Virtue ethics and peer disagreement

Aristotelian ethics is committed to the claim that the virtuous person knows what actions and habits are virtuous and is justified in holding on to that knowledge, and indeed should hold on to it. There is a deep stability to virtue. This means that an Aristotelian virtuous person ought not adopt a conciliationist response to those who disagree as to what is virtuous, suspending judgment over the disagreed-upon items.

Indeed, one imagines that Aristotle’s virtuous person could say of those who disagree: “They are not virtuous, and hence do not see the truth about moral matters.” Aristotle’s virtuous person would reject the idea that someone who disagrees with them about virtue could be an epistemic peer. Virtuous habits give epistemic access to moral (and not only moral) truth.

Of course, the disagreer may think themselves virtuous as well, and may think the same thing about the virtuous person as the virtuous person thinks about them. But that does not shake the Aristotelian virtuous person.

This means that if Aristotelian virtue ethics is correct, there is a clear thing that a Christian can say about religious disagreement. The Christian thinks faith is a virtue, albeit an infused rather than natural one. As such, faith gives epistemic access, and someone lacking faith is simply not an epistemic peer, since they lack a source of truth. The fact that a person lacking faith thinks they have the virtue of faith should not move the person who actually has the virtue.

Of course, one might turn all this around and use it as an argument against virtue ethics. But I think Aristotle’s picture seems exactly correct as to the kind of firmness of moral knowledge that the virtuous person exhibits, the kind of spine that lets them say, without pride or vanity, to vast numbers of others that they are simply wrong.

Saints and faith

There are many ways of life that people claim to be virtuous. A central thesis of Aristotelian ethics is:

1. The virtuous person knows what is the virtuous human form of life, at least insofar as this is relevant to her own circumstances.

She knows this by living virtuously, which enables a from-the-inside appreciation of the virtue of the virtuous life she lives. This is a mysterious thing, but it means that the virtuous person does not need to worry sceptically about the fact that other people disagree with her about this way of life being virtuous (maybe they say to her: “You should have a stronger preference for people of your country over foreigners”, and she just knows that her preference should not be stronger). These other people are not virtuous, and hence lack that from-the-inside view on what it is to live a virtuous life, and hence they are not her epistemic peers with respect to virtue.

Suppose we accept (1). Now imagine that Therese leads a kind of life L that is deeply intertwined with a particular religion R, in such a way that clearly L would be unlikely to be virtuous if R were false, but is very likely to be virtuous if R is true.

It is easy to imagine cases like this. Perhaps most religious and non-religious views other than R would object to significant aspects of L—perhaps, L includes forms of activism that R praises but most other religious and non-religious views look down on, or lacks forms of activity that most religious and non-religious views other than R think are required for a fulfilling human life. The life of a good contemplative Catholic nun is like that: most non-Catholic views will see it as a waste.

Suppose, further, that Therese is in fact virtuous. Then she knows that L is virtuous, and this gives her significant evidence that R is true because of how much L is bound up with R.

One may have a Christian worry about what I just said. What about humility? Would Therese know that she is living a virtuous life? But she might: true self-insight is compatible with humility. However, my argument does not assume that Therese knows that she is living a virtuous life. All that (1) says is that Therese knows that L is a virtuous life—but she need not know that she is in fact living out L. She knows the model of the virtuous life by living it, but she may not know that she is living it. (Aristotle wouldn’t like that.)

Now, suppose that Therese’s virtue in fact comes from God’s grace. Then Therese has a deep reason to know R on the basis of grace: the grace leads to virtue, and the virtue leads to knowledge of what is virtuous.

So, we have a model for how saints of the true religion can know the truths of their faith, because their radical forms of life are so tightly bound up with their religion that their knowledge that this way of life is virtuous (a knowledge compatible with certain ways of agonizing about whether they are in fact living that way) yields knowledge of their religion.

