What is wrong with this argument?

Foundations of physics and/or philosophy of physics, and in particular, posts on unresolved or controversial issues

What is wrong with this argument?

Consider a spreadsheet with N = 4 000 rows, and just 4 columns.
Place a +/-1, however you like, in every single one of the 16 000 positions.

Give the columns names: A1, A2, B1, B2.

Independently of one another, and independently for each row, toss two fair coins.

Define two new columns S and T containing the outcomes of the coin tosses, encoded as follows: heads = 1, tails = 2.

Define two new columns A and B defined (rowwise) as follows: A = A1 if S = 1, otherwise A = A2; B = B1 if T = 1, otherwise B = B2.

Our spreadsheet now has eight columns, named: A1, A2, B1, B2, S, T, A, B.

Define four "correlations" as follows:

rho11 is the average of the product of A and B, over just those rows with S = 1 and T = 1.
rho12 is the average of the product of A and B, over just those rows with S = 1 and T = 2.
rho21 is the average of the product of A and B, over just those rows with S = 2 and T = 1.
rho22 is the average of the product of A and B, over just those rows with S = 2 and T = 2.

I claim that the probability that rho11 - rho12 - rho21 - rho22 is larger than 2.5, is smaller than 0.005 (5 pro mille, or half of one percent).

Proof: in the last displayed formula

(in the appendix) of http://arxiv.org/abs/1207.5103,
set N = 4000, delta = 0.035, and epsilon = 1/16.

Notation: $_{obs}$ is defined to be equal to the average of A times B over those rows for which S = 1 and T = 1. Similarly for the other three averages. The four averages are taken over four disjoint subsets of rows of the whole table.

Code: Select all
delta <- 0.035N <- 4000epsilon <- 1/16pb <- 4 * ( exp(-2 * N * delta^2) + exp(-2 * (0.25 - delta) * N * epsilon^2) )pb

I want to know if Michel Fodje, Fred Diether, or anyone else, disagrees with this claim. If so, why?

If not, I am happy to go on from here and elaborate on the relevance of this little bit of elementary probability theory for the analysis of computer simulation programs of local hidden variable models.

After that, we could possibly go on to discuss the possible relevance of all that to the metaphysics of quantum mechanics and EPR-B type experiments.

PS. Notice that I am just getting a bound to a probability, using some convenient and simple probability inequalities (due to W. Hoeffding) for tails of the binomial and hypergeometric distributions, together with two applications of Boole's inequality. Replacing the Hoeffding bounds with exact computations for the binomial and the hypergeometric gives a sharper bound: the probability in question is actually smaller than 1 in 10 thousand. And this is before optimizing the choice of delta. Obviously, what the probability is exactly, depends on exactly how the spreadsheet is filled with +/- 1 's. Suffice it to say, the probability is so small that I'm willing to bet several hundred Euro's against the event's occurrence. If someone wants me to wager several thousand Euro's against the event occuring, I would ask for a bit larger N (e.g. 10 000 or 40 000).
gill1109
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Re: What is wrong with this argument?

Sorry: "positions" should have been "cells"

And: "rho11 - rho12 - rho21 - rho22" should have been "rho11 + rho12 + rho21 - rho22"

Actually, by interchanging the roles of settings 1 and 2 (separately, on each side of the experiment), and of outcomes +1 and -1 (separately, on each side of the experiment), and of Alice and Bob, one gets a whole family of probability inequalities.

There are 2^5 = 32 of such interchanges possible, but it turns out that we only get 8 distinct inequalities. Essentially, one compares any one of the four correlations with the sum of the other three - the difference must be smaller in absolute value than 2, ie smaller than +2, and larger than -2. Altogether, just 4 x 2 = 8 inequalities.
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Re: What is wrong with this argument?

gill1109 wrote:Consider a spreadsheet with N = 4 000 rows, and just 4 columns.

That is what is wrong with the argument. You still do not get it, and I doubt that you ever will.

minkwe wrote:1) In your LG paper, you admit that the CHSH and Bell inequalities are only valid if the same ensemble [aka a single spreadsheet with just 4 columns] is used to calculate every correlation:
Larsson & Gill wrote:The problem here is that the ensemble on which the correlations are evaluated changes with the settings, while the original Bell inequality requires that they stay the same.

