On Sun, Mar 8, 2020 at 7:59 PM Russell Standish <[email protected]> wrote:
> On Sun, Mar 08, 2020 at 07:08:25PM +1100, Bruce Kellett wrote: > > On Sun, Mar 8, 2020 at 6:14 PM Russell Standish <[email protected]> > wrote: > > > > On Thu, Mar 05, 2020 at 09:45:38PM +1100, Bruce Kellett wrote: > > > On Thu, Mar 5, 2020 at 5:26 PM Russell Standish < > [email protected]> > > wrote: > > > > > > But a very large proportion of them (→1 as N→∞) will report > being > > > within ε (called a confidence interval) of 50% for any given > ε>0 > > > chosen at the outset of the experiment. This is simply the law > of > > > large numbers theorem. You can't focus on the vanishingly small > > > population that lie outside the confidence interval. > > > > > > > > > This is wrong. > > > > Them's fighting words. Prove it! > > > > > > I have, in other posts and below. > > You didn't do it below, that's why I said prove it. What you wrote > below had little bearing on what I wrote. > I outlines the proof in my reply to your other post tonight. Besides, the proof is in Kent's paper arXiv:0905.0624. > > In the binary situation where both outcomes occur for every > > > trial, there are 2^N binary sequences for N repetitions of the > > experiment. This > > > set of binary sequences exhausts the possibilities, so the same > sequence > > is > > > obtained for any two-component initial state -- regardless of the > > amplitudes. > > > > > You appear to assume that the natural probability in this > situation is p > > = 0.5 > > > and, what is more, your appeal to the law of large numbers applies > only > > for > > > single-world probabilities, in which there is only one outcome on > each > > trial. > > > > I didn't mention proability once in the above paragraph, not even > > implicitly. I used the term "proportion". That the proportion will be > > equal to the probability in a single universe case is a frequentist > > assumption, and should be uncontroversial, but goes beyond what I > > stated above. > > > > > > Sure. But the proportion of the 2^N sequences that exhibit any > particular p > > value (proportion of 1's) decreases with N. > > > > So what? You claim that the proportion reporting p ~ 0.5 goes to one as N --> oo. That is manifestly false. The absolute number increases with the number of trials, but the proportion of the 2^N copies at the end of the N trials decreases as 1/sqrt(N). > > In order to infer a probability of p = 0.5, your branch data must > have > > > approximately equal numbers of zeros and ones. The number of > branches > > with > > > equal numbers of zeros and ones is given by the binomial > coefficient. For > > large > > > even N = 2M trials, this coefficient is N!/M!*M!. Using the > Stirling > > > approximation to the factorial for large N, this goes as > 2^N/sqrt(N) > > (within > > > factors of order one). Since there are 2^N sequences, the > proportion with > > n_0 = > > > n_1 vanishes as 1/sqrt(N) for N large. > This is the nub of the proof you wanted. > I wasn't talking about that. I was talking about the proportion of > > sequences whose ratio of 0 bits to 1 bits lie within ε of 0.5, rather > > than the proportion of sequences that have exactly equal 0 or 1 > > bits. That proportion grows as sqrt N. > > > > > > > > No, it falls as 1/sqrt(N). Remember, the confidence interval depends on > the > > standard deviation, and that falls as 1/sqrt(n). Consequently deviations > from > > equal numbers of zeros and ones for p to be within the CI of 0.5 must > decline > > as n becomes large > > > > The value ε defined above is fixed at the outset. It is independent of > N. Maybe I incorrectly called it a confidence interval, although it is > surely related. > Calling it a confidence interval certainly threw me. If it is meant to be a fixed interval independent of N, then OK. But that is not a useful concept. For any fixed interval around p = 0.5, relative frequencies at the limits of that interval will eventually estimate p values for which the CI does not include 0.5. So they can no longer infer that the probability is 0.5. Since the CI decreases with N, the proportion of the total who infer any particular value for the probability decreases with N. > The number of bitstrings having a ratio of 0 to 1 within ε of 0.5 > grows as √N. > > IIRC, a confidence interval is the interval of a fixed proportion, ie we > can be 95% confident that strings will have a ratio between 49.5% and > 51.5%. That interval (49.5% and 51.5%) will decrease as √N for fixed > confidence level (95%). > > > > > > > > Now sequences with small departures from equal numbers will still > give > > > probabilities within the confidence interval of p = 0.5. But this > > confidence > > > interval also shrinks as 1/sqrt(N) as N increases, so these > additional > > > sequences do not contribute a growing number of cases giving p ~ > 0.5 as N > > > increases. > > > > The confidence interval ε is fixed. > > > > > > No, it is not. The width of, say the 95% CI, decreases with N since the > > standard deviation falls as 1/sqrt(N). > > Which only demonstrates my point. An increasing number of strings will > lie in the fixed interval ε. I apologise if I used the term "confidence > interval" in a nonstandard way. > An increasing number of strings will lead to a probability estimate of p = 0.5, sure. But since the total number of strings grows more quickly, the proportion seeing any particular value for p will decline with increasing N. That is really the heart of the matter. The other argument against probabilities in Everett is that the the set of all bit strings of length N is fixed -- it has 2^N members. And those same bit strings will be obtained whatever the coefficients of the initial superposition. So both, the concept of probability has no 3p meaning, and the 1p probabilities are necessarily unrelated to superposition amplitudes. Bruce -- You received this message because you are subscribed to the Google Groups "Everything List" group. To unsubscribe from this group and stop receiving emails from it, send an email to [email protected]. To view this discussion on the web visit https://groups.google.com/d/msgid/everything-list/CAFxXSLTnBKc%2Bc2mF8tv3-ARm-3sva4vr8ONeYebUzwFDfChyow%40mail.gmail.com.
