More on full conditional probabilities and comparative probabilities

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(B|A∪B).

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.

Three invariance arguments

Suppose we have two infinite collections of items L_n and R_n indexed by integers n, and suppose we have a total preorder ≤ on all the items. Suppose further the following conditions hold for all n, m and k:

1. L_n > L_n − 1

2. R_n > R_n + 1

3. If L_n ≤ R_m, then L_n + k ≤ R_m + k.

Theorem: It follows that either L_n > R_m for all n and m, or R_n > L_m for all n and m.

(I prove this in a special case here, but the proof works for the general case.)

Here are three interesting applications. First, suppose that an integer X is fairly chosen. Let L_n be the event that X ≤ n and let R_n be the event that X ≥ n. Let our preorder be comparison of the probabilities of events: A ≤ B means that A is no less likely than B. Intuitively, it is less likely that X is less than n − 1 than that it is less than n, so we have (1), and similar reasoning gives (2). Claim (3) says that the relationship between L_n and R_m is the same as that between L_n + k ≤ R_m + k and that seems right, too.

So all the conditions seem satisfied, but the conclusion of the Theorem seems wrong. It just doesn’t seem right to think that all the left-ward events (X being less than or equal to something) are more likely than all the right-ward events (X being bigger than or equal to something), nor that it be the other way around.

I am inclined to conclude that countable infinite fair lotteries are impossible.

Second application. Suppose that for each integer n, a coin is tossed. Let L_n be the event that all the coins ..., n − 2, n − 1, n are heads. Let R_n be the event that all the coins n, n + 1, n + 2, ... are heads. Let ≤ compare probabilities in reverse: bigger is less likely. Again, the conditions (1)–(3) all sound right: it is less likely that ..., n − 2, n − 1, n are heads than that ..., n − 2, n − 1 are heads, and similarly for the right-ward events. But the conclusion of the theorem is clearly wrong here. The rightward all-heads events aren’t all more likely, nor all less likely, than the leftward ones.

I am inclined to conclude that all the L_n and R_n have equal probability (namely zero).

Third application. Supppose that there is an infinite line of people, all morally on par, standing on numbered positions one meter apart, with their lives endangered in the same way. Let L_n be the action of saving the lives of the people at positions ...., n − 2, n − 1, n and let R_n be the action of saving the lives of the people at positions n, n + 1, n + 2, .... Let ≤ measure moral worseness: A ≤ B means that B is at least as bad as A. Then intuitively we have (1) and (2): it is worse to save fewer people. Moreover, (3) is a plausible symmetry condition: if saving one group of people beats saving another group of people, shifting both groups by the same amount doesn’t change that comparison. But again the conclusion of the theorem is clearly wrong.

I am less clear on what to say. I think I want to deny the totality of ≤, allowing for cases of incommensurability of actions. In particular, I suspect that L_n and R_m will always be incommensurable.

More on benefiting infinitely many people

Once again let’s suppose that there are infinitely people on a line infinite in both directions, one meter apart, on positions numbered in meters. Suppose all the people are on par. Fix some benefit (e.g., saving a life or giving a cookie). Let L_n be the action of giving the benefit to all the people to the left of position n. Let R_n be the action of giving the benefit to all the people to the right of position n.

Write A ≤ B to mean that action B is at least as good as action A, and write A < B to mean that A ≤ B but not B ≤ A. If neither A ≤ B nor B ≤ A, then we say that A and B are noncomparable.

Consider these three conditions:

• Transitivity: If A ≤ B and B ≤ C, then A ≤ C for any actions A, B and C from among the {L_k} and the {R_k}.

• Strict monotonicity: L_n < L_n + 1 and R_n > R_n + 1 for all n.

• Weak translation invariance: If L_n ≤ R_m, then L_n + k ≤ R_m + k and if L_n ≥ R_m, then L_n + k ≥ R_m + k, for any n, m and k.

Theorem: If we have transitivity, strict monotonicity and weak translation invariance, then exactly one of the following three statements is true:

1. For all m and n, L_m and R_n are incomparable

2. For all m and n, L_m < R_n

3. For all m and n, L_m > R_n.

In other words, if any of the left-benefit actions is comparable with any of the right-benefit actions, there is an overwhelming moral skew whereby either all the left-benefit actions beat all the right-benefit actions or all the right-benefit actions beat all the left-benefit actions.

