Chance, credence, and centers

Rational credence should match the expectation of objective chance. Here I will have a brief look at what happens to this connection between credence and chance on the assumption that credence is centered and chance is not.


1. Fixing the time. Both credences and chances evolve over time. When a coin is tossed twice, the chance of two heads may initially be 1/4; after the first toss has come up heads, it is 1/2. So when your beliefs should match the assumed chance, it can only match the chance you assume to obtain at some particular time. At what time?

The simplest proposal (made in Lewis 1980) is that your beliefs at time t should match your expectation of the chance at that same time t. Let 'ch_t=f' say that f plays the chance role at t. We can then express this proposal as follows.

PP1. $m[1].

In words: your degree of belief at time t in any proposition A conditional on the assumption that f plays the chance role at t should equal the value of f for A. Pretending that there are only countably many candidate chances, PP1 is equivalent to

$m[1]

To illustrate, suppose all you know about the chances at time t is that 'two heads' has probability 1/4. Then any chance function you take seriously assigns 1/4 to this outcome, and hence the weighted average of their verdict is also 1/4. Despite appearance, PP1 therefore does not require that you have opinions concerning the chance of every single event in the universe.

If beliefs are centered, we cannot take it for granted that everyone always knows what time it is. What if it is Tuesday but you mistakenly believe it is Monday? PP1 says your credence should match your expectation of the chance on Tuesday. But shouldn't it rather match your expectation of the chance on Monday?

It should. But it is worth noting that your expectation of the chance on Tuesday may well coincide with your expectation of the chance on Monday, in which case it doesn't matter which of the two we choose. Indeed, PP1 itself ensures this coincidence: it entails that if your credence matches your expectation of the chances at t1, then it also matches your expectation of the chances at any later time t2.

Here is why. Assume, by PP1, that for any A,

1) $m[1].

Let 'h' range over the space of possible histories up to t2, the later time. By the law of total probability,

2) $m[1].

By (1), for each h,

3) $m[1].

(2) and (3) entail that

4) $m[1].

For the next step, we assume that chances evolve in a particular way, namely by conditioning on history. Lewis derived this fact from PP1, but it should be independently plausible. For example, the chance of 'two heads' after the first toss equals the previous conditional chance of 'two heads', conditional on the first toss coming up heads. The right-hand side in (4) is therefore just the expectation of A's chance at t2, conditional on ch_t1=f. That is,

5) $m[1].

By (4) and (5),

6) $m[1],

and by (6) and (1),

7) $m[1].

(7) entails

8) $m[1].

By the law of total probability,

9) $m[1].

(8) and (9) yield

10) $m[1].

Thus your credence at t1 will match your expectation of the chance at t2.

As an illustration, take the 'two heads' example, and suppose the first toss takes place at 3am on Monday, the second at 3am on Tuesday. On Sunday, the chance of 'two heads' is 1/4, so your expectation of the Sunday chance is also 1/4. The Monday chance of 'two heads' is either zero or 1/2, depending on whether the first coin landed tails or heads. What is your credence in these two possibilities? If you align your credence with the known chances on Sunday, it is 1/2 for each. So your expectation of the Monday chance of 'two heads' is 1/2*0 + 1/2*1/2 = 1/4 as well.

We can't extend this argument in the other direction, to past times, because earlier chances do not result from later ones by conditioning on history. Indeed, once you've found out, on Tuesday, that both coin flips landed heads, your credence in 'two heads' will hardly match your expectation of the Sunday chance of 'two heads', which is still 1/4.

It is in this direction where we find reasons to give up PP1. Suppose it is Monday, but you mistakenly believe it is Tuesday. You see that today's coin has landed heads, and you know that the Monday chance of 'heads on Tuesday' is 1/2. By PP1, your credence in 'heads on Tuesday' should also be 1/2. But that can't be right: if you're fairly certain that it is Tuesday and that today's coin has landed heads, you have to be fairly certain that the Tuesday coin has landed heads.

The obvious repair is to use not the actual time when the believing takes place, but the time at which the believer locates herself. More generally,

PP2. $m[1].

In terms of expectation, this is

$m[1].

In the coin toss example, we have

$m[1].

Since you know that today's coin has landed heads and past events are no longer chancy, the chance of 'heads on Tuesday' can only be one if it is actually Tuesday. If it is Monday, the coin you see is the Monday coin, and the chance of heads on Tuesday is 1/2. So

P(heads on Tue) = P(Mon) 1/2 + P(Tue) 1,

which looks correct.


