I am applying a theory of probability that I’m taking as reasonable until I reach a contradiction that shows that it isn’t (proof by contradiction). If P is reasonable, it should give a value to statements like P(T | awake), and that value should follow the normal Bayes rules.
I don’t think I ever injected a specific value that wasn’t already there in the assumptions?
You say: “being in room 1 is equally consistent with or . So while .” Write this out in equations using Bayes law and you’ll find you need to assume a value of P(T | awake).
Here it is with plugging in an unknown t instead of assuming a value of Q(T|awake):
For this to equal your value of 0.5, you must be assuming that t := Q(T|awake) = 2⁄3. I suspect you heuristically did “room 1 is ‘equally consistent’ with T and H meaning the the likelihoods are equal and so their posteriors are equal”. However, the likelihoods are not the same: room 1 is more consistent with H than with T in the sense of likelihoods and Bayes rule uses this. If you still think I’m wrong, please write out exactly the computation you’re claiming, which you haven’t yet done.
To get the values of Q(T| awake, room 2) and Q(T| awake, room 1), I am applying “simple Bayes” because upon seeing awake and room 1, there is no longer any anthropic uncertainty as to who you are in the world.
Then, formally, I am assuming that Q is a well defined probability in general (obeying all the Bayesian equations, amongst others) and aiming to get a contradiction from this assumption.
So I set up an equation involving terms like Q(room 1 | awake) in order to infer what these would have to be. I’ve used simple Bayes to pin down some of the terms, and then I’m using the assumption that Q is well defined to set up an equation between the known terms and the unknown terms.
And then, plugging in the known value of Q(T | awake, room 1), one deduces Q(T | awake) = 2⁄3. I haven’t assumed that value; I’ve deduced it, under the “Q is reasonable, martingale, and simple Bayes works” assumptions (actually, all that I can deduce—and all that I need—is that Q(T | awake) > 1⁄2; getting it to be 2⁄3 requires slightly stronger symmetry assumptions).
I can also deduce Q(T | awake) = 1⁄2 (martingale from Sunday to Monday), and thus get a contradiction. One of my assumptions must be wrong. Q cannot simultaneously obey Bayesian updating, obey the martingale, and obey simple Bayes.
“plugging in the known value of Q(T | awake, room 1),”
But that is not known! That’s the value I am contesting your derivation of. You have asserted that it is 0.5 supposedly due to Bayes law but I am telling you that the math does not work out that way. If you want to convince me that Bayes law forces Q(T|awake, room 1) to be 0.5, then you need to write out that application of Bayes law and derive the number 0.5, not simply plug it in.
The condition I’m calling “simple Bayes” is that, if there is no doubt as to which agent you are in the world, then proceed by taking the prior over worlds and updating on the evidence “there exists a person in this world who have made this observation” (in non-anthropic situations, “there exists a person who has observed history H” and “I have observed history H” contain the same information).
On Monday, there is uncertainty are to which agent SB is. On Tuesday there is not: the two agents in the Tails world can tell each other apart, based on room number observation.
So simple Bayes applies to Sunday and Tuesday, but not to Monday.
On Sunday, the observations are independent of the coin flip, so P(“I saw Sunday” | w_T) = P(“I saw Sunday” | w_H) = 1.
If h_0 = (“I saw Sunday”), then P(w_T | h_0) = P(h_0 | w_T)P(w_T)/((P(h_0 | w_T)P(w_T) + P(h_0 | w_H)P(w_H)) = 1*(1/2)/(1/2+1/2)=1/2. Same thing for P(w_H | h_0).
On Tuesday, in the tails world, there exists, with certainty, an agent who has observed: h_1=(“it’s Sunday”, “I’m awake on Monday”, “it’s Tuesday and my room number is 1”). Similarly, in the heads world, there exists, with certainty, an agent (the only agent) with the same observation.
Since both exist with certainty, then the formula is the same as before: P(w_T | h_1) = P(h_1 | w_T)P(w_T)/((P(h_1 | w_T)P(w_T) + P(h_1 | w_H)P(w_H)) = 1*(1/2)/(1/2+1/2)=1/2, and the same for P(w_H | h_1).