Proposition 1 in this paper is a special case of the above theorem, but the proof of the theorem proceeds in basically the same way. For a reductio, assume that (i) is false. Then either L_m ≥ R_n or L_m ≤ R_n for some m and n. First suppose that L_m ≥ R_n. Then the second and third paragraphs of the proof of Proposition 1 show that (iii) holds. Now suppose that L_m ≤ R_n. Let L_k^* = R_−k and R_k^* = L_−k. Say that A≤^*B iff A^* ≤ B^*. Then transitivity, strict monotonicity and weak translation invariance hold for ≤^*. Moreover, we have L_m ≤ R_n, so R_−m≤^*L_−n. Applying the previous case with  − m and  − n in place of n and m respectively we conclude that we always have L_j>^*R_k and hence that we always have L_j < R_k, i.e., (ii).

I suppose the most reasonable conclusion is that there is complete incomparability between the left- and right-benefit actions. But this seems implausible, too.

Again, I think the big conclusion is that human ethics has limits of applicability.

I hasten to add this. One might reasonably think—Ian suggested this in a recent comment—that decisions about benefiting or harming infinitely many people (at once) do not come up for humans. Well, that’s a little quick. To vary the Pascal’s Mugger situation, suppose a strange guy comes up to you on the street, and tells you that there are infinitely many people in a line drowning in a parallel universe, and asks you if you want him to save all the ones to the left of position 123 or all the ones to the right of position  − 11, because he can magically do either one, and nothing else, and he needs help in his moral dilemma. You are, of course, very dubious of what he is saying. Your credence that he is telling the truth is very, very small. But as any good Bayesian will tell you, it shouldn’t be zero. And now the decision you need to make is a real one.

Probability for truly fair infinite lotteries

Long ago, in correspondence with Plantinga and me, Peter van Inwagen suggested that the only way to model a countably infinite fair lottery is by assigning probability zero to every finite set of tickets, probability one to every co-finite set of tickets (a subset A of a set B is co-finite [relative to B] provided that the set of members of B that are not in A is finite), and an undefined probability to every other subset.

Van Inwagen’s proposal has been growing on me. In a truly fair lottery, the ordering of ticket numbers is irrelevant. Therefore, if Ω is the set of tickets and π is any permutation of Ω, the probability of A should be the same as that of πA, with each defined if and only if the other is. In other words the probability function should be permutation-invariant.

Proposition. There are only two finitely-additive real-valued probabilities invariant under all permutations of a countably infinite set Ω: the trivial probability that assigns 0 to the empty set, 1 to Ω and is undefined for all other subsets, and van Inwagen’s probability that assigns 0 to every finite set, 1 to every co-finite set and is undefined for all other subsets.

There is something very appealing about van Inwagen’s proposal: it’s the only finitely-additive real-valued probability that really captures the idea of a countably infinite fair lottery. I can't remember if van Inwagen had the above proposition in the correspondence, but he might have.

Proof of Proposition: For any two subsets X and Y that are neither finite nor co-finite, there is a permutation of Ω mapping X onto Y. Thus, by permutation invariance, either all sets that are neither finite nor co-finite have a probability or none do. Suppose first that all do. In that case, they all have equal probability. Let A be the evens, B be the odds, C the numbers equal to 0 modulo 4 and D the numbers equal to 2 modulo 4. They all must have equal probability. But A is the disjoint union of C and D, so by finite additivity, if all three have equal probability, all three must have probability zero. And so does B. Thus, P(Ω) = P(A) + P(B) = 0, a contradiction.

So, only sets that are neither finite nor co-finite have a probability. If the only subsets that have a probability are ⌀ and Ω, we are done. Suppose some other subset has a probability. If that subset is co-finite, its complement will have to have a probability too, so in either case there is a finite non-empty subset A that has a probability. Let A′ be a finite non-empty subset that has the same cardinality as A but intersects A in only one element. By permutation invariance, A′ has a probability. Thus, so does the intersection of A and A′. Hence, at least one singleton has a probability. Hence by permutation invariance all singletons have a probability. By finite additivity, that probability must be zero. It follows that all finite sets have probability, and that probability is zero, and all co-finite sets have probability, and that probability is one.

Remark 1: Suppose that we allow the probabilities to take values in some non-Archimedean ordered field. Then there are more possibilities. Specifically, for any positive infinitesimal α, we can define a probability that assigns to every finite set the probability nα where n is the set’s cardinality and to every co-finite set the probability 1 − nα where n is the cardinality of the set’s complement. And these are the only extra possibilities.

