L{ f, 2*w}
That establishes that the function is exponential in w, but that's a
far cry from proving the uniqueness of the formula I gave. I should
not have asserted so boldly...
--Abram
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agi
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