Can this help those of us who are not saints? I think so. It is possible to see the virtue of another’s form of life even when one does not have much virtue. And then the tight intertwining between the saint’s life of virtue and the saint’s religion provides one with evidence of the truth of their religion.

(Note the similarities to the line of thought in van Inwagen's deeply moving "Quam Dilecta".)

Is this immune to sceptical worries in the way that the virtuous person’s knowledge of the virtue of the form of life she follows is? I don’t know. I think there is room for some proper-functionalism here: we may have a faculty of recognition of a virtuous form of life.

Note, finally, that there are multiple virtuous forms of life, some less radical than others. The more radical ones are likely to be more tightly bound up with their religion, and hence provide more evidence—even if they are not necessarily more virtuous. Perhaps the difference is in how specific a religion is testified to by the virtue of the way of life. Thus, the contemplative cloistered saint’s life may give strong evidence of Catholicism, or at least of the disjunction of Catholicism and Eastern Orthodoxy, while the life of a married saint as seen from the outside may “only” give strong evidence of Christianity.

Conciliationism and natural law epistemology

Suppose we have a group of perfect Bayesian agents with the same evidence who nonetheless disagree. By definition of “perfect Bayesian agent”, the disagreement must be rooted in differences in priors between these peers. Here is a natural-sounding recipe for conciliating their disagreement: the agents go back to their priors, they replace their priors by the arithmetic average of the priors within the group, and then they re-updated on all the evidence that they had previous got. (And in so doing, they lose their status as perfect Bayesian agents, since this procedure is not a Bayesian update.)

Since the average of consistent probability functions is a consistent probability function, we maintain consistency. Moreover, the recipe is a conciliation in the following sense: whenever the agents previously all agreed on some posterior, they still agree on it after the procedure, and with the same credence as before. Whenever the agents disagreed on something, they now agree, and their new credence is strictly between the lowest and highest posteriors that the group assigned prior to conciliation.

Here is a theory that can give a justification for this natural-sounding procedure. Start with natural law Bayesianism which is an Aristotelian theory that holds that human nature sets constraints on what priors count as natural to human beings. Thus, just as it is unnatural for a human being to be ten feet tall, it is unnatural for a human being to have a prior of 10⁻¹⁰⁰ for there being mathematically elegant laws of nature. And just as there is a range of heights that is natural for a mature human being, there is a range of priors that is natural for the proposition that there are mathematically elegant laws.

Aristotelian natures, however, are connected with the actual propensities of the beings that have them. Thus, humans have a propensity to develop a natural height. Because of this propensity, an average height is likely to be a natural height. More generally, for any numerical attribute governed by a nature of kind K, the average value of that attribute amongst the Ks is likely to be within the natural range. Likely, but not certain. It is possible, for instance, to have a species whose average weight is too high or too low. But it’s unlikely.

Consequently, we would expect that if we average the values of the prior for a given proposition q over the human population, the average would be within the natural range for that prior. Moreover, as the size of a group increases, we expect the average value of an attribute over the group to approach the average value the attribute has in the full population. Then, if I am a member of the group of disagreeing evidence-sharing Bayesians, it is more likely that the average of the priors for q amongst the members of the group lies within the natural human range for that prior for q than it is that my own prior for q lies within the natural human range for q. It is more likely that I have an unnatural height or weight than that the average in a larger group is outside the natural range for height or weight.

Thus, the prior-averaging recipe is likely to replace priors that are defectively outside the normal human range with priors within the normal human range. And that’s to the good rationally speaking, because on a natural law epistemology, the rational way for humans to reason is the same as the normal way for humans to reason.

It’s an interesting question how this procedure compares to the procedure of simply averaging the posteriors. Philosophically, there does not seem to be a good justification of the latter. It turns out, however, that typically the two procedures give the same result. For instance, I had my computer randomly generate 100,000 pairs of four-point prior probability spaces, and compare the result of prior- to posterior-averaging. The average of the absolute value of the difference in the outputs was 0.028. So the intuitive, but philosophically unjustified, averaging of posteriors is close to what I think is the more principled averaging of priors.