2. Factoring out the centers. The next problem is what to do when A is a centered proposition, assuming that chances are uncentered. I suggest that we should treat chance as an expert who has opinions about what kind of world she inhabits, but no idea about where in the world she is.

Right now, this gives the wrong results about centered propositions. For what would such an expert say about whether today is Monday? Since she is completely lost in time, her credence in it being Monday will be 1/7 (pretending weekdays are defined for every spacetime point in every world). Both PP1 and PP2 thus entail that your credence in Monday should also be 1/7. But this seems wrong, especially if you've just looked at the calendar.

What we want is a principle of partial deference: you should defer to the chance expert on every subject matter except on where in the world you are.

A subject matter is a partition of logical space, partitioned by the relation 'sameness with respect to that subject matter'. Let S be such a partition, and let 's' range over its members. Defering to an expert except with respect to S can then be explicated as satisfying

Partial Deference. $m[1].

To see why this makes sense, suppose A is entirely about the subject matter S; that is, A carves up logical space along the borders of the partition S. Then A is a conjunction of members of S. For each such s, f(s|s)=1. Hence if you satisfy Partial Deference, your credence in A conditional on the expert having credence function f will equal your unconditional credence in A -- you effectively ignore the expert's opinion about A. On the other hand, if A is entirely independent of S (by the lights of the expert), then \sum_s P(s) f(A|s) = f(A) and you fully defer to the expert's opinion.

Strictly speaking, since we consider the probability of A conditional on the expert's credence function being f, we should also take the expectation of the expert's judgement conditional on this assumption. That is, our rule should say that

$m[1].

Applying this to PP1 yields

$m[1].

or, equivalently,

$m[1].

PP2 already has 'now=t' in the condition on the left-hand side, so all we have to do is add it to the right:

PP3. $m[1].

There is another motivation for moving from PP2 to something like PP3: one should never defer to an expert's opinion if one has information they might lack. In such a case, one should at most defer to the expert's conditional opinion, conditional on that further information. More generally, your credence in A conditional on C should equal your expectation not of the expert's unconditional credence in A, but of the expert's credence in A conditional on C. Thus PP2 should anyway be replaced by

PP4. $m[1].

The addition of 'ch_t=f' on the right becomes important when we consider 'undermining futures'. I won't do so in this post, so I will mostly ignore it.

In moving from PP2 to PP3, I have assumed that the subject matter of self-location is characterized by propositions of the form 'now=t'. In fact, we should presumably add further coordinates such as place, individual and orientation. To keep the formulas short, let 'now' abbreviate '<now,here,I,front,...>', and let 't' range over corresponding tuples of times, places, individuals, etc.

PP4 is still not very useful if our best theories of chance deliver only uncentered chances -- if they don't specify a chance for such things as 'today's coin landing heads' conditional on 'it is Monday'. But notice that conditional on 'it is Monday', 'today's coin lands heads' is equivalent to 'Monday's coin lands heads'. In general, let 'at t' be an operator such that 'now=t -> (A <-> A at t)' is a tautology. Hence f(A | ch_t=f & now=t) = f(A at t | ch_t=f & now=t), and PP4 turns into

$m[1].

'A at t' is an uncentered proposition, and so our chance expert has an opinion about it. Moreover, 'A at t' is, by her lights, independent of 'now=t'. This is not completely trivial because even though 'now=t' is entirely about self-location, it is not exclusively about self-location. It also entails that a certain time and place and individual exist somewhere in the universe. However, it is plausible that at t, the chance of these things existing is 1 anyway. (The chance, at 2008, that the world ends before 2008, is zero.) So the uncentered information contained in 'now=t' does not affect the chance at t of uncentered propositions such as 'A at t'. Hence we can drop 'now=t' from the right-hand side. We arrive at

PP5. $m[1].

This is what we wanted: a credence-to-chance link that works for arbitrary centered and uncentered propositions without assuming that chances are centered. All propositions on the right-hand side are uncentered.

In expectation terms, PP5 is

$m[1].

I have ignored the possibility of inadmissible evidence. As I argued in the previous post, I am skeptikal about whether inadmissible evidence is possible. If it isn't, then PP5 holds unrestrictedly, no matter what information the subject may have. Nevertheless, it might be useful to have an even more general principle that allows for arbitrary additional assumptions. This is straightforward:

PP6. $m[1].

