Re: practical reasoning and strong SSA
On Mon, Jun 07, 1999 at 11:15:52PM -0700, [EMAIL PROTECTED] wrote: > Yes, I think it is the former. The reference set is those observers who > are having identical observations to me. We all see the same display. > In some worlds Pi actually has that value, and in other worlds it has a > different value, but all see the same mathematica display. What is the > problem with this line of reasoning? That seems to contradict what you said earlier: > They would reason something like: > > P (I observe "N[Pi]=3.14159" | PI == 3.14159 AND I am here/now, doing > this) is very high. > P (I observe "N[Pi]=3.14159" | PI != 3.14159 AND I am here/now, doing > this) is very low. If the reference class only consist of people who observe "N[Pi]=3.14159" then both probabilities would be one and you wouldn't be able to make any conclusions. Also I don't understand what you mean by "in other worlds it [Pi] has a different value". Pi always has the same value since it is a mathematical constant not a physical constant. > Nick Bostrom and Robin Hanson had a debate on extropians, which I was > not able to follow very well. One of the issues seemed to be whether > the reference set should include rocks. That is interesting that the > SIA can be seen to follow from the assumption that the reference set > should be expanded like this. > > So does your Bayesian reasoning example work OK with either the strong > SSA and the "extra strong" SSA? Or are you saying now that the latter > is the only consistent position to allow you to derive the kinds of > implications in your mathematica example? My example works ok with both. However I think there are reasons to prefer the extra strong SSA over the strong SSA. I gave these reasons earlier.
Re: practical reasoning and strong SSA
Wei Dai, <[EMAIL PROTECTED]>, writes: > But where does P (I observe "N[Pi]=3.14159" | PI == 3.14159 AND I am > here/now, doing this) come from (how is it defined) without the Strong > SSA? I don't see that the SSSA would play a role, because you are explicity excluding the possibility that you are someone else in the specific conditional probability under consideration. The strong SSA says that you should consider yourself a random selection from among all observers. But we are stipulating in the conditional probability that you are, in fact, yourself, here and now. Whether you might have been someone else is therefore irrelevant to the calculation and to the definition. The way I have seen this calculation justified is to consider that you are randomly chosen from among all possible observers who are consistent with what you are experiencing here/now. This is more limited than the SSA because the reference set is just those observers who share identical mental states. Each person does his probability calculations with regard to that reference set. From his perspective, he is in one of many possible universes, constrained by what he observes and knows. But there is no chance that he is a tentacled alien on alpha centauri. Consider the "extra strong" SSA where the reference set is all subsystems of all universes, not just conscious ones, so that you consider the possibility that you might have been a rock in your calculations. You will get the same answer for the probability above under this assumption as well as the conventional SSSA, because in both cases we exclude from consideration any other possibility than that you are you. So this extra strong SSA should be just as good as the strong SSA in terms of defining the conditional probability, so therefore the strong SSA must not be a necessary philosophical underpinning. Hal
Re: practical reasoning and strong SSA
Wei Dai, <[EMAIL PROTECTED]>, writes: > But where does P (I observe "N[Pi]=3.14159" | PI == 3.14159 AND I am > here/now, doing this) come from (how is it defined) without the Strong > SSA? I don't see that the SSSA would play a role, because you are explicity excluding the possibility that you are someone else in the specific conditional probability under consideration. The strong SSA says that you should consider yourself a random selection from among all observers. But we are stipulating in the conditional probability that you are, in fact, yourself, here and now. Whether you might have been someone else is therefore irrelevant to the calculation and to the definition. The way I have seen this calculation justified is to consider that you are randomly chosen from among all possible observers who are consistent with what you are experiencing here/now. This is more limited than the SSA because the reference set is just those observers who share identical mental states. Each person does his probability calculations with regard to that reference set. From his perspective, he is in one of many possible universes, constrained by what he observes and knows. But there is no chance that he is a tentacled alien on alpha centauri. Consider the "extra strong" SSA where the reference set is all subsystems of all universes, not just conscious ones, so that you consider the possibility that you might have been a rock in your calculations. You will get the same answer for the probability above under this assumption as well as the conventional SSSA, because in both cases we exclude from consideration any other possibility than that you are you. So this extra strong SSA should be just as good as the strong SSA in terms of defining the conditional probability, so therefore the strong SSA must not be a necessary philosophical underpinning. Hal
Re: practical reasoning and strong SSA