The procedure also has an obvious generalization from the case where the agents share the same evidence to the case where they do not. What’s needed is for the agents to make a collective list of all their evidence, replace their priors by averaged priors, and then update on all the items in the collective list.

Conciliationism with and without peerhood

Conciliationists say that when you meet an epistemic peer who disagrees with you, you should alter your credence towards theirs. While there are counterexamples to conciliationism here is a simple argument that normally something like conciliationism is correct without the assumption of epistemic peerhood:

1. That someone’s credence in a proposition p is significantly below 1/2 is normally evidence against p.

2. Learning evidence against a proposition typically should lower one’s credence.

3. So, normally, learning that someone’s credence is significantly below 1/2 should lower one’s credence.

In particular, if your credence is above 1/2, then learning that someone else’s is significantly below 1/2 should normally lower one’s credence. And there are no assumptions of peerhood here.

The crucial premise is (1). Here is a simple thought: Normally, people’s credences are responsive to evidence. So when their credence is low, that’s likely because they had evidence against a proposition. Now the evidence they had either is or is not evidence you also have. If you know it is not evidence you also have, then learning that they have additional evidence against the proposition should normally provide you with evidence against it, too. If it is evidence you also have, that evidence should normally make no difference. You don’t know which of these is the case, but still the overall force of evidence is against the proposition.

One might, however, have a worry. Perhaps while normally learning that someone’s credence is significantly below 1/2 should lower one’s credence, when that someone is an epistemic peer and hence shares the same evidence, it shouldn’t. But actually the argument of the preceding paragraph shows that as long as you assign a non-zero probability to the person having more evidence, their disagreement should lead you to lower your credence. So the worry only comes up when you are sure that the person is a peer. It would, I think, be counterintuitive to think you should normally conciliate but not when you are sure the other person is a peer.

And I think even in the case where you know for sure that the other person has the same evidence you should lower your credence. There are two possibilities about the other person. Either they are a good evaluator of evidence or not. If not, then their evaluation of the evidence is normally no evidence either for or against the proposition. But if they are good evaluators, then their evaluating the evidence as being against the proposition normally is evidence that the evidence is against the proposition, and hence is evidence that you evaluated badly. So unless you are sure that they are a bad evaluator of evidence, you normally should conciliate.

And if you are sure they are a bad evaluator of evidence, well then, since you’re a peer, you are a bad evaluator, too. And the epistemology of what to do when you know you’re bad at evaluating evidence is hairy.

Here's another super-quick argument: Agreement normally confirms one's beliefs; hence, normally, disagreement disconfirms them.

Why do I need the "normally" in all these claims? Well, we can imagine situations where you have evidence that if the other person disbelieves p, then p is true. Moreover, there may be cases where your credence for p is 1.

Peers and twins

I just realized something that I should have known earlier. Suppose I have a doppelganger who is just like me and goes wherever I go—by magic, he can occupy a space that I occupy—and who always sees exactly what I see and who happened always to judge and decide just as I do. What I’ve just realized is that the doppelganger is not my epistemic peer, even though he is just like me.

He is not my peer because he has evidence that I do not and I have evidence that he does not. For I know what experiences I have and he knows what experiences he has. But even though my experiences are just like his, they are not numerically the same experiences. When he sees, it is through his eyes and when I see, it is through my eyes.

Suppose that on the basis of a perception of a distant object that looked like a dog I formed a credence of 0.98 that the object is a dog, and my doppelganger did the same thing. And suppose that suddenly a telepathic opportunity opens up and we each learn about the other’s existence and credences.

