> In my last message I explained how a conditional probability P(R|Q) can be
> expressed equivalently as a belief function. The restriction of the
> Bayesian approach is due to the fact that in general, many different belief
> functions are possible over the product space of two variables RxQ. In
> contrast, a conditional probability forces the belief function to be of one
> specific form.
The fact that many different belief functions can correspond to any one
conditional probability (not surprising, given that belief functions use a
two-dimensional representation of belief) is an advantage only if you can show
me a problem that actually needs the extra distinctions afforded by belief
functions. I'm still waiting to hear from you what practical problem is solved
-- not mere aesthetic issues addressed -- by the use of belief functions, but
not by Bayesian probability theory.
> Another weakness is the requirement of prior probabilities. For example,
> let P(R|Q)=0.9 and P(R|¬Q)=0.8. In such a case, knowing that either Q or ¬Q
> is true, belief functions already allow to deduce Bel(R)=0.72 and
> Pl(R)=0.98
This example argues against belief functions. By simply sticking to probability
theory and not introducing any extraneous new notions, I obtain stronger bounds
on P(R) (0.8 <= P(R) <= 0.9) [1] than your theory can obtain. And I also obtain
these bounds without having to know what P(Q) is.
[1] See my post, "Deriving belief in R from P(R|Q) and P(R|~Q)."