There is no canonical way to define a regular comparative probability in terms of a full conditional probability

I claim that there is no general, straightforward and satisfactory way to define a total comparative probability with the standard axioms using full conditional probabilities. By a “straightforward” way, I mean something like:

1. A ≲ B iff P(A−B|AΔB) ≤ P(B−A|AΔB)

or:

2. A ≲ B iff P(A|A∪B) ≤ P(A|A∪B) (Pruss).

The standard axioms of comparative probability are:

3. Transitivity, reflexivity and totality.

4. Non-negativity: ⌀ ≤ A for all A

5. Additivity: If A ∪ B is disjoint from C, then A ≲ B iff A ∪ C ≲ B ∪ C.

A “straightforward” definition is one where the right-hand-side is some expression involving conditional probabilities of events definable in a boolean way in terms of A and B.

To be “satisfactory”, I mean that it satisfies some plausible assumptions, and the one that I will specifically want is:

6. If P(A|C) < P(B|C) where A ∪ B ⊆ C, then A < B.

Definitions (1) and (2) are straightforward and satisfactory in the above-defined senses, but (1) does not satisfy transitivity while (2) does not satisfy the right-to-left direction of additivity.

Here is a proof of my claim. If the definition is straightforward, then if A ≲ B, and A′ and B′ are events such that there is a boolean algebra isomorphism ψ from the algebra of events generated by A and B to the algebra of events generated by A′ and B′ such that ψ(A) = A′, ψ(B) = B′ and P(C|D) = P(ψ(C)|ψ(D)) for all C and D in the algebra generated by A and B, then A′ ≲ B′.

Now consider a full conditional probability P on the interval [0,1] such that P(A|[0,1]) is equal to the Lebesgue measure of A when A is an interval. Let A = (0,1/4) and suppose B is either (1/4,1/2) or (1/4, 1/2]. Then there is an isomorphism ψ from the algebra generated by A and B to the same algebra that swaps A and B around and preserves all conditional probabilities. For the algebra consists of the eight possible unions of sets taken from among A, B and [0,1] − (A∪B), and it is easy to define a natural map between these eight sets that swaps A and B, and this will preserve all conditional probabilities. It follows from my definition of straightforwardness that we have A ≲ B if and only if we have B ≲ B. Since the totality axiom for comparative probabilities implies that either A ≲ B or B ≲ A, so we must have both A ≲ B and B ≲ A. Thus A ∼ B. Since this is true for both choices of B, we have

7. (0,1/4) ∼ (1/4,1/2) ∼ (1/4, 1/2].

But now note that ⌀ < {1/2} by (3) (just let A = ⌀, B = {1/2} and C = {1/2}). The additivity axiom then implies that (1/4,1/2) < (1/4, 1/2], a contradiction.

I think that if we want to define a probability comparison in terms of conditional probabilities, what we need to do is to weaken the axioms of comparative probabilities. My current best suggestion is to replace Additivity with this pair of axioms:

8. One-Sided Additivity: If A ∪ B is disjoint from C and A ≲ B, then A ∪ C ≲ B ∪ C.

9. Weak Parthood Principle: If A and B are disjoint, then A < A ∪ B or B < A ∪ B.

Definition (2) satisfies the axioms of comparable probabilities with this replacement.

Here is something else going for this. In this paper, I studied the possibility of defining non-classical probabilities (full conditional, hyperreal or comparative) that are invariant under a group G of transformations. Theorem 1 in the paper characterizes when there are full conditional probabilities that are strongly invariant. Interesting, we can now extend Theorem 1 to include this additional clause:

6. There is a transitive, reflexive and total relation ≲ satisfying (4), (8) and (9) as well as the regularity assumption that ⌀ < A whenever A is non-empty and that is invariant under G in the sense that gA ∼ A whenever both A and gA are subsets of Ω.

To see this, note that if there is are strongly invariant full conditional probabilities, then (2) will define ≲ in a way that satisfies (vi). For the converse, suppose (vi) is true. We show that condition (ii) of the original theorem is true, namely that there is no nonempty paradoxical subset. For to obtain a contradiction suppose there is a non-empty paradoxical subset E. Then E can be written as the disjoint union of A₁, ..., A_n, and there are g₁, ..., g_n in G and 1 ≤ m < n such that g₁A₁, ..., g_mA_m and g_m + 1A_m + 1, ..., g_nA_n are each a partition of E.

A standard result for additive comparative probabilities in Krantz et al.’s measurement book is that if B₁, ..., B_n are disjoint, and C₁, ..., C_n are disjoint, with B_i ≲ C_i for all i, then B₁ ∪ ... ∪ B_n ≲ C₁ ∪ ... ∪ C_n. One can check that the proof only uses One-Sided Additivity, so it holds in our case. It follows from G-invariance that A₁ ∪ ... ∪ A_m ∼ E ∼ A_m + 1 ∪ ... ∪ A_n. Since E is the disjoint union of A₁ ∪ ... ∪ A_m with A_m + 1 ∪ ... ∪ A_n, this violates the Weak Parthood Principle.