Good's Theorem, perfect rationality, and conditioning on zero probability events

Recently, I found myself puzzled by the difficulty in applying “classical” evidential decision theory to a perfectly rational agent. The problem was that the rational agent decides whether to do A or B based on a comparison between the conditional expectations E(U|A) and E(U|B) of the utility function U. But supposing that in fact E(U|A) > E(U|B), the perfectly rational agent has no chance of doing B, so P(B) = 0, and hence E(U|B) is undefined.

But then I thought this isn’t a big deal, because we aren’t perfectly rational agents, so we always have a chance of screwing up and hence P(B) > 0 even if E(U|B) is much less than E(U|A).

I am not entirely satisfied with this. After all, you might think: “I may be pretty imperfect, but if I am choosing between a donut D and a year of torture T, I have zero chance of choosing the year of torture. But then E(U|T) is undefined, so how am I being rational in this choice? Maybe that’s a good objection, maybe not.

But here is another reason why the “We’re imperfect” solution isn’t completely ideal. We want to say that Good’s Theorem tells us something important about rationality—namely, that more information makes rational agents make better decisions. Good’s Theorem is usually interpreted as saying that under some independence conditions, the expected value of a perfectly rational choice given more information is no less than that of a perfectly rational choice given less information. Notice that this is obviously false in the case of an imperfectly rational agent. Thus, we have to make sense of “What a perfectly rational agent would choose” to make sense of the standard interpretation of Good’s Theorem. Moreover, in the setting of Good’s Theorem, the perfectly rational agent has to be choosing based on expected utilities—and that’s precisely what generates the zero-probability-conditioning problem.

Now, the Theorem is still true as an abstract bit of mathematics. But the application is difficult if we can’t make sense of a perfectly rational agent who is certain to maximize expected utility.

Likely we can extend Good’s Theorem to talk about the limiting case of imperfect agents getting more and more perfect. But it would be nice if we didn’t have to.

A problem with perfectly rational agents and decision theory

Suppose I am perfectly rational in the decision theoretic sense. A coin is about to be tossed, and I will get five dollars on heads (H) and one dollar on tails (T). I have a choice whether to leave the coin fair (F) or load it (L) in favor of tails so that the probability of tails is 3/4.

It is obvious what I do. I calculate the expected utilities of my options F and L as follows.

• EU(F) = P(H|F) ⋅ $5 + P(T|F) ⋅ $1 = (1/2) ⋅ $5 + (1/2) ⋅ $1 = $3

and

• EU(L) = P(H|L) ⋅ $5 + P(T|L) ⋅ $1 = (1/4) ⋅ $5 + (3/4) ⋅ $1 = $2.

And then I choose F.

Except it’s not so simple. For I am perfectly rational. But since, as we just saw, the perfectly rational agent has to choose F, it follows that P(L) = 0, and so P(H|L) and P(T|L) are undefined. So I can’t decide! So now there is no guarantee how I will act, and P(H|L) and P(T|L) once again make sense. And then again they don’t. Oops!

What can be done? Causal decision theorists will note that I reasoned like an evidential decision theorist above. But this makes no difference in this case. The causalist’s story will be a bit more complicated but will end up with the same problem.

We might want to introduce primitive conditional probabilities like Popper functions that let you conditionalize on events with zero probability, and then have P(H|L) = 1/4 and P(T|L) = 3/4, even though P(L) = 0. But that is introducing a lot of complications. Primitive conditional probabilities are not unproblematic.

What should we do? Maybe we should suppose something like primitive suppositional decision theory, where what we are primitively given are the suppositional probabilities P^F and P^L, without them being defined in terms of conditional and unconditional credences as in evidential and causal decision theories. But this seems problematic. Do we have to suppose that in addition to conditional and unconditional credences, we have suppositional credences? Maybe.

Or perhaps decision theory only applies to agents that have non-zero credences of going for all the options.

Probabilities of regresses of chickens

Suppose we have a backwards-infinite sequence of asexually reproducing chickens, ..., c₋₃, c₋₂, c₋₁, c₀ with c_n having a chance p_n of producing a new chicken c_n + 1 (chicken c₀ may or may not have succeeded; the earlier ones have succeeded). Suppose that the p_n are all strictly between 0 and 1, and that the infinite product p₋₁p₋₂p₋₃... equals some number p strictly between 0 and 1.

Intuitively, we should be surprised that chicken c₀ exists if p is low and not surprised if p is high. If we have observed c₀ and are considering theories as to what the chances p_n are, other things being equal, we should prefer the theories on which the product p is high to ones on which it’s low.