Then our credences that the distant object is a dog will go up slightly, because we will each have learned that someone else’s experiences matched up with ours. Given that the other person in this case is just like me, this doesn’t give me much new information. It is very likely that someone just like me looking in the same direction would see things the same way. But it is not certain. After all, my perception could still be due to a random error in my eyes. So could my doppelganger’s be. But the fact that our perceptions match up rules makes it implausible to suppose the random error hypothesis, and hence it raises the credence that the object really is a dog. Let’s say our credences will go up to 0.985.

Now suppose that instead this is a case of slight disagreement: His credence that there is a dog there is 0.978 and mine is 0.980, this being the first time we deviate in our whole lives. I think the closeness to me of the other’s judgment is still evidence of correctness. So I think my credence, and his as well, should still go up. Maybe not to 0.985, but maybe 0.983.

Peer disagreement and models of error

You and I are epistemic peers and we calculate a 15% tip on a very expensive restaurant bill for a very large party. As shared background information, add that calculation mistakes for you and me are pretty much random rather than systematic. As I am calculating, I get a nagging feeling of lack of confidence in my calculation, which results in $435.51, and I assign a credence of 0.3 to that being the tip. You then tell me that you you’re not sure what the answer is, but that you assign a credence of 0.2 to its being $435.51.

I now think to myself. No doubt you had a similar kind of nagging lack of confidence to mine, but your confidence in the end was lower. So if all each of us had was their own individual calculation, we’d each have good reason to doubt that the tip is $435.51. But it would be unlikely that we would both make the same kind of mistake, given that our mistakes are random. So, the best explanation of why we both got $435.51 is that we didn’t make a mistake, and I now believe that $435.51 is right. (This story works better with larger numbers, as there are more possible randomly erroneous outputs, which is why the example uses a large bill.)

Hence, your lower reported credence of 0.2 not only did not push me down from my credence of 0.3, but it pushed me all the way up into the belief range.

Here’s the moral of the story: When faced with disagreement, instead of moving closer to the other person’s credence, we should formulate (perhaps implicitly) a model of the sources of error, and apply standard methods of reasoning based on that model and the evidence of the other’s credence. In the case at hand, the model was that error tends to be random, and hence it is very unlikely that an error would result in the particular number that was reported.

Conciliationism is false or trivial

Suppose you and I are adding up a column of expenses, but our only interest is the last digit for some reason. You and I know that we are epistemic peers. We’ve both just calculated the last digit, and a Carl asks: Is the last digit a one? You and I speak up at the same time. You say: “Probably not; my credence that it’s a one is 0.27.” I say: “Very likely; my credence that it’s a one is 0.99.”

Concialiationists now seem to say that I should lower my credence and you should raise yours.

But now suppose that you determine the credence for the last digit as follows: You do the addition three times, each time knowing that you have an independent 1/10 chance of error. Then you assign your credence as the result of a Bayesian calculation with equal priors over all ten options for the last digit. And since I’m your epistemic peer, I do it the same way. Moreover, while we’re poor at adding digits, we’re really good at Bayesianism—maybe we’ve just memorized a lot of Bayes’ factor related tables. So we don’t make mistakes in Bayesian calculations, but we do at addition.

Now I can reverse engineer your answer. If you say your credence in a one is 0.27, then I know that of your three calculations, one of them must have been a one. For if none of your calculations was a one, your credence that the digit was a one would have been very low and if two of your calculations yielded a one, your credence would have been quite high. There are now two options: either you came up with three different answers, or you had a one and then two answers that were the same. In the latter case, it turns out that your credence in a one would have been fairly low, around 0.08. So it must be that your calculations yielded a one, and then two other numbers.

And you can reverse engineer my answer. The only way my credence could be as high as 0.99 is if all three of my calculations yielded a one. So now we both know that my calculations were 1, 1, 1 and yours were 1, x, y where 1, x, y are all distinct. So now you aggregate this data, and I do the same as your peer. We have six calculations yielding 1, 1, 1, 1, x, y. A Bayesian analysis, given the fact that the chance of error in each calculation is 0.9, yields a posterior probability of 0.997.

