On 2/16/2020 9:48 PM, Bruce Kellett wrote:
On Mon, Feb 17, 2020 at 4:13 PM 'Brent Meeker' via Everything List
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On 2/16/2020 2:17 PM, Bruce Kellett wrote:
That is where the proof given by Kent comes into play. If in the
N trials you observe pN zeros and (1-p)N ones, you estimate the
probability for zero to be p, within certain confidence limits
that depend on the number of trials. Note that this is precisely
the 1p perspective, one person taking his actual data and making
some estimates. This person then considers that some other person
might have obtained r zeros, rather than the pN that he obtained.
Applying the binomial theorem, he estimates the probability for
this to occur as p^r(1-p)^{N-r}. This goes to zero in the limit
as N becomes very large, so our original observer believes that
he has the correct probability, since the probability of results
significantly deviant from his goes to zero as N becomes large.
The problem, of course, is that this reasoning applies equally
well for all the inhabitants (from their individual first-person
perspectives), whatever relative frequency p they see on their
branch. All of them conclude that their relative frequencies
represent (to a very good approximation) the branch weights. They
clearly can't all be right, so either there is no actual
probability underlying the events and their calculations are
misguided, or the theory itself is incoherent.
But exactly the same reasoning applies for any given true value of
p. There will be different estimates by different experimenters
and they can't all be right. Each will infer that any proportion
other than the one he observed will have zero measure in the limit
N->oo.
Exactly right. That is what my example of spin measurements on an
ensemble of equally prepared spin states comes into play. If all 2^N
bit strings are realized for one orientation of the S-G magnet, then
exactly the same 2^N bit strings are realized for every other
orientation.
?? Suppose the ensemble is equally prepared in spin-up. What does it
mean to say all 2^N bit strings are realized for the S-G oriented
left/right? We may expect they will be for any number of trials >>N.
But certainly not for the S-G oriented up/down.
Consequently, the coefficients in the expansion play no role in
determining the data, and it makes no sense to talk of "the true value
of p". There is no such true value if all values are realized.
In Kent's thought experiment, if you consider the self-location as
probabilistic then it's exactly the same as taking a sample of N
from an ensemble for which p=0.5 is the true proportion. I think
you prove too much by saying the estimate of any proportion of the
other observers has zero measure in the limit, therefore everybody
is wrong.
That is a strange thing to say -- I prove too much by showing that the
whole thing makes no sense?
No you prove too much by showing that everybody is necessarily wrong.
If you take a sample of N from an ensemble with true proportion
p=0.5????? The trouble is that you get the same ensemble even if the
true proportion is 0.99, or 0.01. or any other value.
I don't understand that last remark. If I take a sample of N from and
ensemble with true proportion 0.5, then with high probability I get a
sample with proportion near 0.5. I don't know what you mean by "you get
the same ensemble"?
If instead you estimate how many other experimenters will get
estimates which are consistent with yours by being of high
probability in your posterior Bayesian distribution, with high
probability you will find that most of them will.
Exactly. Even if you estimate p=0.01, you will dismiss branches with
approximately equal numbers of zeros and ones as highly unlikely, and
you expect other experimenters to verify your results.
But if you estimate p=0.01 you will find that no one agrees with you
even approximately. And the bigger is N the more singular will seem
your result. Every particular sequence of observed 1s and 0s is equally
rare, the proportion of sequences with equal numbers of 1s and 0s is
large. So if you estimate p=0.5 you will have lots of agreeable
replications.
If the number of trials N is large, there are N(N-1)/2 branches with
exactly 2 zeros and N-2 ones. The probability for N/2 zeros and N/2
ones is (2/N)^N/2*(1-2/N)^N/2 ~ N^{-N/2}, which goes to zero very
rapidly for large N.
There is no "intrinsic probability" in your scenario.
If there is no probability, what do you expect when you are
still in Helsinki. If you predict that you die, then you
reject Mechanism (assumed here). If you predict P(W) = 1, the
city in Moscow will understand that the prediction was wrong.
If you predict that your history is the development of PI,
then only 1/2^N will be be confirmed, etc.
