Sure Loss and Logical Ignorance

Here’s something I’ve been thinking about. The basic idea is to wonder what consequences follow from relaxing the standard assumption that Bayesian agents are logically omniscient.

Bayesian epistemology and decision theory typically assume that the ideal agents are logically omniscient. Borrowing a practice from I.J. Good, I refer to my putative ideal agent as “you”. That is, they assume that if $\phi$ and $\psi$ are logically equivalent, then you should believe them to the same degree.

This seems to have some normative weight. What reason could you have for believing equivalent propositions to different degrees except for some failure of insight on your part?

But consider “believe all and only the truths”. Likewise, this has some normative force. The only reason to fail to fully believe a truth or to fully disbelieve a falsehood, is some lack of insight or evidence. So does this mean that we should force you to have only trivial 0/1 degrees of belief? Of course not! Bayesianism is all about how to behave when you can’t live up to the regulative ideal of truth.

I think we need to make the same kind of move in the case of logical omniscience. Of course logical omniscience is a kind of regulative ideal (just like truth is) but we should think about how to model agents who fail to live up to this ideal. So how do you model logical ignorance? With great difficulty, it turns out.

Let’s start with a basic discussion of simple Bayesianism, and find out where the assumptions about logical omniscience creep it. Here’s a first attempt. The objects of belief are sets of possible worlds. These sets of possible worlds form an algebra and … Whoah. Wait a minute. If beliefs are about sets of possible worlds, then we’ve already baked in logical omniscience! If $\phi$ and $\psi$ are logically equivalent, then they will be true in all the same possible worlds. So you can’t have different beliefs about them.

Here’s our first interesting result. Logical omniscience is built into Bayesianism at a very early stage. This makes it really quite difficult to drop or weaken the assumption.

Let’s try to start again in such a way that it is at least clear where logical omniscience enters. So, the objects of beliefs are sentences in a language: syntactic things. Now $\phi$ and $\psi$ can be different syntactic objects but still be logically equivalent. So that’s nice.

However, even if there’s only one elementary letter in your language, you now have infinitely many sentences to have beliefs over! $a,a\wedge a,a \wedge a \wedge a\dots$ This is awkward, to say the least. Can we do anything to manage this awkward infinity of sentences? There must be some kinds of things we can do, since logically omniscient agents managed to have beliefs over finite algebras.

How to we build in logical omniscience? We take equivalence classes of sentences using the logical equivalence relation “$\equiv$“. That is, we identify all the propositions logically equivalent to $\phi$ and we call this $\overline{\phi}$. We build up an algebra of equivalence classes of sentences. This is called the “Lindenbaum algebra”. This is then the thing logically omniscient agents have beliefs over.

So now we’ve seen that by taking this “quotient algebra“ we can strip down the number of sentences you need to have beliefs over. So, are there equivalence relations other than classical logical equivalence that will give us “well-behaved” quotient algebras that don’t build in logical omniscience? Yes there are. What you need is for your relation to be a “congruence relation”. This means basically that it “plays nicely” with the logical structure.

Where does this get us? Remember we had our norm that said “logically equivalent sentences should be believed equally strongly”. We might argue that a weaker, but still reasonable norm would only demand that “sentences known to be logically equivalent should be believed equally strongly”.

What can we say about this idea of “known equivalence”? Well, if you were being a logically omniscient Bayesian you would say that the right relation to use to “quotient up” your space of sentences is the classical equivalence. We might instead argue that some other congruence which is less stringent is a more reasonable requirement. We still require that the relation is a congruence, which is no small requirement. But it’s a start!

So, I assume there are sentences $\phi$ and $\psi$ that express the same proposition, but that are believed to different degrees. This means that the odds you would offer on one are different from the odds on the other. Since in the standard Dutch book setting, the opponent is allowed to choose the “direction” of the bet, she can choose to bet against the one with the longer odds and on the one with the shorter odds. In either case exactly one bet wins. If she arranges her bets such that they would win the same, then her outlay will be less than her winnings: she will have Dutch booked you!

Consider bets of the following form:

  • Win $S (1-q)$ if $\phi$, lose $S q$ otherwise.
  • Win $T (1-p)$ if $\psi$, lose $T p$ otherwise.

The $S$ is the stake, and the $q$ is the betting quotient. Likewise for $T$ and $p$. The idea behind the Dutch book argument is that if your opponent can set the size of $S$ and decide whether it is positive or negative, then unless your $q$s have the structure of a probability measure, you are guaranteed to lose money: you will be “Dutch booked”. What I want to show here is that a logically ignorant punter can be Dutch booked. That is, if $q \neq p$ then an agent who knows that the two propositions are logically equivalent can guarantee herself a profit.