So the naive direct application of Bayes in non-anthropic situations gives us these values. The impossibility result is just that, given these values for Sunday and Tuesday, there are no values on Monday (the time-slice where we can’t use simple Bayes because there is genuine uncertainty as to which agent SB is) that allow the martingale condition to extend between Sunday and Tuesday.
Note that I’m not saying that SIA or SSA are wrong or that you can’t do anthropic probability. I’m saying that if you do do anthropic probability, you have to drop some intuitive properties possessed by standard probability.
For instance SIA drops the martingale condition from Sunday to Monday (indeed SIA always obeys simple Bayes). SSA (which is less uniquely defined) either drops the martingale condition from Monday to Tuesday or drops simple Bayes on Tuesday (using “centered worlds” is an explicit acknowledgement of dropping simple Bayes).
Thank you for the explanation. We need some better notation for this: write I(x) to mean that I observed x (or will observe it). Write E(x) to mean that I know that there exists with certainty someone who observed x (or will observe it).
First, in general, we don’t expect P(I(x)) = P(E(x)). For example, P(I(room 1)|I(room 2)) = 0 but P(E(room 1) | I(room 2)) = 1.
You define “simple Bayes” to mean:
P(y | I(x)) = P(y | E(x)) for any y and x
I would argue that you really need to pick a totally different name for this property since it doesn’t have anything to do with Bayes’ law. (If anything, I would call it “nonstandard Bayes”.) And your definition of it in your article needs to be clearer: as stated, it’s just about I(x) without mentioning E(x).
This is also the same crux that the paradox has always revolved around: does finding out who you are give you information? I think that’s closer to the standard phrasing and makes it more clear what you’re being asked to give up or not. You’re not being asked to give up Bayes’ law: that’s always true.
I am applying a theory of probability that I’m taking as reasonable until I reach a contradiction that shows that it isn’t (proof by contradiction). If P is reasonable, it should give a value to statements like P(T | awake), and that value should follow the normal Bayes rules.
I don’t think I ever injected a specific value that wasn’t already there in the assumptions?
You say: “being in room 1 is equally consistent with
Here it is with plugging in an unknown t instead of assuming a value of Q(T|awake):
Q(T| awake, room 1) = Q(room 1 | T, awake) Q (T | awake) / Q(room 1 | awake)
If Q(T|awake) = t, then Q(H | awake) = 1-t and Q(room 1| T, awake) = 0.5 etc. give that Q(room 1 | awake) = t /2 + (1-t) = 1 - t/2
So:
Q(T| awake, room 1) = (1/2) (t) / (1-t/2) = t/(2 - t)
For this to equal your value of 0.5, you must be assuming that t := Q(T|awake) = 2⁄3. I suspect you heuristically did “room 1 is ‘equally consistent’ with T and H meaning the the likelihoods are equal and so their posteriors are equal”. However, the likelihoods are not the same: room 1 is more consistent with H than with T in the sense of likelihoods and Bayes rule uses this. If you still think I’m wrong, please write out exactly the computation you’re claiming, which you haven’t yet done.
To get the values of Q(T| awake, room 2) and Q(T| awake, room 1), I am applying “simple Bayes” because upon seeing awake and room 1, there is no longer any anthropic uncertainty as to who you are in the world.
Then, formally, I am assuming that Q is a well defined probability in general (obeying all the Bayesian equations, amongst others) and aiming to get a contradiction from this assumption.
So I set up an equation involving terms like Q(room 1 | awake) in order to infer what these would have to be. I’ve used simple Bayes to pin down some of the terms, and then I’m using the assumption that Q is well defined to set up an equation between the known terms and the unknown terms.
So one can write:
Q(T| awake, room 1) = (1/2) (t) / (1-t/2) = t/(2 - t)
And then, plugging in the known value of Q(T | awake, room 1), one deduces Q(T | awake) = 2⁄3. I haven’t assumed that value; I’ve deduced it, under the “Q is reasonable, martingale, and simple Bayes works” assumptions (actually, all that I can deduce—and all that I need—is that Q(T | awake) > 1⁄2; getting it to be 2⁄3 requires slightly stronger symmetry assumptions).
