On 3/4/2020 7:54 PM, Bruce Kellett wrote:
On Thu, Mar 5, 2020 at 2:02 PM 'Brent Meeker' via Everything List <[email protected] <mailto:[email protected]>> wrote:

    On 3/4/2020 6:45 PM, Bruce Kellett wrote:
    On Thu, Mar 5, 2020 at 1:34 PM 'Brent Meeker' via Everything List
    <[email protected]
    <mailto:[email protected]>> wrote:

        On 3/4/2020 6:18 PM, Bruce Kellett wrote:

        But one cannot just assume the Born rule in this case -- one
        has to use the data to verify the probabilistic predictions.
        And the observers on the majority of branches will get data
        that disconfirms the Born rule. (For any value of the
        probability, the proportion of observers who get data
        consistent with this value decreases as N becomes large.)

        No, that's where I was disagreeing with you.  If "consistent
        with" is defined as being within some given fraction, the
        proportion increases as N becomes large.  If the probability
        of the an even is p and q=1-p then the proportion of events
        in N trials within one std-deviation of p approaches 1/e and
        N->oo and the width of the one std-deviation range goes down
        at 1/sqrt(N). So the distribution of values over the ensemble
        of observers becomes concentrated near the expected value,
        i.e. is consistent with that value.



    But what is the expected value? Does that not depend on the
    inferred probabilities? The probability p is not a given -- it
    can only be inferred from the observed data. And different
    observers will infer different values of p. Then certainly, each
    observer will think that the distribution of values over the 2^N
    observers will be concentrated near his inferred value of p. The
    trouble is that that this is true whatever value of p the
    observer infers -- i.e., for whatever branch of the ensemble he
    is on.

    Not if the branches are unequally weighted (or numbered), as
    Carroll seems to assume, and those weights (or numbers) define the
    probability of the branch in accordance with the Born rule.  I'm
    not arguing that this doesn't have to be put in "by hand".  I'm
    arguing it is a way of assigning measures to the multiple worlds
    so that even though all the results occur, almost all observers
    will find results close to the Born rule, i.e. that self-locating
    uncertainty will imply the right statistics.


But the trouble is that Everett assumes that all outcomes occur on every trial. So all the branches occur with certainty -- there is no "weight" that differentiates different branches. That is to assume that the branches occur with the probabilities that they would have in a single-world scenario. To assume that branches have different weights is in direct contradiction to the basic postulates the the many-worlds approach. It is not that one can "put in the weights by hand"; it is that any assignment of such weights contradicts that basis of the interpretation, which is that all branches occur with certainty.

All branches occur with certainty so long as their weight>0. Yes, Everett simply assumed they all occur.  Take a simple branch counting model.  Assume that at each trial a there are a 100 branches and a of them are |0> and b are |1> and the values are independent of the prior values in the sequence.  So long as a and b > 0.1 every value, either |0> or |1> will occur at every branching.  But almost all observers, seeing only one sequence thru the branches, will infer P(0)~|a|^2 and P(1)~|b|^2.

Do you really disagree that there is a way to assign weights or probabilities to the sequences that reproduces the same statistics as repeating the N trials many times in one world?  It's no more than saying that one-world is an ergodic process.

Brent


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