2) In your paper, you admit that when 4 different ensembles are being used the inequalities therefore only applies to the common part of the 4 different ensembles:
Larsson & Gill wrote: In eﬀect,the Bell inequality only holds on the common part of the four diﬀerent ensembles
minkwe

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Re: What is wrong with this argument?

minkwe wrote:
gill1109 wrote:Consider a spreadsheet with N = 4 000 rows, and just 4 columns.

That is what is wrong with the argument. You still do not get it, and I doubt that you ever will.

Hear, hear.

(You have to be British to really understand what this expression means.)
Joy Christian
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Re: What is wrong with this argument?

Zen wrote:How are you, Richard? May I check your definitions thoroughly just to make sure that I understand your argument?

We have a random array $\{(A_i,A'_i,B_i,B'_i,S_i,T_i)\}_{i=1}^N$ for which the random variables $A_i,A_i',B_i,B'_i\in\{-1,1\}$, for $i=1,\dots,N$.

For each line $i=1,\dots,N$, we have that $S_i$ and $T_i$ are independent random variables with distributions

$P(S_i=1)=P(S_i=2)=1/2 \, ,$

$P(T_i=1)=P(T_i=2)=1/2 \, .$

The lines of the array are supposed to be independent.

Do we have to sate anything else about the distributions of these random objects?

Is it necessary in your proofs to suppose that $S_i$ is independent of $A_i,A'_i,B_i,B'_i$ (mutatis mutandis for $T_i$)?

Now, we define

$\left< AB \right> := \frac{1}{N} \sum_{i=1}^N A_i B_i \, ,\qquad \left< A'B \right> := \frac{1}{N} \sum_{i=1}^N A'_i B_i \, ,$

$\left< AB' \right> := \frac{1}{N} \sum_{i=1}^N A_i B'_i \, , \qquad \left< A'B' \right> := \frac{1}{N} \sum_{i=1}^N A'_i B'_i \, .$

Finally, we define

$\left< AB \right>_\text{obs} := \frac{\sum_{i=1}^N A_i B_i \, \mathbb{I}_{\{(1,1)\}}(S_i,T_i)}{\sum_{i=1}^N \mathbb{I}_{\{(1,1)\}}(S_i,T_i)} \, ,\qquad \left< AB' \right>_\text{obs} := \frac{\sum_{i=1}^N A_i B'_i \, \mathbb{I}_{\{(1,2)\}}(S_i,T_i)}{\sum_{i=1}^N \mathbb{I}_{\{(1,2)\}}(S_i,T_i)}\, ,$

$\left< A'B \right>_\text{obs} := \frac{\sum_{i=1}^N A'_i B_i \, \mathbb{I}_{\{(2,1)\}}(S_i,T_i)}{\sum_{i=1}^N \mathbb{I}_{\{(2,1)\}}(S_i,T_i)} \, ,\qquad \left< A'B' \right>_\text{obs} := \frac{\sum_{i=1}^N A'_i B'_i \, \mathbb{I}_{\{(2,2)\}}(S_i,T_i)}{\sum_{i=1}^N \mathbb{I}_{\{(2,2)\}}(S_i,T_i)}\, ,$

in which $\mathbb{I}_D$ is the indicator function of the set $D$, defined by $\mathbb{I}_D(x)=1$, if $x\in D$, and $\mathbb{I}_D(x)=0$, if $x\notin D$.

If any of the above denominators is zero, define the correspondent fraction to be equal to zero.

I'm very well, thank you, Zen!

The Nx4 array of numbers A_i, A'_i, B_i, B'_i are given. Fixed. Somebody comes along with an Nx4 spreadsheet of numbers +/-1. They are not random variables. (They might be outcomes of some earlier random experiment). Maybe I could better have used small letters, but I wrote using typical spread-sheet or computer program language.

The 2N coin fair tosses S_i, T_i are all independent of one another. All probability statements I make refer to the 2^2N equally likely outcomes of the 2N coin tosses, for given initial Nx4 spreadsheet.

Your interpretations are correct, including the convention 0/0=0 for use in the very unlikely event that one of the four subsamples would be empty.
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Re: What is wrong with this argument?

minkwe wrote:
gill1109 wrote:Consider a spreadsheet with N = 4 000 rows, and just 4 columns.

That is what is wrong with the argument. You still do not get it, and I doubt that you ever will.