But what exactly does p measure? It seems to be some kind of a chance of us getting c₀. But it doesn’t measure the unconditional probability of getting an infinite sequence of chickens leading up to c₀. For that is very tiny indeed, since it is extremely unlikely that the world would contain chickens at all. It seems to be a kind of conditional probability. Let q_n be the proposition that chicken c_n exists. Then P(q₀∣q_n) = p₀p₋₁p₋₂...p_n, and so p is the limit of the conditional probabilities P(q₀∣q_n). It is plausible thus to think of p as a conditional probability of q₀ on q_−∞, which is the infinite disjunction of all the q_n.

But q_−∞ is a rather odd proposition. It is grounded in q_n for every finite n, assuming that a disjunction, even an infinite one, is grounded in its true disjuncts. Thus every one of the q_n is explanatorily prior to q_−∞. But this means that P(q₀∣q_−∞) is actually a conditional probability of q₀ on something that isn’t explanatorily prior to q₀—indeed, that is explanatorily posterior to q₀. This challenges the interpretation of p as a chance of getting chicken c₀.

I am not quite sure what conclusion to draw from the above argument. Maybe it offers some support for causal finitism, by suggesting that things are weird when you have a backwards infinite causal sequence?

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.

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.

Defining comparative probabilities in terms of conditional probabilities

Suppose we have a full conditional probability P(A∣B) defined for all pairs of events (stipulating that P(A∣⌀) = 1 if we wish). I've proposed two methods for defining a probability comparison using conditional probabilities:

1. A ⪅ B iff P(A∣A∪B) ≤ P(B∣A∪B).

2. A ⪅ B iff P(A−B∣AΔB) ≤ P(B−A∣AΔB), where AΔB = (A−B) ∪ (B−A) is the symmetric difference.

In a footnote in a paper, I wrote about the second ordering, which I incorrectly attributed to de Finetti: “This ordering has the advantage that if A is a proper subset of B, then A < B, but it is somewhat harder to prove transitivity”.

Well, that was an understatement! It’s not just harder to prove transitivity: it’s impossible.

Define:

• Ω: all integers

• E₀: non-negative even integers

• E: positive even integers

• D: positive odd integers.

Let P be a full conditional probability such that:

3. P(E₀∣E₀∪D) = 1/2 = P(D∣E₀∪D)

4. P(D∣D∪E) = 1/2 = P(E|D∪E).

It is easy to see from (3) and (4) that because E₀ and D are disjoint, and so are D and E, then by definition (2) we have E₀ ⪅ D and D ⪅ E. (For disjoint A and B, the definition (2) of A ⪅ B is equivalent to thedefinition (1).) However, E₀ − E = {0}, E − E₀ = ⌀, and E₀ΔE = {0}, so P(E−E₀∣E₀ΔE) = 0 while P(E₀−E∣E₀ΔE) = 1, and thus we cannot have E₀ ⪅ E.

The only question is whether there actually is a full conditional probability satisfying (3) and (4). If there is, then (2) is not transitive in our case.

There is such a full conditional probability. Let Q_n(A) =  ∣ A ∩ [−n,n] ∣ /(2n+1), where $B$ is the cardinality of a set B. Then Q_n is a probability. Let Q be a limit of the Q_n along an ultrafilter. This is a finitely additive hyperreal probability which is non-zero for all non-empty sets. Define P(A∣B) as the standard part of Q(A∩B)/Q(B) for B non-empty. This is a full conditional probability. Moreover, P(A∣B) = lim Q_n(A∩B)/Q_n(B) whenever the latter limit is defined. That limit is defined in the cases of the events involved in (3) and (4), and it is easy to evaluate the limits and see that (3) and (4) are true.

I don’t know what I was thinking when I wrote that footnote. My guess is that I had in my mind a proof sketch that doesn’t work (I have some idea what that might have been).

Whew! I noticed this afternoon that Theorem 1 of this paper of mine was incompatible with the transitivity of comparison (2). This made me really worried that my Theorem 1 was false. But since the comparison (2) isn’t transitive, I can relax.

This raises an interesting potential research problem. The Pruss definition of ⪅ does not satisfy the additivity axiom for comparative probabilities, namely that if C is disjoint from A ∪ B, then A ≲ B if and only if A ∪ C ≲ B ∪ C (it only preserves the left-to-right implication). Definition (2) does satisfy the additivity axiom, which is what I liked about it.

I suspect there is no good definition of comparative probabilities in terms of full conditional probabilities that satisfies the additivity axiom. (One reason for this intuition has to do with the fact that in Figure 1 here there are entries with a Yes in column 3 and a No in column 5.)

So, I now wonder: Is there some good combination of a definition of comparative probabilities in terms of full conditional probabilities with some weakened version of the additivity axiom?

Conditionalizing on classically null events

Some events have probability zero in classical probability. For instance, if you spin a continuous and fair spinner, the probability of its landing on any specific value is classically zero.

Some philosophers think we should be able to conditionalize on possible events that classically have zero probability, say by assigning non-zero infinitesimal probabilities to such events or by using Popper functions. I think there is very good reason to be suspicious of this.