I can also deduce Q(T | awake) = 1⁄2 (martingale from Sunday to Monday), and thus get a contradiction. One of my assumptions must be wrong. Q cannot simultaneously obey Bayesian updating, obey the martingale, and obey simple Bayes.
“plugging in the known value of Q(T | awake, room 1),”
But that is not known! That’s the value I am contesting your derivation of. You have asserted that it is 0.5 supposedly due to Bayes law but I am telling you that the math does not work out that way. If you want to convince me that Bayes law forces Q(T|awake, room 1) to be 0.5, then you need to write out that application of Bayes law and derive the number 0.5, not simply plug it in.
Ok, got the issue. Thanks!
The condition I’m calling “simple Bayes” is that, if there is no doubt as to which agent you are in the world, then proceed by taking the prior over worlds and updating on the evidence “there exists a person in this world who have made this observation” (in non-anthropic situations, “there exists a person who has observed history H” and “I have observed history H” contain the same information).
On Monday, there is uncertainty are to which agent SB is. On Tuesday there is not: the two agents in the Tails world can tell each other apart, based on room number observation.
So simple Bayes applies to Sunday and Tuesday, but not to Monday.
On Sunday, the observations are independent of the coin flip, so P(“I saw Sunday” | w_T) = P(“I saw Sunday” | w_H) = 1.
If h_0 = (“I saw Sunday”), then P(w_T | h_0) = P(h_0 | w_T)P(w_T)/((P(h_0 | w_T)P(w_T) + P(h_0 | w_H)P(w_H)) = 1*(1/2)/(1/2+1/2)=1/2. Same thing for P(w_H | h_0).
On Tuesday, in the tails world, there exists, with certainty, an agent who has observed: h_1=(“it’s Sunday”, “I’m awake on Monday”, “it’s Tuesday and my room number is 1”). Similarly, in the heads world, there exists, with certainty, an agent (the only agent) with the same observation.
Since both exist with certainty, then the formula is the same as before: P(w_T | h_1) = P(h_1 | w_T)P(w_T)/((P(h_1 | w_T)P(w_T) + P(h_1 | w_H)P(w_H)) = 1*(1/2)/(1/2+1/2)=1/2, and the same for P(w_H | h_1).
So the naive direct application of Bayes in non-anthropic situations gives us these values. The impossibility result is just that, given these values for Sunday and Tuesday, there are no values on Monday (the time-slice where we can’t use simple Bayes because there is genuine uncertainty as to which agent SB is) that allow the martingale condition to extend between Sunday and Tuesday.
Note that I’m not saying that SIA or SSA are wrong or that you can’t do anthropic probability. I’m saying that if you do do anthropic probability, you have to drop some intuitive properties possessed by standard probability.
For instance SIA drops the martingale condition from Sunday to Monday (indeed SIA always obeys simple Bayes). SSA (which is less uniquely defined) either drops the martingale condition from Monday to Tuesday or drops simple Bayes on Tuesday (using “centered worlds” is an explicit acknowledgement of dropping simple Bayes).
Thank you for the explanation. We need some better notation for this: write I(x) to mean that I observed x (or will observe it). Write E(x) to mean that I know that there exists with certainty someone who observed x (or will observe it).
First, in general, we don’t expect P(I(x)) = P(E(x)). For example, P(I(room 1)|I(room 2)) = 0 but P(E(room 1) | I(room 2)) = 1.
You define “simple Bayes” to mean:
P(y | I(x)) = P(y | E(x)) for any y and x
I would argue that you really need to pick a totally different name for this property since it doesn’t have anything to do with Bayes’ law. (If anything, I would call it “nonstandard Bayes”.) And your definition of it in your article needs to be clearer: as stated, it’s just about I(x) without mentioning E(x).
This is also the same crux that the paradox has always revolved around: does finding out who you are give you information? I think that’s closer to the standard phrasing and makes it more clear what you’re being asked to give up or not. You’re not being asked to give up Bayes’ law: that’s always true.