First of all, I think that the g and f functions should be the same. My reason is that to be completely satisfied with who gets the 20, both people should update their probabilities to the same value, so f(p,q) seems like the probability that EY and NB should both walk away with saying is the probability that EY deserved the 20. Since EY and NB trust each other, this should be the same if there is a single person with both beliefs learning they are equivalent. Some of the features I want in f are easier to justify in the g example, which might have caused some of my mistakes.
I think this question is inherently about how much we treat p and q as coming from independent sources of information. If we say that the sources are independent, then #4 is the only reasonable answer. However, the dependency of the evidence is not known.
I think f should have the following properties:
A: f(p,p)=p—If we wanted f to be a general rule for how to take two probabilities and output a probability that is an agreement of the two, then there is a danger in setting g(p,p) to anything other than p. In that we can repeatedly apply our rule and get different answers. We update p and q to g(p,q) and g(p,q), but then these are our new probabilities, so we will update them both to g(g(p,q),g(p,q)). Our new answer is not consistent under reflection. Therefore, I think g(p,p) should be p, and this makes me believe that f(p,p) should also be p. (Maybe I am working by analogy, but I still believe this.)
B: f(p,q)=1-f(1-q,1-p) -- This is just saying that the answer is symmetric with respect to swapping EY and NB.
C: f(p,q)=f(q,p) -- This is saying that the answer is symmetric with respect to swapping who says what. This property seems really necessary in the g problem.
D: f(1,q)=1 -- If EY knows he put the 20 in, he should get it back. In the g problem, if A is a theorem, and we learn B if and only if A, then we can prove B also.
E: p1>=p2, then f(p1,q)>=f(p2,q) -- If EY is more sure he should get the money, he shouldn’t get less money.
F: f(f(p,q),f(r,s))=f(f(p,r),f(q,s)) -- This relation doesn’t mean much for f, but for g, it is saying that the order in which we learn conjectures are equivalent shouldn’t change the final answer.
G: f is continuous—A small change in probability shouldn’t have a huge effect on the decision.
I think that most people would agree with B, C, and E, but looking at the comments, A and D are more controversial. F doesn’t make any sense unless you believe that f=g. I am not sure how people will feel about G. Notice that if B, C, and D together imply that f(1,0) is not defined, because it would have to be both 1 and 0, I think this is okay. You can hardly expect EY and NB to continue trusting each other after something like this, but it is necessary to say to be mathematically correct.
Now, to critique the proposed solutions.
1 Violates C and D (I didn’t check F)
2 Violates D
4 Violates A
3 Does not violate any of my features.
I did not do a good job of describing where #3 came from, so let me do better. #3 chooses the unique value f, such that if we wanted to update EYs probability to f and update NBs probability to f, this would take the same amount of evidence. It satisfies A because if they are both already the same, then it doesn’t take any evidence. It satisfies D, because no finite amount of evidence can bring you back from certainty.
We do not have enough information to say that #3 is the unique solution. If we were to try to, it would look like roughly like this:
If we think about the problem by looking at p =log(p/1-p). Then #3 just finds the f =(p +q )/2. E and A together tell us that f should be somewhere between p and q but it is not immediately clear that the arithmetic mean is the best compromise. However, I believe this implies that we can apply some monotone function h, such that h(f ) is the always the arithmetic mean of h(p ) and h(q ). B tells us that this monotone function must be an odd function (h(-x)=-h(x)).
From here, #3 assumes that h(x)=x, but if we were to take h(x)=x^3 for example, we would still meet all of our properties. We have freedom from the fact that we can weigh probabilities with different distances from 1⁄2 differently.
Now that I understand 4 (thanks again for the explanation!), this seems to be the key:
I think this question is inherently about how much we treat p and q as coming from independent sources of information. If we say that the sources are independent, then #4 is the only reasonable answer. However, the dependency of the evidence is not known.
I’m not sure that it makes sense for the general rule to do anything other than make an estimate of the amount of dependence in the evidence and update accordingly. You would of course need some kind of prior for that, but a Bayesian already needs a prior for everything anyway. Does that approach seem problematic for reasons I’m not thinking of?
Because you have two people, they might disagree on how independent their sources are, and further disagree on how independent their sources which told them their sources were independent are. This infinite regress must be stopped at some point, since they don’t have infinite time to compare all of their notes, and when it stops, the original question must be answered.
For me though, I understand that we cant do Bayesian updates completely rigorously unless we account for all the information, but since we are not perfect Bayesianists, I think the question of how well we can do on a first approximation is an important one.