Consider these very plausible claims:

1. For y equal to 1 or 2, let H_y be a hypothesis about the production of the random variables X such that the conditional distribution of X on H_y is uniform over the interval [0, y). Suppose H₁ and H₂ have non-zero priors. Then the fact that the value of X is x supports H₁ over H₂ if x < 1.

2. If two claims are logically equivalent and they can be conditionalized on, and one supports a hypothesis over another hypothesis, so does the other.

3. If a random variable is independent of each of two hypotheses, then no fact about the value of the random variable supports either hypothesis over the other.

But (1) yields a counterexample to the method of conditionalization by infinitesimal probabilities. For suppose a random variable Z is uniformly randomly chosen in [0, 1) by some specific method. Suppose further that a fair coin, independent of Z, was flipped, and on heads we let X = Z and on tails we let X = 2Z. Let H₁ be the heads hypothesis and let H₂ be the tails hypothesis. Then X is uniformly distributed over [0, y) conditionally on H_y for y = 1, 2.

But now let E be the fact that X = 0, and suppose we can conditionalize on E. By (1), E supports H₁ over H₂ as 0 < 1. But E is logically equivalent to the fact that Z = 0. By (2), then Z = 0 supports H₁ over H₂. But Z is independent of H₁ and of H₂. So we have a contradiction to (3).

I think this line of thought undercuts my toy model argument in my last post.

Full conditional probabilities and the Axiom of Choice

Here’s a claim that turns out to be equivalent to the Axiom of Choice:

1. Given any non-empty set Ω and a collection M of [0,∞]-valued finitely additive measures on the powerset of Ω such that for any non-empty E ⊆ Ω there is a μ ∈ M with 0 < μ(E) < ∞, there is a full conditional probability P on the powerset of Ω definable in terms of the measures in M in the sense that for each non-empty E there is a μ ∈ M such that 0 < μ(E) < ∞ and for all A we have P(A|E) = μ(A∩E)/μ(E).

The easy direction of proof is from (1) to AC. Let M be the collection of all the finitely additive probability measures on Ω that assign probability one to some singleton. Clearly M has the desired properties. Now, for any non-empty E ⊆ Ω, there will be a μ ∈ M such that P(A|E) = μ(A∩E)/μ(E) and 0 < μ(E) < ∞. Thus, the point at which μ is concentrated must be in E. Moreover, it is clear that for each E, the measure μ must be unique. Let f(E) be the point at which μ is concentrated. This is a choice function for all subsets of Ω. Since Ω is an arbitrary non-empty set, we have AC.

The other direction follows by the method of proof of Lemma 3 here.

A weird space for non-classical probability values

Consider the proper class V of formal expressions of the form xϵ^y where x is a non-negative real number that is permitted to be zero only if y = 0, y is a non-negative surreal number, and ϵ is a formal symbol to be thought of as “something very small”. (If we want to be rigorous, we let V be the set of ordered pairs (y,x).) Stipulate:

1. x = xϵ⁰ for real x

2. xϵ^y ≤ x′ϵ^y′ iff either (a) y > y′ or (b) y = y′ and x ≤ x′

3. xϵ^y + x′ϵ^y′ equals (x+x′)ϵ^y if y = y′ and otherwise equals the greater of xϵ^y and x′ϵ^y′

4. if xϵ^y ≤ x′ϵ^y′ and they’re not both zero, then (xϵ^y/x′ϵ^y′) = (x/x′)ϵ^{y − y′}

5. Std xϵ^y equals x if y = 0 and equals 0 othewise.

We can then define finitely-additive probabilities with values in V in the same way that we do so for reals, and we can then define conditional probabilities using the standard formula P(A∣B) = P(A∩B)/P(B).

Say that a V-valued probability P is regular iff 0 < P(A) whenever A is non-empty.

Now here is a fun fact. Given a V-valued probability P, we can define a real-valued full conditional probability as the standard part (Std) of P. Conversely, and less trivially, any real-valued full conditional probability can be obtained this way (this follows from the fact that any linear order can be embedded in the surreals).

So far this doesn’t mark any advantage of using V instead of hyperreals as the values of our probabilities. But there is an advantage. Specifically, if our probability space Ω is acted on by a supramenable group G of symmetries (any Abelian group is supramenable)—for instance, Ω might be a circle acted on by the group of rotations—then there is a V-valued regular G-invariant probability defined for all subsets of Ω. But if we have hyperreal (or surreal, for that matter) values, then the existence of a regular probability invariant under G requires significantly stricter conditions, ones that won’t be met in the case where Ω is the circle and G is rotations.

