On 3/5/2020 1:57 PM, Bruce Kellett wrote:
On Fri, Mar 6, 2020 at 8:08 AM 'Brent Meeker' via Everything List
<[email protected]
<mailto:[email protected]>> wrote:
On 3/5/2020 2:45 AM, Bruce Kellett wrote:
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. So, again
within factors of order unity, the proportion of sequences
consistent with p = 0.5 decreases without limit as N increases.
So it is not the case that a very large proportion of the binary
strings will report p = 0.5. The proportion lying outside the
confidence interval of p = 0.5 is not vanishingly small -- it
grows with N.
I agree with you argument about unequal probabilities, in which
all the binomial sequences occur anyway leading to inference of
p=0.5. But in the above paragraph you are wrong about the how the
probability density function of the observed value changes as
N->oo. For any given interval around the true value, p=0.5, the
fraction of observed values within that interval increases as
N->oo. For example in N=100 trials, the proportion of observers
who calculate an estimate of p in the interval (0.45 0.55) is
0.68. For N=500 it's 0.975. For N=1000 it's 0.998.
Confidence intervals are constructed to include the true value
with some fixed probability. But that interval becomes narrower
as 1/sqrt(N).
So the proportion lying inside and outside the interval is
relatively constant, but the interval gets narrower.
I think I am beginning to see why we are disagreeing on this. You are
using the normal approximation to the binomial distribution for a
large sequence of trials with some fixed probability of success on
each trial. In other words, it is as though you consider the 2^N
binary strings of length N to have been generated by some random
process, such as coin tosses or the like, with some prior fixed
probability value. Each string is then constructed as though the
random process takes place in a single word, so that there is only one
outcome for each toss.
Given such an ensemble, the statistics you cite are undoubtedly
correct: as the length of the string increases, the proportion of each
string within some interval of the given probability increases -- that
is what the normal approximation to the binomial gives you. And as N
increases, the confidence interval shrinks, so the proportion within a
confidence interval is approximately constant. But note these are the
proportions within each string as generated with some fixed
probability value. If you take an ensemble of such strings, the the
result is even more apparent, and the proportion of strings in which
the probability deviates significantly from the prior fixed value
decreases without limit.
That is all very fine. The problem is that this is not the ensemble of
strings that I am considering!
The set of all possible bit strings of length N is not generated by
some random process with some fixed probability. The set is generated
entirely deterministically, with no mention whatsoever of any
probability. Just think about where these strings come from. You
measure the spin of a spin-half particle. The result is 0 in one
branch and 1 in the other. Then the process is repeated, independently
in each branch, so the 1-branch splits into a 11-branch and a
10-branch; and the 0-branch splits into a 01-branch and a 00-branch.
This process goes on for N repetitions, generating all possible bit
strings of length N in an entirely deterministic fashion. The process
is illustrated by Sean Carroll on page 134 of his book.
Given the nature of the ensemble of bit strings that I am considering,
the statistical results I quote are correct, and your statistics are
completely inappropriate. This may be why we have been talking at
cross purposes. I suspect that Russell has a similar misconception
about the nature of the bit strings under consideration, since he
talked about statistical results that could only have been obtained
from an ensemble of randomly generated strings.
Yes, I understand that. And I understand that you have been talking
about Everett's original idea in which at each split both results
obtain, one in each branch...with no attribute of weight or probability
or other measure. It's just 0 and 1. Which generates all strings of
zeros and ones. This ensemble of sequences has the same statistics as
random coin flipping sequences, even though it's deterministic. But it
doesn't have the same statistics as flipping an unfair coin, i.e. when
a=/=b. So to have a multiple world interpretation that produces
statistics agreeing with the Born rule one has to either assign weights
to each of the two worlds at a split and then interpret those as
probability amplitudes OR postulate that the splits are into many copies
so that the branch count gives the Born statistics. Consider the many
copies case as an ensemble and it will reproduce the Born statistics
even though it is deterministic. This is easy to see because every
sequence a single observer has seen is the result of a random choice at
the split of which path you call "that observer".
Brent
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