Remark 2: If we drop the countability condition on Ω, and assume the Axiom of Choice, then in the setting of the Proposition we can prove that P(A) is 0 or 1 for every subset A for which P(A) is defined.

Domination and uniform spinners

About a decade ago, I offered a counterexample to the following domination principle:

1. Given two wagers A and B, if in every state B is at least as good as A and in at least one state B is better than A, then one should choose B over A.

But perhaps (1) is not so compelling anyway. For it might be that it’s reasonable to completely ignore zero probability outcomes. If a uniform spinner is spun, and on A you get a dollar as long as the spinner doesn’t land at 90^∘ and on B you get a dollar no matter what, then (1) requires you to go for B, but it doesn’t seem crazy to say “It’s almost surely not going to land at 90^∘, so I’ll be indifferent between A and B.”

But now consider the following domination principle:

2. Given two wagers A and B, if in every state B is better than A, then one should choose B over A.

This seems way more reasonable. But here is a potential counterexaple. Consider a spinner which uniformly selects a point on the circumference of a circle. Assume x is any irrational number. Consider a function u such that u(z) is a real number for any z on the circumference of the circle. Imagine two wagers:

• A: After the spinner is spun and lands at z, you get u(z) units of utility

• B: After the spinner is spun, the spinner is moved exactly x degrees counterclockwise to yield a new landing point z′, and you get u(z′) units of utility.

Intuitively, it seems absurd to think that B could be preferable to A. But it turns out that given the Axiom of Choice, we can define a function u such that:

3. For any z on the circumference of the circle, if z′ is the result of rotating z by x degrees counterclockwise around the circle, then u(z′) > u(z).

And then if we take the states to be the initial landing points of the spinner, B always pays strictly better than A, and so by the domination principle (2), we should (seemingly absurdly) choose B.

Remarks:

• The proof of the existence of u requires the Axiom of Choice for collections of countable sets of reals). In my Infinity book, I argued that this version of the Axiom of Choice is true. However, arguments similar to those in the book’s Axiom of Choice chapter suggest that the causal finitist has a good way out of the paradox by denying the implementability of the function u.

• Some people don’t like unbounded utilities. But we can make sure that u is bounded if we want (if the original function u is not bounded, then replace u(z) by arctan u(z)).

• Of course the function u is Lebesgue non-measurable. To see this, replacing u by its arctangent if necessary, we may assume u is bounded. If u were measurable and bounded, it would be integrable, and its Lebesgue integral around the circle would be rotation invariant, which is impossible given (3).

It remains to prove the existence of u. Let ∼ be the relation for points on the (circumference of the circle) defined by z ∼ z′ if the angle between z and z′ is an integer multiple of x degrees. This is an equivalence relation, and hence it partitions the circle into equivalence classes. Let A be a choice set that contains exactly one element from each of the equivalence classes. For any z on the circle, let z₀ be the point in A such that z₀ ∼ z. Let u(z) be the (unique!) integer n such that rotating z₀ counterclockwise around the circle by an angle of nx degrees yields z. Then for any z, if z′ is the result of rotating z′ by x degrees around the circle, then u(z′) = u(z) + 1 > u(z) and so we have (3).

Symmetry, regularity and qualitative probability

Let ⪅ be a qualitative probability comparison for some collection F of subsets of a space Ω. Say that A ≈ B iff A ⪅ B and B ⪅ A, and that A < B provided that A ⪅ B but not B ⪅ A. Minimally suppose that ⪅ is a partial preorder (i.e., transitive and reflexive). Say it’s total provided that for all A and B either A ⪅ B or B ⪅ A. Suppose that G is a group of symmetries acting on Ω, and that F is G-invariant in the sense that gA ∈ F for all g ∈ G. Then we can define:

1. ⪅ is strongly G-invariant provided that for all A in G and all g in G we have A ≈ gA, and

2. ⪅ is weakly G-invariant provided that for all A and B in G and all g in G we have A ⪅ B iff gA ⪅ gB.

There is some reason to be suspicious of strong G-invariance. For in some interesting cases, say where Ω is a circle that G is the set of all rotations, there will be cases where gA is a proper subset of A, and by regularity we would expect to have gA < A rather than gA ≈ A. But weak G-invariance seems harder to question.

Say that g ∈ G is of order n provided that gⁿ = e, where e is identity. However, we also have:

Lemma 1. If ⪅ is total, g is of order 2, and A ⪅ B implies gA ⪅ gB for all A and B, then A ≈ gA for all A.