However, the advantage comes from the fact that V one to have a + b = a even though b > 0, so that one can have weak regularity—the condition that 0 < P(A) whenever A is nonempty—without strong regularity—the condition that P(A) < P(B) whenever A ⊂ B. If one wants strong regularity, using V instead of the hyperreals doesn’t have the same advantage.

Error in "Non-classical probabilities invariant under symmetries"

Yesterday, I discovered an error in the proof of “Theorem 1” of this recent paper of mine (arxiv version). The error occurs in the harder direction of Lemma 2. I do not know how to fix the error. Here’s what I know to remain of the “Theorem”. The proof that (i) implies (ii)–(v) is unaffected. The proof that (iv) implies (ii)–(v) is also unaffected, and likewise unaffected is the equivalence of (ii), (iii) and (v).

But I no longer know if any of (ii)–(v) imply (i). However, (i) is true under the stronger assumption that G is supramenable or that there exist invariant hyperreal probabilities.

The above remarks suffice for almost all the philosophical points in the paper (the philosophical point that behavior for countable sets is decisive is no longer supported in the full conditional probability case), and all the applications I mention in the paper.

I do not know if “Theorem 1” is true. This is an interesting mathematical question.

Update: The error has been fixed and Theorem 1's proof now works.

Non-uniqueness of "uniform" full conditional probabilities

Consider a fair spinner that uniformly chooses an angle between 0 and 360^∘. Intuitively, I’ve just fully described a probabilistic situation. In classical probability theory, there is indeed a very natural model of this: Lebesgue probability measure on the unit circle. This model’s probability measure can be proved to be the unique function λ on the subsets of the unit circle that satisfies these conditions:

1. Kolmogorov axioms with countable additivity

2. completeness: if λ(B) is zero and A ⊆ B, then λ is defined for A

3. rotational invariance

4. at least one arc on the circle of length greater than zero and less than 360^∘ has an assigned probability

5. minimality: any other function that satisfies 1-4 agrees with λ on the sets where λ is defined.

In that sense “uniformly chooses” can be given a precise and unique meaning.

But we may be philosophically unhappy with λ as our probabilistic model of the spinner for one of two reasons. First, but less importantly, we may want to have meaningful probabilities for all subsets of the unit circle, while λ famously has “non-measurable sets” where it is not defined. Second, we may want to do justice to such intuitions as that it is more likely that the spinner will land exactly at 0^∘ or 180^∘ than that it will land exactly at 0^∘. But λ as applied to any finite (in fact, any countable) set of positions yields zero: there is no chance of the spinner landing there. Moreover, we want to be able to update our probabilities on learning, say, that the spinner landed on 0^∘ or 180^∘—presumably, after learning that disjunction, we want 0^∘ and 180^∘ to have probability 1/2—but λ provides no guidance how to do that.

One way to solve this is to move to probabilities whose values are in some field extending the reals, say the hyperreals. Then we can assign a non-zero (but in some cases infinitesimal) probability to every subset of the circle. But this comes with two serious costs. First, we lose rotational invariance: it is easy to prove that we cannot have rotational invariance in such a context. Second, we lose uniqueness: there are many ways of assigning non-zero probabilities, and we know of no plausible set of conditions that makes the assignment unique. Both costs put in serious question whether we have captured the notion of “uniform distribution”, because uniformity sure sounds like it should involve rotational invariance and be the kind of property that should uniquely determine the probability model given some plausible assumptions like (1)–(5).

There is another approach for which one might have hope: use Popper functions, i.e., take conditional probabilities to be primitive. It follows from results of Armstrong and the supramenability of the group of rotations on the circle that there is a rotation-invariant (and, if we like, rotation and reflection invariant) finitely-additive full conditional probability on the circle, which assigns a meaningful real number to P(A|B) for any subsets A and B with B non-empty. Moreover, if Ω is the whole circle, then we can further require that P(A|Ω) = λ(A) if λ(A) is defined. And now we can compare the probability of two points and the probability of one point. For although P({x,y}|Ω) = λ({x,y}) = 0 = λ({x}) = P({x}|Ω) when x ≠ y, there is a natural sense in which {x, y} is more likely than {x} because P({x}|{x,y}) = 1/2.

Unfortunately, the conditional probability approach still doesn’t have uniqueness, and this is the point of this post. Let’s say that what we require of our conditional probability assignment P is this:

6. standard axioms of finitely-additive full conditional probabilities

7. (strong) rotational and reflection invariance

8. being defined for all pairs of subsets of the circle with the second one non-empty

9. P(A|Ω) = λ(A) for any Lebesgue-measurable A.

Unfortunately, these conditions fail to uniquely define P. In fact, they fail to uniquely define P(A|B) for countably infinite B.

