Conditionalization
Extended version of article:
Introduction
Letβs now look at the equivalent of a conditional probability measure for our theory of epistemic probability based on the EPL Theorem. As one would expect, the product rule holds for arbitrary finite premises π and queries π΄ and π΅:
A small bit of algebra then gives the familiar identity for conditional probabilities,
when ππ³(π΄ | π) > 0.
In our theory the premise π plays the same role as a probability measure π does in the standard theory of probability, and the conjunction π΄ β§ π plays the same role as the conditional probability measure π( β | [π΄]), where [π΄] is the set of truth assignments on β³ satisfying π΄; therefore it is natural to define βπ conditional on π΄β to be the premise π΄ β§ π, with the requirement that ππ³(π΄ | π) > 0 so that π΄ β§ π is satisfiable.
To extend this to generalized premises π³ = (πα΅’) it seems obvious that one should define βπ³ conditional on π΄β to be the the sequence (π΄ β§ πα΅’), but there are some complications. First, it may be that ππ³(π΄ | πβ) = 0 for some indices π, in which case π΄ β§ π³ is not a valid generalized premise. If this occurs for only a finite number of premises πβ, we may solve the problem by simply discarding a sufficiently large (but finite) initial prefix of the sequence, since
for any query π΅ and finite π β β.
Second, even if ππ³(π΄ | πβ) > 0 for all π, and the sequence (ππ³(π΅ | π΄ β§ πα΅’)) converges for all queries π΅, so that π΄ β§ π³ is in fact a valid generalized premise, there is still a problem if ππ³(π΄ | π³) = 0 (which occurs if the sequence (ππ³(π΄ | πα΅’)) converges to 0.) In this case we can always construct an infinite number of generalized premises π΄ that are each topologically equivalent to π³ (yielding the same probabilities) and yet for which π΄ β§ π΄ is topologically distinct from π΄ β§ π³:
(The extended version of this article shows how to construct these premises π΄.)
Since we need our operations on generalized premises to preserve topological equivalence (and in fact be continuous), we therefore must restrict the domain of (π΄ β§ β ) to those π³ β π« for which ππ³(π΄ | π³) > 0.
Definitions
Now for the formal definitions.
Definition. π«β¨π΄β© is the set of generalized premises for which query π΄ has nonzero probability:
Remark. The set π«β¨ π΄β© is open because, for any π > 0, it is equal to the basic open set
Definition. If π³ = (πα΅’) is a generalized premise and π β β, then π³[π:] is the sequence (πα΅’) with the first π elements removed.
Remark. For any generalized premise π³ and π β β, π³[π:] is a generalized premise. Furthermore, ππ³(π΄ | π³[π:]) = ππ³(π΄ | π³) for every query π΄, and therefore π³ and π³[π:] are topologically equivalent. These follow from the fact that removing any finite number of elements from a convergent sequence leaves its convergence and its limits unchanged.
Definition. For any query π΄ and generalized premise π³ = (πα΅’) β π«β¨π΄β©,
That is, we replace each πβ with π΄ β§ πβ and drop initial elements of the resulting sequence until all that remain are satisfiable. Since π³ β π«β¨π΄β© only a finite number of elements will be discarded.
For symmetry we likewise define π³ β§ π΄ β π΄ β§ π³.
Results
First we prove that π³ β π«β¨π΄β© is a sufficient condition for π΄ β§ π³ to be a generalized premise, and to ensure that the familiar formula for conditional probabilities still holds for generalized premises.
Theorem. π΄ β§ π³ β π« for any query π΄ and generalized premise π³ β π«β¨π΄β©; furthermore, for every query π΅ we have
Proof. Let π³ = (πα΅’) and let π be the smallest nonnegative integer such that ππ³(π΄ | πβ) > 0 for all π β₯ π. Since ππ³(π΄ | π³) > 0 and division is a continuous function, we have for any query π΅ that
and since we get this convergence for any query π΅, the sequence (π΄ β§ π³) = (π΄ β§ πα΅’)[π:] defines a generalized premise, with
β
The other important property is that (π΄ β§ β ) is continuous on its domain (and hence preserves topological equivalence).
Theorem. For any query π΄, the function (π΄ β§ β ) : π«β¨π΄β© β π« is continuous.
Proof. Let (π³α΅’) be a sequence in π«β¨π΄β© converging to π΄ β π«β¨π΄β© . Then for any query π΅ we have
with the second line following from the continuity of ππ³(π΄ β§ π΅ | β ) and ππ³(π΄ | β ) and the fact that ππ³(π΄ | π΄) > 0. Therefore
and thus (π΄ β§ β ) is sequentially continuous, and hence (topologically) continuous. β