That is my assumption. In this thread, we discuss the argument which comes after the assumption. Do you agree with the theorem? The theorem is about an N x4 spreadsheet and 2N coin tosses. The spreadsheet is given, fixed. The coin tosses are fair, completely random.

I am not asking if you think that the theorem has any relevance to quantum mechanics, or whatever. I told you that after discussing the theorem, we will either decide it is true, or that it is not true.

If we decide it is true, then we can discuss its relevance (if any) to computer programs like epr-simple. Your prior opinion that it has no relevance, is well known. I believe that you are trapped by your own (false) prior beliefs and some misconceptions (prejudices) about probability and statistics in a circular argument. Running round in circles, you can't see outside your little prison. Time to open your mind.

If we decide it is false, I will retract my paper.

By the way, de Raedt, Hess and Michielsen got the point. They endorsed the Larsson-Gill paper and they use our ideas, very effectively. De Raedt and Khrennikov (who works closely with Giullaume Adenier, who later also worked with Jan-Ake) have invited me to the Vaxjo meeting in June. I will talk about your computer programs.
Last edited by gill1109 on Tue Mar 11, 2014 11:50 pm, edited 4 times in total.
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Re: What is wrong with this argument?

Joy Christian wrote:
minkwe wrote:
gill1109 wrote:Consider a spreadsheet with N = 4 000 rows, and just 4 columns.

That is what is wrong with the argument. You still do not get it, and I doubt that you ever will.

Hear, hear.

(You have to be British to really understand what this expression means.)

I am British.

We see that Joy Christian and Michel Fodje do not get the point, and probably never will. And instead of honestly saying whether they think a simple mathematical theorem is true or false, they prefer to shout abuse. Very British - just look at Prime Minister's question time on BBC, in the House of Commons. Very "common". Yob mentality.

No scientific discourse. No open exchange of ideas. Just cheap debating tricks to make political points to impress a stupid public.
gill1109
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Re: What is wrong with this argument?

minkwe wrote:
gill1109 wrote:Consider a spreadsheet with N = 4 000 rows, and just 4 columns.

That is what is wrong with the argument. You still do not get it, and I doubt that you ever will.

Hear, hear.

(You have to be British to really understand what this expression means.)
Joy Christian
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Re: What is wrong with this argument?

Joy Christian wrote:
minkwe wrote:
gill1109 wrote:Consider a spreadsheet with N = 4 000 rows, and just 4 columns.

That is what is wrong with the argument. You still do not get it, and I doubt that you ever will.

Hear, hear.

(You have to be British to really understand what this expression means.)

The aim is to discuss the argument here (the theorem), not its premisses.

Michel and Joy's responses are off topic, and they do not even constitute a scientific discussion of another topic.

They are transparent attempts to disrupt a discussion.
gill1109
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Re: What is wrong with this argument?

Here is a computer simulation illustration of the theorem.

http://rpubs.com/gill1109/Fred3

The R script reads a spreadsheet from internet. You can also download it yourself: http://www.math.leidenuniv.nl/~gill/fred3.csv. The spreadsheet has 100 rows and 7 columns. The first column contains just a sequence number ("run", which runs from 1 to 100). The next four columns are named A1, A2, B1, B2. This part of the spreadsheet (100 x 4) is the part which my theorem is about. The last two columns contain a particular realization of the settings S and T (called here Sa and Tb since "T" is also a name for "TRUE" in R).

The code computes the correlations and the CHSH quantity both the example realisation as well as for 100 new sets of random coin tosses. You'll see that the supplied coin tosses Sa and Tb were in fact rather lucky ... or the result of some kind of reverse engineering - not the result of fair coin tosses at all.

You can experiment by fillling the Nx4 part in different ways, and you can wonder how I managed to create fred3.csv. It was created by another R script, including some generation of random numbers but with the initial seed known, so that it can be independently reproduced on other computers. You can find that script here: http://www.math.leidenuniv.nl/~gill/fred.R.

(I later changed the order of the columns and the names of two of the columns. The script "fred.R" creates a spreadsheet called "fred.csv". It has the two special settings, which are there called SA and SB, in front of the A1, A2, B1, B2 columns, instead of behind them. The very first column, the one with the run number, isn't given a name).
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Re: What is wrong with this argument?

Zen,

You forgot about the role played by S and T here. Being random coin tosses, they generate the required randomness so that the first expression you mention is not necessarily equal to either zero or one.