Here’s why. Let E be a countably infinite subset of the circle with the following property: for any non-identity isometry ρ of the circle (combination of rotations and reflections), E ∩ ρE is finite. (One way to generate E is this. Let E₀ be any singleton. Given E_n, let G_n be the set of isometries ρ such that ρx = y for some x, y in E. Then G_n is finite. Let z be any point not in {ρx : ρ ∈ G_n, x ∈ E}. Let E_n + 1 = E_n ∪ {z} (since z is not unique, we’re using the Axiom of Dependent Choice, but a lot of other stuff depends on stronger versions of Choice anyway). Let E be the union of the E_n. Then it’s easy to see that E ∩ ρE contains at most one point for any non-identity isometry ρ.)

Let μ be any finitely additive probability on E that assigns zero to finite subsets. Note that μ is not unique: there are many such μ. Now define a finitely additive measure ν on Ω as follows. If A is uncountable, let ν(A) = ∞. Otherwise, let ν(A) = ∑_ρμ(E∩ρA), where the sum is taken over all isometries ρ. The condition that E ∩ ρE is finite for non-identity ρ and that μ is zero for finite sets ensures that if A ⊆ E, then ν(A) = μ(A). It is clear that ν is isometrically invariant.

Let λ^* be any invariant extension of Lebesgue measure to a finitely additive measure on all subsets of the circle. By Armstrong’s results (most relevantly Proposition 1.7), there is a full conditional probability P satisfying (6)–(8) and such that P(A|E) = μ(A∩E) and P(A|Ω) = λ^*(A) (here we use the fact that ν(A) = ∞ whenever λ^*(A) > 0, since λ^*(A) > 0 only for uncountable A). Since μ wasn’t unique and E is countable, conditions (6)–(9) fail to uniquely define P for countably additive conditions.

Anti-Bayesian update and scoring rules in infinite spaces

Bayesian update on evidence E is transitioning from a credence function P to the credence function P(⋅∣E). Anti-Bayesian update on E is moving from P to P(⋅∣E^c) (where E^c is the complement of E). Whether one thinks that Bayesian update is rationally required, it is clear that Bayesian update is better than anti-Bayesian update.

But here is a fun fact (assuming the Axiom of Choice). For any scoring rule on an infinite space, there is a finitely additive probability function P and an event E such that 0 < P(E) < 1 where P(⋅∣E) and P(⋅∣E^c) get exactly the same score everywhere in the probability space. It follows that when dealing with finitely additive probabilities on infinite spaces, a scoring rule will not always be able to distinguish Bayesian update from anti-Bayesian update. This is a severe limitation of scoring rules as a tool for evaluating the accuracy of a credence function in infinite cases.

Here’s a proof of the fun fact. Let s be a scoring rule. Say that two credence functions are maximally opinionated provided that they assign 0 or 1 to every event. It is known that then there are two different maximally opinionated finitely additive probability functions p and q such that s(p) = s(q) everywhere. Let P = (p+q)/2 be their average. Let E be an event such that p(E) = 1 and q(E) = 0 (such an event exists because p and q are maximally opinionated and yet different). Then P(⋅∣E) = p and P(⋅∣E^c) = q while P(E) = 1/2. Hence conditionalization on E and E^c has exactly the same score.

One might take this as some evidence that finite additivity is not good enough.

Investigative scoring rules

Let s be an accuracy scoring rule on a finite probability space. Thus, s(P) is a random variable measuring how close a probability assignment P is to the truth. Here are two reasonable conditions on the rule (the name for the second is made up):

1. Propriety: E_Ps(P)≤E_Ps(Q) for any distinct probability assignments P and Q.

2. Investigativeness: E_Ps(P)≤P(A)E_PAs(P_A)+P(A^c)E_PA^cs(P_A^c) whenever 0 < P(A)<1.

where E_P is expected value with respect to P, P_A is short for P(⋅|A), and A^c is the complement of A. Propriety says that if we are trying to maximize expected accuracy, we will never have reason to evidencelessly switch to a different credence. Investigativeness says that expected accuracy maximization never requires one to close one’s eyes to evidence because the expected accuracy after conditionalizing on learning whether A holds is at least as good as the currently expected accuracy. And we have strict versions of the two conditions provided the inequalities are always strict.

It is well-known that propriety implies investigativeness, and ditto for the strict variants.

One might guess that the other direction holds as well: that investigativeness implies propriety. But (perhaps surprisingly) not! In fact, strict investigativeness does not imply propriety.

Let s(P) be the following score: s(P)(w)=|{A : P(A)=1 and w ∈ A}|. In other words, s(P) measures how many true propositions P assigns probability 1 to. It is easy to see that s(P_A)≥s(P) everywhere on A, and ditto for A^c in place of A, so the right-hand side in (2) is at least as big as P(A)E_PAs(P)+P(A^c)E_PA^cs(P)=E_Ps(P).