3. Applications. As an application, consider the doomsday argument. We have two hypotheses: that humans will go extinct before 2100, and that they will live for much longer. Suppose both of these currently have an objective chance of 1/2. The first hypothesis ('doom') entails that we live before the year 2100; the second does not. The fact that we find ourselves living before 2100 therefore supports doom. To calculate the degree of support, let's assume that on the no-doom hypothesis, we will not go extinct for another 50 million years. Assuming we could just as well have lived at any time in the history of mankind, the probability of living in 2008 conditional on the truth of doom is therefore twice as large as the probability of living in 2008 conditional on no-doom. Hence, says the doomsday argument, our credence in doom should be 2/3, and our credence in no-doom 1/3.

The argument assumes that before taking into account the fact that we live in 2008, our credence should match the known objective chance, yielding P(doom) = 1/2. The fact that we live in 2008 is then considered to be 'inadmissible information', moving P(doom) to 2/3. One might echo Lewis on Sleeping Beauty: "when we find out that it is 2008, we are getting evidence -- centered evidence -- about the future: namely that we are not now in it".

Let's see what our principle says. Before taking into account that it is 2008, we get, by PP5,

$m[1].

We have to consider, for each time t, a) our 'prior' credence that we live at t, and b) our opinion about the objective chance of doom given that we live at t. Now we have to choose whether to hold fixed the information that the chance of doom is currently 1/2.

Since the doom hypothesis concerns an event in 2100, its chance at any time after 2100 is either zero or one. If we assume that the chance of doom is 1/2, we therefore have to assume that we live before 2100. Dividing our credence evenly between all times before 2100, we reach P(doom) = 1/2. But we also reach P(doom | now=2008) = 1/2. Since we've held fixed that the chance of doom is 1/2, the information that we live before 2100 is not news, let alone inadmissible news.

What if instead we hold fixed that before 2100, the chance of doom is 1/2? This doesn't tell us when we live. Let t be a time in, say, 2208. What is the objective chance of doom given that we live in 2208? It is either zero or one. Indeed, it must be zero, for if doom were true, there would be no humans living in 2208. So for t > 2100, the chance of doom given that we live at t is zero, while for t < 2100, it is 1/2. Dividing our credence evenly between all these times, we reach P(doom) = 1/3. When we now take into account the fact that we live before 2100, P(doom) rises to 1/2.

Either way, the doomsday argument fails.

As another (and final) application, we can strengthen the case for thirding in the Sleeping Beauty problem. Remember that when Sleeping Beauty wakes up on Monday, she knows that if a certain fair coin lands heads, she is awakened only on Monday, while if it lands tails, she is awakened both on Monday and Tuesday, with her memories of Monday erased before Tuesday. Now she is awake and doesn't know whether it is Monday or Tuesday. By PP5,

$m[1].

As before, we have to consider, for each time t, a) Beauty's degree of belief that it is t, and b) her opinion about the objective chance of heads given that it is t. There are two relevant times, Monday morning and Tuesday morning. Suppose the coin is tossed on Monday evening. Then Beauty knows that the chance of heads, given that it is Monday, is 1/2. On Tuesday, the chance of heads is either zero or one, depending on how the coin actually landed. If it landed heads, Beauty would not be awake. So conditional on it being Tuesday, Beauty can be certain that the chance of heads is now 0. Putting all this together, we get

P(Heads) = P(Mon) 1/2 + P(Tue) 0 = P(Mon) 1/2

On the thirder distribution, P(Mon)=2/3 and P(Heads)=1/3, which nicely fits this result. The 'sequential halfer' distribution that I have defended also fits: on this account, P(Mon)=1 and P(Heads)=1/2. What doesn't fit is the traditional halfer distribution (what I have called the 'branching halfer' distribution): on this account, P(Mon)=3/4 and P(Heads)=1/2, but 3/4 times 1/2 is not 1/2.

Interestingly, both the thirder account and the traditional halfer account satify PP5 if the coin toss takes place before the Monday awakening. The chance of heads on Monday is then either zero or one, depending on the actual outcome, and all we get from PP5 is

P(Heads) = P(Mon) P(Heads|Mon) + P(Tue) 0,

which fits both the thirder's 1/3 = 2/3*1/2 and the halfer's 1/2 = 3/4*2/3.

The upshot is that Lewis was wrong when he claimed that halfing preseves the Principal Principle, and that Beauty's higher credence in heads after learning that it is Monday is due to inadmissible information. Whatever happens later, she already violates the principle on Monday morning, at least if the coin is tossed afterwards.

Nevertheless, I think the traditional halfer solution gives the correct answer to the branching version of Sleeping Beauty. This is because branching, like chance, is a source of uncertainty and imposes constraints on rational credence. PP5 ignores this constraint and therefore yields wrong results in situations that involve branching.

More on this next year. For now, merry Christmas.

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