And hopefully we won't get bogged down in a discussion about notation here. I find Richard's notation perfectly understandable.
Heinera

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Re: What is wrong with this argument?

Zen wrote:Thanks, Heinera! You're right. I thought we were given the whole spreadsheet initially, and not just the first four columns. My mistake. I will fix it and post my analysis of Gill's proof asap.

In the mathematical sense, there is "given" an Nx4 array of numbers +/-1. Alice and Bob then toss coins. From each row of the array, Alice gets to see the entry from column A1 (S = heads) or from column A2 (S = tails). Similarly for Bob. Alice and Bob then get together to calculate four correlations each based on a different (disjoint) subset of rows.

We start with an Nx4 array with the numbers A1, A2, B1, B2. From now on it is fixed, given. Independently of this we do independent fair coin tosses, S, T; think of them as filling another Nx2 table. The two tables are combined (and reduced) to a new table with columns S, T, A_obs, B_obs. The four correlations are calculated from the third table: correlations between A_obs and B_obs for each combination of values of S and T.

I hope this is slowly getting crystal clear! Improvements to my notation and presentation are surely possible. My aim is to tell a story which science journalists, and high school students, and your grandmum and granddad, can *all* understand.

The link to Bell is that A1, A2, B1, B2 stand for (local functions of) the local hidden variables. S and T stand for Alice and Bob's freely chosen settings. A(obs) and B(obs) are the actually observed outcomes of Alice and Bob's measurements. Because of local realism, or because of local hidden variables, the experiment is "as if" Alice and Bob just randomly pick a predetermined outcome from one of two "preexisting" values. Note the "as if". I'm not saying it really is that way. I'm saying that the final results - what we finally get to see - is mathematically indistinguishable from the final results described here.

We can later discuss why this "as if" is certainly valid for computer simulations like Michel Fodje's "epr-simple". And once one has got the idea, one can extend further to e.g. "epr-clocked". But first I want to see if there is agreement on this little bit of elementary mathematics about randomly picking rows from a spreadsheet.

Think of it as "creating facts on the ground". But in fact, they are not being created by force - they already exist. It seems that Michel and Joy don't want to see them, but they are there, all right.
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Re: What is wrong with this argument?

Interestingly, we have now seen a second discussion "locked".

I wanted to mention to Michel that I am sure I will be able to explain to him my answers to his questions (1) to (12) once he has worked through the proof of my theorem. So far it seems he refuses to do this, because he doesn't like the assumption of the theorem. The theorem is about an Nx4 spreadsheet. He might think, at the moment, that it is irrelevant to our main quest. Fine. If he's right, he has absolutely nothing to fear! I just want to hear whether he understands the theorem, and whether he thinks it's true or false. If he thinks it is false, I'd like to know why. Obviously I want to immediately retract any false mathematical claim I might have made in the past.

The same question, I put to Fred. If you're right that my theorem is *irrelevant* then it's truth can't hurt you.

And to Joy. If you're right that my theorem is *irrelevant* then it's truth can't hurt you.

Pythagoras's theorem doesn't hurt science. No one is scared of it. No-one ignores it.
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Re: What is wrong with this argument?

Zen wrote:Not really related to this thread, but in one of your papers you say that your current position is kind of "keep locality, give up on realism". Can you talk a little about how this position "explains" / "deals with" the existence of perfect anti-correlations when the two detectors in Aspect's lab are properly aligned?

Nice question! This position doesn't *explain* perfect anti-correlations. It just refuses to make the step, which Einstein did take, of deducing from perfect anti-correlation that Alice's outcome for any measurement she might have made "exists" in reality, in (or at) the particle, at that given time and place.

In our present context, "realism" is actually a kind of idealistic point of view. Unperformed measurements also have outcomes and those outcomes are moreover "located" in space-time in exactly the same place where the actual outcome of the actual measurement comes to be, after it is done.

If you have a local hidden variables theory, then you can do that. You can think of all the A(a, lambda) (hidden variable lambda fixed, Alice's possible settings a varying) as all living in the same place, all "existing" simultaneously. In fact they are all simply encoded by the value of lambda.

It can be thought to be a cheap way out. Just playing with words.

I think the more subtle position to take is that QM is different from classical physics. It allows things which classically would be impossible (like violating CHSH). It forbids things which classical physics in principle can allow. It is non-deteministic. The future is *not* determined by the past.