But propriety does not hold as long as our probability space has at least two points. For let P be any regular probability—one that assigns a non-zero value to every non-empty set—and let Q be any probability concentrated at one point w₀. Then s(P)=1 everywhere (the only subset P assigns probability 1 to is the whole space) while E_Ps(Q)≥1 + P({w₀}) > 1 (since Q assigns probability 1 to {w} and to the whole space), and so we don’t have propriety.

If we want strict investigativeness, just replace s with s + ϵs′ where s′ is a Brier score and ϵ is small and positive. Then we will have strict investigativeness for s′, and hence for s + ϵs′ as well, but if ϵ is sufficiently small, we won’t have propriety.

It is interesting to think if investigativeness plus some additional plausible condition might imply propriety. A very plausible further condition is that if P is at least as close to the truth as Q for every event, then P gets a no-worse score. Another plausible condition is additivity. But my examples satisfy both conditions. I don’t see other plausible conditions to add, besides propriety as such.

Preprint: Conditional, Regular Hyperreal and Regular Qualitative Probabilities Invariant Under Symmetries

Abstract: Classical countably additive real-valued probabilities come at a philosophical cost: in many infinite situations, they assign the same probability value---namely, zero---to cases that are impossible as well as to cases that are possible. There are three non-classical approaches to probability that can avoid this drawback: full conditional probabilities, qualitative probabilities and hyperreal probabilities. These approaches have been criticized for failing to preserve intuitive symmetries that can easily be preserved by the classical probability framework, but there has not been a systematic study of the conditions under which these symmetries can and cannot be preserved. This paper fills that gap by giving complete characterizations under which symmetries understood in a certain "strong" way can be preserved by these non-classical probabilities, as well as by offering some results to make it plausible that the strong notion of symmetry here may be the right one. Philosophical implications are briefly discussed, but the main purpose of the paper is to offer technical results to inform more sophisticated further philosophical discussion.

Preprint here.

Weak invariance of full conditional probabilities

In two papers (here and here), I explored two different concepts of symmetry for conditional probabilities. The concept of strong invariance says that P(gA|B)=P(A|B) for a symmetry g as long as A and gA are subsets of B. The concept of weak invariance says that P(gA|gB)=P(A|B) for a symmetry g. In some special cases, the weak concept implies the strong concept.

Anyway, here’s an interesting thing: the weak concept does not capture our symmetry intuitions. Take perhaps the simplest case, a lottery on the set of integers Z, and say that the symmetries are shifts. It turns out that there is a weakly shift-invariant full conditional probability P such that:

1. P({m}|{m, n}) = P({n}|{m, n}) (singleton fairness)

2. P(A|A ∪ B)=0 and P(B|A ∪ B)=1 whenever B has infinitely many positive integers and A has finitely many positive integers.

Condition (2) implies that it is more likely that the winning ticket is a power of two than that that is a negative integer. So weak shift invariance is very far from strong invariance.

(And in fact one can have strong invariance for the lottery on Z if one wants. One can even have have strong invariance under shifts and reflections if one wants.)

The proof is a modification of West's proof of a result for qualitative probabilities.

Complete Probabilistic Characterizations

Consider the concept of a complete probabilistic characterization (CPC) of an experiment. It’s a bit of a fuzzy concept, but we can get some idea about it. For instance, if I have a coin loaded in favor of heads, then saying that heads is more likely than tails is not a CPC. Minimally, the CPC will give exact numbers where the probabilities have exact numbers. But the CPC may go beyond giving numerical probabilities. For instance, if you toss infinitely main fair coins, the numerical probability that they are all heads is zero as is the probability that all the even numbered ones are heads. But intuitively it is more likely that the even numbered ones are heads than that all of them are heads. If there is something to this intuition, the CPC will include the relevant information: it may do that by assigning different infinitesimal probabilities to the two events, or by giving conditional probabilities conditioned on various zero-probability events.

A deep question that has sometimes been discussed by philosophers of probability is what CPCs are like. Here are three prominent candidates:

1. classical real-valued probabilities

2. hyperreal probabilities assigning non-zero (but perhaps infinitesimal) probability to every possible event

3. primitive conditional probabilities allowing conditioning on every possible event.

The argument against (1) and for (2) and (3) is that (1) doesn’t distinguish things that should be distinguished—like the heads case above. I want to offer an argument against (2) and (3), however.

Here is a plausible principle:

4. If X and Y are measurements of two causally independent experiments, then the CPC of the pair (X, Y) is determined by the CPCs of X and Y together with the fact of independence.

If (4) is true, then a challenge for a defender of a particular candidate for CPC is to explain how the CPC of the pair is determined by the individual CPCs of the independent experiments.

In the case of (1), the challenge is easily met: the pair (X, Y) has as its probability measure the product of the probability measures for X and Y.

In the cases of (2) and (3), the challenge has yet to be met, and there is some reason to think it cannot be met. In this post, I will argue for this in the case of (2): the case of (3) follows from the details of the argument in the case of (2) plus the correspondence between Popper functions and hyperreal probabilities.