As promised, here is my analysis.
First of all, I think that the g and f functions should be the same. My reason is that to be completely satisfied with who gets the 20, both people should update their probabilities to the same value, so f(p,q) seems like the probability that EY and NB should both walk away with saying is the probability that EY deserved the 20. Since EY and NB trust each other, this should be the same if there is a single person with both beliefs learning they are equivalent. Some of the features I want in f are easier to justify in the g example, which might have caused some of my mistakes.
I think this question is inherently about how much we treat p and q as coming from independent sources of information. If we say that the sources are independent, then #4 is the only reasonable answer. However, the dependency of the evidence is not known.
I think f should have the following properties:
A: f(p,p)=p—If we wanted f to be a general rule for how to take two probabilities and output a probability that is an agreement of the two, then there is a danger in setting g(p,p) to anything other than p. In that we can repeatedly apply our rule and get different answers. We update p and q to g(p,q) and g(p,q), but then these are our new probabilities, so we will update them both to g(g(p,q),g(p,q)). Our new answer is not consistent under reflection. Therefore, I think g(p,p) should be p, and this makes me believe that f(p,p) should also be p. (Maybe I am working by analogy, but I still believe this.)
B: f(p,q)=1-f(1-q,1-p) -- This is just saying that the answer is symmetric with respect to swapping EY and NB.
C: f(p,q)=f(q,p) -- This is saying that the answer is symmetric with respect to swapping who says what. This property seems really necessary in the g problem.
D: f(1,q)=1 -- If EY knows he put the 20 in, he should get it back. In the g problem, if A is a theorem, and we learn B if and only if A, then we can prove B also.
E: p1>=p2, then f(p1,q)>=f(p2,q) -- If EY is more sure he should get the money, he shouldn’t get less money.
F: f(f(p,q),f(r,s))=f(f(p,r),f(q,s)) -- This relation doesn’t mean much for f, but for g, it is saying that the order in which we learn conjectures are equivalent shouldn’t change the final answer.
G: f is continuous—A small change in probability shouldn’t have a huge effect on the decision.
I think that most people would agree with B, C, and E, but looking at the comments, A and D are more controversial. F doesn’t make any sense unless you believe that f=g. I am not sure how people will feel about G. Notice that if B, C, and D together imply that f(1,0) is not defined, because it would have to be both 1 and 0, I think this is okay. You can hardly expect EY and NB to continue trusting each other after something like this, but it is necessary to say to be mathematically correct.
Now, to critique the proposed solutions.
1 Violates C and D (I didn’t check F)
2 Violates D
4 Violates A
3 Does not violate any of my features.
I did not do a good job of describing where #3 came from, so let me do better. #3 chooses the unique value f, such that if we wanted to update EYs probability to f and update NBs probability to f, this would take the same amount of evidence. It satisfies A because if they are both already the same, then it doesn’t take any evidence. It satisfies D, because no finite amount of evidence can bring you back from certainty.
We do not have enough information to say that #3 is the unique solution. If we were to try to, it would look like roughly like this:
If we think about the problem by looking at p =log(p/1-p). Then #3 just finds the f =(p +q )/2. E and A together tell us that f should be somewhere between p and q but it is not immediately clear that the arithmetic mean is the best compromise. However, I believe this implies that we can apply some monotone function h, such that h(f ) is the always the arithmetic mean of h(p ) and h(q ). B tells us that this monotone function must be an odd function (h(-x)=-h(x)).
From here, #3 assumes that h(x)=x, but if we were to take h(x)=x^3 for example, we would still meet all of our properties. We have freedom from the fact that we can weigh probabilities with different distances from 1⁄2 differently.
Now that I understand 4 (thanks again for the explanation!), this seems to be the key:
I’m not sure that it makes sense for the general rule to do anything other than make an estimate of the amount of dependence in the evidence and update accordingly. You would of course need some kind of prior for that, but a Bayesian already needs a prior for everything anyway. Does that approach seem problematic for reasons I’m not thinking of?
Because you have two people, they might disagree on how independent their sources are, and further disagree on how independent their sources which told them their sources were independent are. This infinite regress must be stopped at some point, since they don’t have infinite time to compare all of their notes, and when it stops, the original question must be answered.
For me though, I understand that we cant do Bayesian updates completely rigorously unless we account for all the information, but since we are not perfect Bayesianists, I think the question of how well we can do on a first approximation is an important one.