So, your credence did go up. But mine went up too. Thus we can have cases where the aggregation of a high credence with a low credence results in an even higher credence.

Of course, you may say that the case is a cheat. You and I are not epistemic peers, because we don’t have the same evidence: you have the evidence of your calculations and I have the evidence of mine. But if this counts as a difference of evidence, then the standard example conciliationists give, that of different people splitting a bill in a restaurant, is also not a case of epistemic peerhood. And if the results of internal calculations count as evidence for purposes of peerhood, then there just can’t be any peers who disagree, and conciliationism is trivial.

Conciliationism and another toy model

Conciliationism holds that in cases of peer disagreement the two peers should move to a credence somewhere between their individual credences. In a recent post I presented a toy model of error of reasoning on which conciliationism was in general false. In this post, I will present another toy model with the same property.

Bayesian evidence is additive when instead of probability p one works with log-odds λ(p)=log(p/(1 − p)). From that point of view, it is natural to model error in the evaluation of the force of evidence as the addition of a normally-distributed term with mean zero to the log-odds.

Suppose now that Alice and Bob evaluate their first-order evidence, which they know they have in common, and come to the individual conclusions that the probability of some Q is α and β respectively. Moreover, both Alice and Bob have the above additive model of their own error-proneness in the evaluation of first-order evidence, and in fact they assign the same standard deviation σ to the normal distribution. Finally, we assume that Alice and Bob know that their errors are independent.

Alice and Bob are good Bayesians. They will next apply a discount for their errors to their first-order estimates. You might think: “No discount needed. After all, the error could just as well be negative as well as positive, and the positive and negative possibilities cancel out, leaving a mean error of zero.” That’s mistaken, because while the normal distribution is symmetric, what we are interested in is not the expected error in the log-odds, which is indeed zero, but the mean error in the probabilities. And once one transforms back from log-odds to probabilities, the normal distribution becomes asymmetric. A couple of weeks back, I worked out some formulas which can be numerically integrated with Derive.

First-order probability	σ	Second-order probability
0.80	1.00	0.76
0.85	1.00	0.81
0.90	1.00	0.87
0.95	1.00	0.93
0.80	0.71	0.78
0.80	0.71	0.83
0.90	0.71	0.88
0.95	0.71	0.94

So, for instance, if Alice has a first-order estimate of 0.90 and Bob has a first-order estimate of 0.95, and they both have σ = 1 in their error models, they will discount to 0.87 and 0.93.

Let the discounted credences, after evaluation of the second-order evidence, be α^* and β^* (the value depends on σ).

Very good. Now, Alice and Bob get together and aggregate their final credences. Let’s suppose they do so completely symmetrically, having all information in common. Here’s what they will do. The correct log-odds for Q, based on the correct evaluation of the evidence, equals Alice’s pre-discount log-odds log(α/(1 − α)) plus an unknown error term with mean zero and standard deviation σ, as well as equalling Bob’s pre-discount log-odds log(α/(1 − α)) plus an unknown error term with mean zero and standard deviation σ.

Now, there is a statistical technique we learn in grade school which takes a number of measurements of an unknown quantity, with the same normally distributed error, and which returns a measurement with a smaller normally distributed error. The technique is known as the arithmetic mean. The standard deviation of the error in the resulting averaged data point is σ/n^1/2, where n is the number of samples. So, Alice and Bob apply this technique. They back-calculate α and β from their final individual credences α^* and β^*, they then calculate the log-odds, average, and go back to probabilities. And then they model the fact that there is still a normally-distributed error term, albeit one with standard deviation σ/2^1/2, so they adjust for that to get a final credence α^** = β^**.

So what do we get? Do we get conciliationism, so that their aggregated credence α^** = β^** is in between their individual credences? Sometimes, of course, we do. But not always.