We have stipulated that $\phi$ and $\psi$ are logically equivalent. Importantly, this means that they are always true or false together. So there are two possibilities: both are true or both are false.

Now, the clever punter will set $T = -S$ . If $\phi$ (and hence $\psi$) is true, then the two bets together win:

$$ S(1-q) + T(1-p) = S(p - q) $$

If the proposition is false then the outcome for our agent is:

$$ -S q -T p = S(p - q) $$

So in either case the punter’s wealth changes the same amount. Now here’s the thing: as in the standard Dutch book argument, we are allowing the opponent to decide the value of $S$. So depending on whether $q$ or $p$ is bigger, the opponent can set $S$ such that our agent loses money!

The only way to avoid this is to set $q = p$. But that is simply to say that the agent should be logically omniscient!

But who cares about logical ignorance? Surely logical omniscience is a norm of rational belief: what possible reason could you have for believing logically equivalent propositions to different degrees? So in this post I want to step back a bit and wonder about why modelling logical ignorance is important.

First, let’s consider the argument I hinted at above: “why would you believe logically equivalent propositions to different degrees except through some failing of your reasoning capacity?” Is this a good argument? I think it is exactly as good as the following argument: “Why would you have any degree of belief less than full belief in a truth, except through some failing of your evidence?” Now the obvious outcome of taking this second argument too seriously is to have no non-trivial degrees of belief: you must fully believe all and only the truths.

We don’t take this norm seriously because of the “ought implies can” principle. We simply can’t fully believe all and only the truths: we don’t always know which sentences are the truths. Since we can’t conform to this ideal, it can’t be something we ought to do. This idea is still a kind of regulative ideal on belief, but it certainly isn’t a norm. Failing to live up to the ideal of believing all and only the truths doesn’t impugn your rationality; you aren’t irrational for having failed to fully believe a truth. But you still consider it better to believe more truths more strongly.

I want to make the same case for logical omniscience: it is obviously a regulative ideal of belief, but it is not a norm. Rational agents can fail to believe logically equivalent propositions to the same degree (because of some lack of reasoning capacity) but they aren’t thereby irrational.

Bayesianism starts from the idea that there is something interesting to say about rational belief even if you fail to fully believe all and only the truths. That is, there are rational ways to partially believe. In the same way, I want to say that there are rational ways to have logically ignorant beliefs.

Now, I don’t really know what more to say about logically ignorant belief for now. Dutch book arguments don’t really get us very far. It’s all but impossible to rule out the possibility of being Dutch booked. I’m going to write about that in a later post.

We certainly want to demand that sentences known to be logically equivalent ought to be believed to the same degree. So there are some rationality constraints of logically ignorant belief. But I’m not sure how to fully articulate them. If you’ll indulge some highly speculative vague suggestions, I think it would be interesting to see what happens when you throw ideas like proof complexity, or costs of computation into the mix. Being Dutch booked for failing to spot a really easy proof seems worse than failing to, say, find a proof of Goldbach’s conjecture.

The major difficulty with this sort of idea is that any relation of known entailment based on complexity of proof is not going to give you a transitive “entailment” relation, thus it won’t be an equivalence relation in the sense of defining equivalence classes. So all the nice quotient algebra stuff I discussed previously flies out the window!

To finish, let me situate this even more broadly within my interests. Any formal theory of rational belief or decision is going to be, to some extent, idealised. Bayesianism is no exception. It is interesting and worthwhile to see what happens when these idealisations are relaxed. This process of “de-idealising” is something that might be familiar from discussions of idealisations in science. We “idealise away” air resistance when considering cannon ball trajectories. Why? Because we don’t think it has a big effect. We don’t think adding it back in (de-idealising) will substantially affect the results we obtain. This is why the idealisation is legitimate: the decrease in computational cost far outweighs the distortion in the results thereby obtained. (Those familiar with this discussion may find this a caricature of the debate, but it is good enough for my purposes).

So I think we should think about the same sort of de-idealising process in formal theories of belief and decision. Instead of “what happens when we add back in air resistance?” we should be wondering “what happens when we remove logical omniscience?” And we have an extra important purpose in the case of de-idealising theories of belief: we are not ideal agents, so it’s useful to see what norms non-ideal agents such as ourselves should conform to.

As we’ve seen, it’s not trivial to excise logical omniscience: it’s built into the framework at quite a low level. And indeed, even when we don’t build it in, it emerges as a norm.

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Seamus Bradley
Marie Curie Individual Fellow

I’m a philosopher at the University of Leeds.