Wei Dai, <[EMAIL PROTECTED]>, writes: > > I'm not sure I am on the right track here... > > You are. :) Now the point I was trying to make was that without the strong > SSA, P(I observe "N[Pi]=3.14159" | PI == 3.14159) would be ill-defined, so > we could not conclude that P(PI == 3.14159 | I observe "N[Pi]=3.14159") > is high. Do you agree? I think when most people do this, they implicitly assume that they are themselves, so to speak; that they are doing what they are doing and seeing what they are seeing. They would reason something like: P (I observe "N[Pi]=3.14159" | PI == 3.14159 AND I am here/now, doing this) is very high. P (I observe "N[Pi]=3.14159" | PI != 3.14159 AND I am here/now, doing this) is very low. >From this they conclude: P (PI == 3.14159 | I observe "N[Pi]=3.14159" AND I am her/now, doing this) is very high. Does this seem like legitimate reasoning? I'm not sure if the math works out formally but it so this would seem to be an OK way to work it. Hal
Re: practical reasoning and strong SSA
Wei Dai, <[EMAIL PROTECTED]>, writes: > Bayesian analysis in general does not depend on the Strong SSA, but any > Bayesian analysis where you try to compute P(X | I observe Y) does because > you need the Strong SSA to compute P(I observe Y | X) and P(I observe Y | > not X). > > My example is one of classical Bayesian reasoning, but it is slightly > different from the way you put it, because I don't have direct knowledge > that Mathematica outputs a 9 for the sixth digit of Pi. I do know that I > am reading "N[Pi]=3.14159", and Strong SSA is needed to derive the > probability that I am reading "N[Pi]=3.14159" if Pi doesn't begin with > 3.14159. So, are you saying that P(I observe "N[Pi]=3.14159" | PI == 3.14159) should be considered the probability, given that PI has that value, that a randomly chosen observer-moment is of me, sitting at a computer, seeing this value display, out of all observer-moments in all possible universes? In that case the probability would be extremely low, since only a tiny fraction of all observer-moments would consist of me doing that. Maybe it would be something like 1E-100. However, the probability P(I observe "N[Pi]=3.14159" | PI != 3.14159) would be even lower, since not only would the randomly chosen observer moment have to be me running mathematica, but it would also be necessary that mathematica is wrong. Maybe this probability would be 1E-110. We would then run the Bayesian formula with these two extremely low conditional probabilities. From this we would conclude that P(PI == 3.14159 | I observe "N[Pi]=3.14159") would be high, as we expect. Is this how you would work this problem, based on the strong SSA? I'm not sure I am on the right track here... Hal
Re: practical reasoning and strong SSA
On Tue, Jun 01, 1999 at 10:07:05PM -0700, [EMAIL PROTECTED] wrote: > I don't follow where the dependence on SSSA comes from. This is the > assumption that each observer-moment should be considered as a random > selection from all observer-moments in the universe (broadly defined). > > Your example would seem to be classical Bayesian reasoning. A priori > you don't know whether the sixth digit of pi is a 9, so you give that > 1/10 probability. After seeing Mathematica's output, you estimate the > probability that it would say it is a 9 when the actual digit is not a 9 > (i.e. make a mistake), which is very small. You feed that into the Bayes > formula and end up with a strong probability that the sixth digit is 9. > > Are you saying that Bayesian analysis depends on the Strong SSA? Could > you elaborate on this? Bayesian analysis in general does not depend on the Strong SSA, but any Bayesian analysis where you try to compute P(X | I observe Y) does because you need the Strong SSA to compute P(I observe Y | X) and P(I observe Y | not X). My example is one of classical Bayesian reasoning, but it is slightly different from the way you put it, because I don't have direct knowledge that Mathematica outputs a 9 for the sixth digit of Pi. I do know that I am reading "N[Pi]=3.14159", and Strong SSA is needed to derive the probability that I am reading "N[Pi]=3.14159" if Pi doesn't begin with 3.14159.
Re: practical reasoning and strong SSA
Wei Dai writes: > When I learn a new way to thinking I tend to forget how to think the old > way. I just typed into Mathematica "N[Pi]" and it displayed to me > "3.14159". So I think that gives me reason to believe the first 6 digits > in the decimal expansion of Pi is 3.14159 because if it wasn't the case > my current experience would be very atypical. More formally, the > probability that I am reading "N[Pi] = 3.14159" given that the first 6 > digits of Pi is not 3.14159 is very small compared to the probability that > I am reading "N[Pi] = 3.14159" given that the first 6 digits of Pi IS > 3.14159. > This and similar kinds of reasoning depend on the Strong SSA (as defined > by Hal and Nick). I don't follow where the dependence on SSSA comes from. This is the assumption that each observer-moment should be considered as a random selection from all observer-moments in the universe (broadly defined). Your example would seem to be classical Bayesian reasoning. A priori you don't know whether the sixth digit of pi is a 9, so you give that 1/10 probability. After seeing Mathematica's output, you estimate the probability that it would say it is a 9 when the actual digit is not a 9 (i.e. make a mistake), which is very small. You feed that into the Bayes formula and end up with a strong probability that the sixth digit is 9. Are you saying that Bayesian analysis depends on the Strong SSA? Could you elaborate on this? Hal