Observe first what happens if α^* = β^*. “But then there is no disagreement and nothing to conciliate!” True, but there is still data to aggregate. If α^* = β^*, then the error discount will be smaller by a factor of the square root of two. In fact, the table above shows what will happen, because (not by coincidence) 0.71 is approximately the reciprocal of the square root of two. Suppose σ = 1. If α^* = β^* = 0.81, this came from pre-correction values α = β = 0.85. When corrected with the smaller normal error of 0.71, we now get a corrected value α^** = β^** = 0.83. In other words, aggregating the data from one another, Alice and Bob raise their credence in Q from 0.81 to 0.83.

But all the formulas here are quite continuous. So if α^* = 0.8099 and β^* = 0.8101, the aggregation will still yield a final credence of approximately 0.83 (I am not bothering with the calculation at this point). So, when conciliating 0.8099 and 0.8101, you get a final credence that is higher than either one. Conciliationism is thus false.

The intuition here is this. When the two credences are reasonably close, the amount by which averaging reduces error overcomes the downward movement in the higher credence.

Of course, there will also be cases where aggregation of data does generate something in between the two data points. I conjecture that on this toy model, as in my previous, this will be the case whenever the two credences are on opposite sides of 1/2.

Peer disagreement, conciliationism and a toy model

Let suppose that Alice and Bob are interested in the truth of some proposition Q. They both assign a prior probability of 1/2 to Q, and all the first-order evidence regarding Q is shared between them. They evaluate this first-order evidence and come up with respective posteriors α and β for Q in light of the evidence.

Further, Alice and Bob have background information about how their minds work. They each have a random chance of 1/2 of evaluating the evidence exactly correctly and a random chance of 1/2 that a random bias will result in their evaluation being completely unrelated to the evidence. In the case of that random bias, their output evaluation is random, uniformly distributed over the interval between 0 and 1. Moreover, Alice and Bob’s errors are independent of what the other person thinks. Finally, Alice and Bob’s priors as to what the correct evaluation of the evidence will show is uniformly distributed between 0 and 1.

Given that each now has this further background information about their error-proneness, Alice and Bob readjust their posteriors for Q. Alice reasons thus: the probability that my first-order evaluation of α was due to the random bias is 1/2. If I knew that the random bias happened, my credence in Q would be 1/2; if I knew that the random bias did not happen, my credence in Q would be α. Not knowing either way, my credence in Q should be:

1. α^* = (1/2)(1/2)+(1/2)α = (1/2)(1/2 + α).

Similarly, Bob reasons that his credence in Q should be:

2. β^* = (1/2)(1/2 + β).

In other words, upon evaluating the higher-order evidence, both of them shift their credences closer to 1/2, unless they were at 1/2.

Next, Alice and Bob pool their data. Here I will assume an equal weight view of how the data pooling works. There are now two possibilities.

First, suppose Alice and Bob notice that their credences in Q are the same, i.e., α^* = β^*. They know this happens just in case α = β by (1) and (2). Then they do a little Bayesian calculation: there is a 1/4 prior that neither was biased, in which case the equality of credences is certain; there is a 3/4 prior that at least one was biased, in which case the credences would almost certainly be unequal (the probability that they’d both get the same erroneous result is zero given the uniform distribution of errors); so, the posterior that they are both correct is 1 (or 1 minus an infinitesimal). In that case, they will adjust their credences back to α and β (which are equal). This is the case of peer agreement.

Notice that peer agreement results in an adjustment of credence away from 1/2 (i.e., α^* is closer to 1/2 than α is, unless of course α = 1/2).

Second, suppose Alice and Bob notice that their credences in Q are different, i.e., α^* ≠ β^*. By (1) and (2), it follows that their first-order evaluations α and β were also different from one another. Now they reason as follows. Before they learned that their evaluations were different, there were four possibilities:

• EE: Alice erred and Bob erred
• EN: Alice erred but Bob did not err
• NE: Alice did not err but Bob erred
• NN: no error by either.