I turn the tables on you here, Bruno. You are confusing the 1p
and 3p pictures. From each individual's personal perspective, he
concludes, according to above argument, that his are the correct
probabilities. It is only from the outside, third-person
perspective, that we can see that he represents only a small
fraction of the total population of 2^N branches.
What is you prediction, if there is no probability. Keep in
mind that “W” and “M” does not refer to self-localisation,
but to the first person experience. Do you agree that in this
case W and M are incompatible.
I just try to understand.
As I said, I make no prediction, since I do not think that the
concept of probability can be meaningfully applied in cases of
person duplication, such as the WM scenario, or, for that matter,
Everettian quantum mechanics.
I think you are inconsistent in this. You (correctly) emphasize that
science is an inference from data. Everett's QM is an interpretation.
I doesn't change the data. So how does it change the inference? What
prediction will be falsified?
This is also Adrian Kent's objection to MWI, and it will
also nullify any benefit you might seek to gain from the
"frequency operator" -- every "first person" will get a
different eigenvalue in the limit of infinite trials..
That is not correct. If it is the frequency operator which is
measure, it gives the Born Probabilities, at least if the
“simple” derivation is correct.
No,that argument is mistaken, as Kent's general argument in terms
of the binomial expansion shows. All 2^N persons will use the
frequency operator to conclude that their probabilities are the
correct ones. Some will be seriously wrong,
But almost all will intersubjectively agree that p is near 0.5.
Science theories are based on intersubjective agreement...not
personal expriences.
But the true probability,
You said there was no true probability. I said that if there is one, in
an empirical sense, it must be 0.5.
as given by the amplitudes, might be very far from 0.5.
It might be. And I might be even if the experiment were bernoulli
sampling with a true value of 0.5. But given Kent's thought experiment
(or coin flipping), most observers will infer a value near 0.5...and so
we may expect to be one of them. This sort of hypothesis that our
observations are typical in some sense are commonplace in cosmology and
science would be impossible without it.
This shows that the data are independent of the Born amplitudes -- the
Born rule does not apply, and cannot be simply grafted on.
so the frequency operator is not a reliable indicator of probability.
Incidentally, the fact that there are more bit strings in the set
of all 2^N bit strings with approximately equal numbers of 0 and
1 results is a consequence of the binomial expansion when there
are only two possible outcomes, as in the cases we have
considered -- it is no more fundamental than that, and does not
reflect some 3p-preferred probability.
But my question is independent of Everett, so even if Kent is
correct for QM, it remains false for Mechanism. Let us agree
first on the simple Mechanist case, and then come back to
Everett.
Kent posed his argument in terms of completely classical
simulations, so it is precisely parallel to your WM-duplication
scenario. I have applied the argument to Everettian QM because of
the parallels between the two: Everett is just like the classical
duplication case since it is completely deterministic and every
possibility occurs on every trial. The only real difference is
that the different outcomes in QM occur on different branches
which, by decoherence, cannot interact or be aware of each other.
So there is no effective 3p perspective in QM as there is in the
WM-duplication. Arguments about the proportion of individuals who
see particular sets of outcomes in QM are arguments from the 3p
perspective, and it can be argued that in the absence of any
possible 3p observer, such arguments are invalid.
Then the arguments about every estimate being contrary to other
estimates is also invalid.
Why should you think that?
*I* don't think it. But you above make an observation that the
estimations will disagree is an argument from a 3p perspective and hence
invalid.
The estimates of probability that the individual observer makes are
all strictly first person estimates -- they use their own data for
their estimates, they do not take any third person view of the
situation. It is the argument that most observers will lie towards the
centre of the binomial distribution for two results that is an invalid
appeal to the third person view, and an appeal which leads to
manifestly wrong results if the true probability is far from 0.5.
It's not an argument, it is the premise of the thought experiment. The
thought experiment implies the binomial distribution. What is invalid
is to say each observer will infer a p value from one of the unique
binomial sequences (a 3p view) and then deny that most of those
observers will observe a value near 0.5 because it's a 3p view.
Brent
Bruce
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