Consider the case where X and Y are uniformly distributed over the interval [0, 1]. By independence, we want the pair (X, Y) to have a hyperreal finitely additive probability measure P such that P(X ∈ A, Y ∈ B)=P(X ∈ A)P(Y ∈ B) for all events A and B. But it turns out that this requirement on P highly underdetermines P. In particular, it seems to be that for any positive real number r, we can find a hyperreal measure P such that P(X ∈ A, Y ∈ B)=P(X ∈ A)P(Y ∈ B) for all A and B, and such that P(X = Y)=rP(Y = 0). Hence, independence highly underdetermines what value P assigns to the diagonal X = Y as compared to the value it assigns to Y = 0.

Maybe some other conditions can be added that would determine the CPC of the pair. But I think we don’t know what these would be. As it stands, we don’t know how to determine the CPC of the pair in light of the CPC of the members of the pair, if CPCs are of type (2).

Product spaces for hyperreal and full conditional probabilities

I think the following is a consequence of a hyperreal variant of the Horn-Tarski extension theorem for measures on boolean algebras:

Claim: Suppose that <Ω_i, F_i, P_i> for i ∈ I is a finitely additive probability space with values in some field R^* of hyperreals. Then, assuming the Axiom of Choice, there is a hyperreal-valued finitely additive probability space <Ω, 2^Ω, P> where Ω = ∏_i ∈ IΩ_i and where the Ω_i-valued random variables π_i given by the natural projections of Ω to Ω_i are independent and have the distributions given by the P_i.

Note that the values of P might be in a hyperreal field larger than R^*.

Given the Claim, and given the well-known correspondences between hyperreal-valued probabilities and full conditional real-valued probabilities, it follows that we can define meaningful product-space conditional real-valued probabilities.

It would be really nice if the product-space conditional probabilities were unique in the special case where F_i is the power set of Ω_i, or at least if they were close enough to uniqueness to define the same real-valued conditional probabilities.

For a particularly interesting case, consider the case where X and Y are generated by uniform throws of a dart at the interval [0, 1], and we have a regular finitely additive hyperreal-valued probability on [0, 1] (regular meaning that all non-empty sets have positive measure). Let Z be the point (X, Y) in the unit square.

Looking at how the proof of the Horn-Tarski extension theorem works, it seems to me that for any positive real number r, and any non-trivial line segment L along the x = y diagonal in the square [0, 1]², there is a product measure P satisfying the conditions of the Claim (where P₁ and P₂ are the uniform measures on [0, 1]) such that P(L)=rP(H), where H is the horizontal line segment {(x, 0):x ∈ [0, 1]}. For instance, if L is the full diagonal, we would intuitively expect P(L)=2^1/2P(H), but in fact we can make P(L)=100000P(H) or P(L)=P(H)/100000 if we like. It is clear that such a discrepancy will generate different conditional probabilities.

I haven’t checked all the details yet, so this could be all wrong.

But if it is right, here is a philosophical upshot. We would expect there to be a unique canonical product probability for independent random variables. However, if we insist on probabilities that are so fine-grained as to tell infinitesimal differences apart, then we do not at present have any such unique canonical product probability. If we are to have one, we need some condition going beyond independence.

This is part of a larger set of claims, namely that we do not at present have a clear notion of what “uniform probability” means once we make our probabilities more finegrained than classical real-valued probability.

Putative Sketch of Proof of Claim: Embedding R^* in a larger field if necessary, we may assume that R^* is |2^Ω|-saturated. Define a product measure on the cylinder subsets of Ω as usual. The proof of the Horn-Tarski extension theorem for measures on boolean algebras looks to me like it works for |B|-saturated hyperreal-valued probability measures where B is the boolean algebra, and completes the proof of our claim.

Another simple way to see a problem with infinitesimal probabilities

Suppose I independently randomly and uniformly choose X and Y between 0 and 1, not including 1 but possibly including 0. Now in the diagram above, let the blue event B be that the point (X, Y) lies one one of the two blue line segments, and let the red event R be that it lies on one of the two red line segments. (The red event is the graph of the fractional part of 2x; the blue event is the reflection of this in the line y = x.) As usual, a filled circle indicates a point included and an unfilled circle indicates a point not included; the purple point at (0, 0) is in both the red and blue events.

It seems that B is twice as likely as R. For, given any value of X—see the dotted line in the diagram—there are two possible values of Y that put one in B but only one possible value of X that puts one in R.

But of course the situation is completely symmetric between X and Y, and the above reasoning can be repeated with X and Y swapped to conclude that R is about twice as likely as B.

Hmm.

Of course, there is no paradox in classical probability theory where we just say that the red and blue events have zero probability, and twice zero equals zero.