Each of these had equal probability 1/4. Upon learning that their evaluations were different, the last option was ruled out. Moreover, given the various uniform distribution assumptions, the exact values of the errors do not affect the probabilities of which possibility was the case. Thus, the EE, EN and NE options remain equally likely, but now have probability 1/3. If they knew they were in EE, then their credence should be 1/2—they have received no data. If they knew they were in EN, their credence should be β, since Bob’s evaluation of the evidence would be correct. If they knew they were in NE, their credence should be α, since Alice’s evaluation would be correct. But they don’t know which is the case, and the three cases are equally likely, so their new credence is:

3. α^** = β^** = (1/3)(1/2 + α + β) = (1/3)(2α^* + 2β^* − 1/2).

(They can calculate α and β from α^* and β^*, respectively.)

Now here’s the first interesting thing. In this model, the “split the difference” account of peer disagreement is provably wrong. Splitting the difference between α^* and β^* would result in (1/2)(α^* + β^*). It is easy to see that the only case where (3) generates the same answer as splitting the difference is when α^* + β^* = 1, i.e., when the credences of Alice and Bob prior to aggregation were equidistant from 1/2, in which case (3) says that they should go to 1/2.

And here is a second interesting thing. Suppose that α^* < β^*. Standard conciliationist accounts of peer disagreement (of which “split the difference” is an example) say that Alice should raise her credence and Bob should lower his. Does that follow from (3)? The answer is: sometimes. Here are some cases:

• α^* = 0.40, β^* = 0.55, α^** = β^** = 0.47
• α^* = 0.55, β^* = 0.65, α^** = β^** = 0.63
• α^* = 0.60, β^* = 0.65, α^** = β^** = 0.67
• α^* = 0.60, β^* = 0.70, α^** = β^** = 0.70.

Thus just by plugging some numbers in, we can find some conciliationist cases where Alice and Bob should meet in between, but we can also find a case (0.60 and 0.70) where Bob should stand pat, and a case (0.60 and 0.65) where both should raise their credence.

When playing with numbers, remember that by (1) and (2), the possible range for α^* and β^* is between 1/4 and 3/4 (since the possible range for α and β is from 0 to 1).

What can we prove? Well, let's first consider the case where α^* < 1/2 < β^*. Then it's easy to check that Bob needs to lower his credence and Alice needs to raise hers. That's a conciliationist result.

But what if both credences are on the same side of 1/2? Let say 1/2 < α^* < β^*. Then it turns out that:

4. Alice will always raise her credence

5. Bob will lower his credence if and only if β^* > 2α^* − 1/2

6. Bob will raise his credence if and only if β^* < 2α^* − 1/2.

In other words, Bob will lower his credence if his credence is far enough away from Alice’s. But if it’s moderately close to Alice’s, both Alice and Bob will raise their credences.

While the model I am working with is very artificial, this last result is pretty intuitive: if both of them have credences that are fairly close to each other, this supports the idea that at least one of them is right, which in turn undoes some of the effect of the α → α^* and β → β^* transformations in light of their data on their own unreliability.

So what do we learn about peer disagreement from this model? What we learn is that things are pretty complicated, too complicated to encompass in a simple non-mathematical formulation. Splitting the difference is definitely not the way to go in general. Neither is any conciliationism that makes the two credences move towards their mutual mean.

Of course, all this is under some implausible uniform distribution and independence assumptions, and a pretty nasty unreliability assumption that half the time we evaluate evidence biasedly. I have pretty strong intuitions that a lot of what I said depends on these assumptions. For instance, suppose that the random bias results in a uniform distribution of posterior on the interval 0 to 1, but one’s prior probability distribution for one’s evaluation of the evidence is not uniform but drops off near 0 and 1 (one doesn’t think it likely that the evidence will establish or abolish Q with certainty). Then if α (say) is close to 0 or 1, that’s evidence for bias, and a more complicated adjustment will be needed than that given by (1).

So things are even more complicated.