Proof: Since ⪅ is a total order, either A ⪅ gA or gA ⪅ A. Suppose A ⪅ gA. Then gA ⪅ g²A. But g²A = A. Hence gA ≈ A. Similarly, if gA ⪅ A, then g²A ≈ gA. But g²A = A, so gA ≈ A.

So, we have:

Proposition 1. If ⪅ is total and G is a group generated by elements of order 2, then weak G-invariance entails strong G-invariance.

Say that ⪅ is strongly regular provided that if A is a proper subset of B, then A < B. Weak regularity would say that if B is non-empty then ∅ < B. Weak regularity together with an appropriate additivity condition will imply strong regularity (details left to the reader).

Proposition 2. If G is generated by elements of order 2, and ⪅ is total and weakly G-invariant, then if there exists a g ∈ G and A ∈ F such that gA is a proper subset of A, then G is not strongly regular.

Proof: Strong regularity would require that gA < A, but that would contradict strong G-invariance which we have by Proposition 1.

Corollary 1. If F is a collection of subsets of the unit circle containing all countable sets and invariant under all reflections, and ⪅ is a total qualitative probability comparison weakly invariant under all reflections, then ⪅ is not strongly regular.

Proof: The group generated by all reflections includes all rotations. But there is a subset A of the circle and a rotation g such that gA is a proper subset of A. For instance, let A be the set of points at angles r, 2r, 3r, ... in degrees, where r is irrational, and let g be rotation by r. Then rA is the set of points at angles 2r, 3r, 4r, ... in degrees.

Now, imagine an infinite line on which there are infinitely many evenly spaced people, stretching out in both directions, each of whom flips a fair coin. Let Ω be the probability space describing these flips. Let H_n be the event that all the flips starting with person number n (i.e., n, n + 1, n + 2, ...) land heads. Suppose that F contains all the H_n and is invariant under all reflections of the situation (where we reflect the setup either about the point at which some person stands or at a point half-way between two neighboring people).

Corollary 2. If ⪅ is a total qualitative probability comparison weakly invariant under all reflections, then ⪅ is not strongly regular.

Proof: Let G be the group generated by the reflections. This group contains all translations. A non-trivial translation of H_n will either be a proper subset or a proper superset of H_n, depending on the direction of H_n. So by Proposition 2 we cannot have regularity.

Bibliographic note: Lemma 1 and Corollary 2 are analogous to Lemma 2 and Theorem 4 of this paper.

Morality and Relativity Theory

Einstein tells us that basic laws of physics should be invariant under change of reference frame. Is the same true of basic laws of morality? What would that mean? I think it would mean that any law of morality which is not invariant under change of reference frame can only be a consequence of a more general moral law that is invariant together with some conditions explaining why, contingently, things are arranged in some particular manner in space-time as to give rise to the non-invariant law. (Similarly, the non-invariant law about dropped objects near the earth moving in the direction of the center of the earth follows from an invariant Einsteinian law together with contingent facts about how matter is distributed in our vicinity.)

Could this abstract observation have any actual consequences? Suppose Georgina believes that when she works unowned land, by natural law the land becomes hers (cf. Locke), and by natural law she gains mineral rights to what is below the surface of the land she has worked. That doesn't seem right. What counts as being "below the surface of the land she has worked" depends on the reference frame. So it can't just be a basic moral law that one gets whatever unowned stuff is below where one worked. A story must be given explaining the lack of invariance. And probably the easiest way to do this is to say that if there is any such acquisition of mineral rights, it comes from a non-invariant positive law. This isn't very interesting, since I assume we knew that there is no natural acquisition of mineral rights.

There could, however, be some slightly more interesting consequences in other areas. For instance, in sexual ethics, it follows that considerations based on the shapes of organs, as well as ones based on inside-outside distinctions (what is in one reference frame a cup that is red on the outside, green on the inside, with juice within is in another reference frame a cup-shaped object that is green on the outside, red on the inside, with juice adhering to the outside due to odd gravitational fields), should not be of basic relevance, absent some further story. Instead, basic moral rules about sexuality should involve reference-frame invariant concepts such as contact, causation, teleology, intention, and consent. This is helpful—it focuses the philosopher's mind on what the morally relevant features of the activities are. (I've used this in a comment to argue that the use of condoms to prevent HIV transmission within a married couple is unacceptable within a Catholic sexual ethics.)