Being different, it also clashes with our "embodied cognition". Our brains evolved to allow us feed, breed and multiply in a world by always assuming there is a cause for anything that happened.
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Re: What is wrong with this argument?

gill1109 wrote:Interestingly, we have now seen a second discussion "locked".

Richard,
We both agreed that discussion was over, and the moderator acted based on our documented mutual agreement.
I wanted to mention to Michel that I am sure I will be able to explain to him my answers to his questions (1) to (12) .

You already stated in the relevant thread that my points (1) to (12) were nonsense (viewtopic.php?f=6&t=23&start=100#p827), as we agreed, there is nothing more for the two of us to discuss about them. Feel free to discuss with others but I won't participate in that discussion, not because I'm afraid of anything, but because I believe you don't get it, don't want to get it and I never will. So in case you again accuse me of attempting to disrupt your discussions, this will be my last post in this thread. I'm sure you can discuss your theories with others without making occasional snide remarks about me, or forwarding every post you make to my e-mail.
minkwe

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Re: What is wrong with this argument?

minkwe wrote:So in case you again accuse me of attempting to disrupt your discussions, this will be my last post in this thread. I'm sure you can discuss your theories with others without making occasional snide remarks about me, or forwarding every post you make to my e-mail.

Interesting. A scientist who appears to be "allergic" to a mathematical fact. By refusing to talk about it, he does not make it untrue. He simply closes off his mind to some useful information. De Raedt, Hess and Michielsen did not have this attitude; nor did Giullaume Adenier. Even Joy Christian has had useful discussions with Richard Gill, leading on the one hand to very nice improvements to Michel Fodje's simulation model, and on the other hand to an attempt to get a key experiment actually performed.
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Re: What is wrong with this argument?

Off topic; let's get back on topic here.
FrediFizzx
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Re: What is wrong with this argument?

FrediFizzx wrote:Off topic; let's get back on topic here.

The central question of the thread which I started here, is: is the proof of the quoted theorem, correct?

If we come to the conclusion that it seems to be a true theorem, then I will be delighted to open a new topic in which I would like to discuss if and how it can be applied to computer simulation models, QRC and so on.

If the answer is that it seems to be an incorrect theorem, then I will retract the paper in which I planned to publish it. (I have to submit the final version in one week).

The assumptions of the theorem are given, and I hope they are now completely clear. The relevance of the theorem can/will be another topic. (Superfluous if the theorem is false).
Richard Gill wrote:Consider a spreadsheet with N = 4 000 rows, and just 4 columns.
Place a +/-1, however you like, in every single one of the 16 000 positions.

Give the columns names: A1, A2, B1, B2.

Independently of one another, and independently for each row, toss two fair coins.

Define two new columns S and T containing the outcomes of the coin tosses, encoded as follows: heads = 1, tails = 2.

Define two new columns A and B defined (rowwise) as follows: A = A1 if S = 1, otherwise A = A2; B = B1 if T = 1, otherwise B = B2.

Our spreadsheet now has eight columns, named: A1, A2, B1, B2, S, T, A, B.

Define four "correlations" as follows:

rho11 is the average of the product of A and B, over just those rows with S = 1 and T = 1.
rho12 is the average of the product of A and B, over just those rows with S = 1 and T = 2.
rho21 is the average of the product of A and B, over just those rows with S = 2 and T = 1.
rho22 is the average of the product of A and B, over just those rows with S = 2 and T = 2.

I claim that the probability that rho11 + rho12 + rho21 - rho22 is larger than 2.5, is smaller than 0.005 (5 pro mille, or half of one percent)

You can find a proof at http://arxiv.org/abs/1207.5103 (appendix: Proof of Theorem 1 from Section 2), together with remarks by me in the first posting of this thread.
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Re: What is wrong with this argument?

Fred, getting back on topic, I am also interested in the question whether or not *you* agree with my claim: in the situation described, the probability that rho11 + rho12 + rho21 - rho22 will be larger than 2.5, is smaller than 0.005 (5 pro mille, or half of one percent).

I am submitting the final version of my paper later this week. Anyone who believes the main theorem is wrong can still try to explain to me what is wrong about it, and prevent the literature from being polluted yet again by another pro-Bell nonsense propaganda piece.
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Re: What is wrong with this argument?

I would have advise to not publish until you fix your mistakes. We have tried to explain to you why you are wrong but nothing seems to work so we should just drop it as it is just going around in circles.
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