But if we have any probability theory that distinguishes different events that are classically of zero-probability and says things like “it’s more likely that Y is 0.2 or 0.8 than that Y is 0.2” (say because both events have infinitesimal probability, with one of these infinitesimals being twice as big as the other), then the above reasoning should yield the absurd conclusion that B is more likely than R and R is more likely than B.

Technically, there is nothing new in the above. It just shows that when we have a probability theory that distinguishes classically zero-probability events, that probability theory will fail conglomerability. I.e., we have to reject the reasoning that just because conditionally on any value of X it’s twice as likely that we’re in B as in R, therefore it’s twice as likely that we’re in B as in R. We already knew that conglomerability reasoning had to be rejected in such probability theories. But I think this is a really vivid way of showing the point, as this instance of conglomerability reasoning seems super plausible. And I think the vividness of it makes it clear that the problem doesn’t depend on any kind of weird trickery with strange sets, and that no mere technical tweak (such as moving to qualitative or comparative probabilities) is likely to get us out of it.

Yet another variant of the Borel-Kolmogorov paradox

Suppose that a point is uniformly randomly choosen in the unit square. Then you learn that either the point lies on the diagonal y = x (red), or it lies on the horizontal line y = 1/2 (blue). What probability should you assign to its lying on the diagonal?

Answer 1: The diagonal has length 2^1/2 and the horizontal line has length 1. Thus, the total length of the lines where the point might be is 2^1/2 + 1, and the probability that it’s on the diagonal is 2^1/2/(2^1/2 + 1) ≈ 0.59.

Answer 2: We can think of the uniform random choice of a point in the unit square as the choice of two independent coordinates, x and y. Suppose that x has been chosen. Then to be on the diagonal line, y has to equal x, while to be on the horizontal line, y has to equal 1/2. These two things are clearly equally likely, regardless of what x is, so the probability must be 1/2.

Both answers seem reasonable.

Suppose you are attracted to Answer 1 which gives 0.59. Then I can give you an argument for a third answer.

Answer 1b: Here is a way to uniformly choose a point in a square. I first uniformly choose a point in a rectangle whose height is twice its width, and then divide the y coordinate by a factor of two. Being on the diagonal or on the middle horizontal line of the point uniformly chosen in the square are respectively equivalent to being on the diagonal or on the middle horizontal line of the rectangle. But the length of the diagonal of the rectangle is 5^1/2, while the middle horizontal line has the same length 1 as in the square. So applying the reasoning behind Answer 1 to the rectangle case, the probability that the point is on the diagonal is 5^1/2/(5^1/2 + 1) ≈ 0.69.

Thus, if you are attracted to the “geometrical” reasoning behind Answer 1, there are infinitely many other answers available, corresponding to the infinitely many ways of generating a point uniformly in a square by generating it in a rectangle and squashing or stretching.

This might push you to Answer 2, since the reasoning behind Answer 2 seems much more determinate. But there are variants to Answer 2. Here is another way to generate a point uniformly on the unit square. Rotate the unit square by 45 degrees clockwise around the orgin to get a diamond whose size along the x and y axes is 2^1/2. Now choose x with a symmetric triangular probability density between 0 and 2^1/2, choose y₀ uniformly between −1 and 1, and then rescale y₀ to make its range fit within the diamond. Parallel reasoning to that used in Answer 2 will now generate a different answer, indeed an answer making the diagonal be more likely.

Note that while I put the paradox in terms of conditioning on a measure zero (the union of the two line segments), one can also put the paradox in terms of comparing probabilities if one likes to be able to compare zero probability sets.

Lesson: Either there are infinitely many different kinds of “uniform distributions of a point in a square”, or else we shouldn’t compare sets of zero measure.

Independence and probability

Thesis: If we stick to real-numbered probabilities, genuine independence of events A and B cannot be defined in terms of any condition on the conditional probabilities P(X|Y) where X and Y are events can be constructed from A and B by using the boolean operations of union, intersection and complement, even if the conditional probabilities are taken to be primitive rather than defined as ratios.

Argument: Suppose that three genuinely independent darts are thrown uniformly at the interval [0, 1], and consider the events:

• A: the first dart hits [0, 1/2) or the third hits 1/2

• B: the second dart hits [0, 1/2) or the third hits 1/2.

The events A and B are not genuinely independent. The third dart’s hitting 1/2 would guarantee that both events A and B happen. But it is easy to check that the conditional probabilities for any boolean combination of A and B are exactly the same as for the corresponding boolean combination of A′ and B′, where:

• A′: the first dart hits [0, 1/2)

• B′: the second dart hits [0, 1/2).

So, conditional probabilities can’t distinguish the non-genuinely independent pair A and B from the genuinely independent pair A′ and B′.

Nor should we mind this fact. For genuine independence is a concept about causation or rationality, while probabilities give us a statistical concept.