Justo wrote:You just say my
is not weakly objective but you don't say why and where my reasoning fails.
I have. From this paper
https://arxiv.org/pdf/2012.10238v5.pdf Equation (23) is where you list 4 different sums each over the same set of hidden variables. From equation (23) you extract
)
(equation (29) by converting the linear combination of sums into the sum of a linear combination over the same set of lambdas. Think for a second what this operation means. It means you went from adding/subtracting a global property of each individual dataset (ie, the average of the paired product of outcomes), to taking the average of the sum/difference individual properties of the combined dataset. Equation (23) can be evaluated from data stored on 4 independent spreadsheets of pairs of outcomes. All you need is to calculate the average of each spreadsheet and use just the averages to calculate (23). Because of this, the ordering is not that important for (23).
Equation (24) can only be evaluated on a single spreadsheet containing all outcome pairs. This is because you must calculate the individual products and then combine them for each row before doing the summation. You can't do this without re-ordering! This is the implicit assumption. On its face, as a purely mathematical operation, it appears legitimate, however, you then claim that your equations (23 to 29) represent the results of a weakly objective experiment. But for the weakly objective case, the re-ordering required to proceed from (23) is impossible. I've explained it multiple times already using multiple examples and even posted code to demonstrate it.
Let me try explaining a different way. Say we perform a weakly objective experiment and obtain some data from it. Let us assume that the data from this experiment are in fact due to local hidden variables. Let us also assume that we perform N iterations from each of the weakly objective runs, and let us also assume that the fair sampling assumption applies and N is large enough such that the distributions of the hidden variables that produced the data are exactly the same for all 4 sets, except for the fact that each set is a separate random experiment with different orderings of the hidden variables. In this case, we have satisfied all your assumptions leading up to (23).
Our task now is to calculate
similar to your (23) to verify if it agrees with (28).
Per your (20 & 21),
Note that in our weakly objective experiment, the outcomes represented by
do not necessarily represent the results from the same lambda. The only information we have is that all the data in each run result from the same distribution of lambda. The orderings are random for each independent run. It's similar to tossing a coin 100 times in two independent experiments and obtaining 50 heads and 50 tails in each but the ordering of the outcomes will be random and independent between the two sets.
But, take a closer look at equation (24). Let us try to express
in a similar manner to (24). We can't unless we make an assumption about the ability to reorder the data. After reordering the values in all the
columns will match each other not just in relative frequencies but also in the pattern and order, and similarly for
,
and
. If this reordering is impossible, then we can't expect equations (24 to 29) to apply to this data and therefore we can't expect the upper bound of 2 to apply to weakly objective experimental outcomes.
To see that it is impossible to reorder the data, let us try to do the re-orderings required. We can apply the permutation
to
such that
\cong i)
. This creates the match
}^1)
. This takes care of
. To take care of
. We can apply the permutation
to
to obtain the match
}^1)
. Note that every permutation we apply must be applied to both columns of each independent spreadsheet, in order to keep corresponding pairs of values on each row together. Therefore after applying
)
and
)
we now also end up with a the columns
}^2)
and
}^2)
. But we still need to make the
and
columns match
and
, plus whatever permutation we apply to
must be the same permutation we apply to
in order to keep rows together. Unfortunately this is impossible to complete because we need to apply
to
to obtain the match
}^2 \cong B_{\xi(l)}^2)
but we need a different incompatible permutation
to get the match
}^2 \cong A_{\varrho(l)}^2)
