A silly computer experiment ... or, the heart of the matter?

[gill1109]

Michel,

I don't have Python installed here, but from the results you posted your Python translation of Richard's first two R scripts (which I translated to Perl) doesn't give the same results as the R and Perl codes? You get the same CHSH = 2.0 for both the dependent and independent cases?

Best,

Zen.

Zen, why oh why is Michel looking at the maximum? He should draw histograms of the results of the two cases. The point is to *see* the results. All of them. A histogram of a few hundred would do fine.

I think he doesn't realise that when we do experiments, if we say "violate a bound" we really mean "statistically violate a bound" or "statistically significant violation of a bound" ... and even those words, need to be expanded, because people who have never heard of standard errors or p-values won't have any idea what it is supposed to mean.

I think everyone has already taken Michel's point that in experiments, one can easily observe values of CHSH larger than 2. In fact this is well known and has been well known for years and years and years. In a CHSH experiment one shows that one sees correlations close to the negative cosine, and not close to the triangle wave. One looks at just a few points on the curve ... more precisely, one looks at four points on the surface. Is the calculated - observed - measured E(a, b) close to the theoretical rho(a, b) of quantum theory, or to the theoretical rho(a, b) of the standard local hidden variables model?

When we talk about experiments, we need a vocabulary with words like "statistically significant", "standard error", "p-value", "goodness-of-fit", "type 1 error", "type 2 error", "discriminatory power"... We need to distinguish population from sample. Theory from experiment.

[minkwe]

Michel,

I don't have Python installed here, but from the results you posted your Python translation of Richard's first two R scripts (which I translated to Perl) doesn't give the same results as the R and Perl codes? You get the same CHSH = 2.0 for both the dependent and independent cases?

Best,

Zen.

That's because I made a typo. Notice that beta is not being used. Given that no results were presented it is easy to make mistakes. I will fix it soon but it doesn't change my opinion about it one bit.

[gill1109]

Hi Michel,

I've found out why the Python code was giving the wrong results. You've made a typo. It should be:

Code: Select all

B = -numpy.sign(numpy.cos(numpy.radians(beta - lmda)))

You had retyped alpha instead of beta when computing B.

Best,

Zen.

P.S. Did you run it, Rick?

Not yet, Zen. (Michel writes beautiful code, no doubt about that). I want to see a histogram of the results. I'm not interested in the maximum over many many repetitions. Can you draw histograms in Python?

[minkwe]

Rick:I don't think you can quickly explain (to a nonstatistician) the concept of a "sampling distribution of some statistic". Remember the experience with your students... Plotting histograms in Python is made with the library matplotlib. Just import it.

P.S. Yeah, Michel code reads like poetry.

Thanks for the compliment.

Zen,
Do you believe it is possible to produce a 4xN spreadsheet of any size, from any source, introducing as much error as one likes, which produces a CHSH value above 2 ("statistically") by even 0.000000000001.

Richard,
Thanks for the compliment too. I'll post some histograms tomorrow for both your experiment and mine. Note: I did not use maximum values in your experiment. I presented them in mine along side the averages to clearly show which inequality applies in what scenario, all determined from the data.

[gill1109]

Rick:I don't think you can quickly explain (to a nonstatistician) the concept of a "sampling distribution of some statistic". Remember the experience with your students... Plotting histograms in Python is made with the library matplotlib. Just import it.

P.S. Yeah, Michel code reads like poetry.

Thanks for the compliment.

Zen,
Do you believe it is possible to produce a 4xN spreadsheet of any size, from any source, introducing as much error as one likes, which produces a CHSH value above 2 ("statistically") by even 0.000000000001.

Richard,
Thanks for the compliment too. I'll post some histograms tomorrow for both your experiment and mine. Note: I did not use maximum values in your experiment. I presented them in mine along side the averages to clearly show which inequality applies in what scenario, all determined from the data.

Looking forward to seeing your histograms, Michel. We are making progress.

Starting with the 4xN spreadsheet of values of A, A', B, B' it is of course impossible to get above 2, when all four correlations are based on all N rows.

If however the rows are randomly split into four sub-groups of roughly equal size, and each correlation based on a different subgroup, then almost anything can happen. However, most splittings will generate CHSH values which are not much larger than 2. The word "most" can be quantified in probability terms, where the probability depends also how you quantify "much larger". It also depends on N, of course. See Theorem 1 of my paper for a very precise statement.

[minkwe]

Richard, I won't produce any histograms until you answer the quesion:
Do you believe it is possible to produce a 4xN spreadsheet of any size, from any source, introducing as much error as one likes, which produces a CHSH value above 2 ("statistically") by even 0.000000000001.
If you can not answer this simple question I've asked you many times, you can produce your own histograms and present them. For some reason Zen did not want to answer it either. You would think two mathematical statisticians familiar with the CHSH would know he answer immediately.

If you answer that questions, then we would have made progress.

[gill1109]

Richard, I won't produce any histograms until you answer the quesion:
Do you believe it is possible to produce a 4xN spreadsheet of any size, from any source, introducing as much error as one likes, which produces a CHSH value above 2 ("statistically") by even 0.000000000001.
If you can not answer this simple question I've asked you many times, you can produce your own histograms and present them. For some reason Zen did not want to answer it either. You would think two mathematical statisticians familiar with the CHSH would know he answer immediately.
If you answer that questions, then we would have made progress.

I've answered that question time and time again. My answer always stayed the same. It's even written out in my famous paper as "Fact 2". In the simple pure mathematical section 2. Page 4 of the current version (version 4)

[minkwe]

I guess you don't want to answer it then. Too bad.

[gill1109]

I guess you don't want to answer it then. Too bad.

Well we just discovered that you never got past page three of my paper.

My answer to your question is the same as ever: "no".

[gill1109]

Starting with the 4xN spreadsheet of values of A, A', B, B' it is of course impossible to get above 2, when all four correlations are based on all N rows.

Why did you ask the question? Can't you read?

Now please ask your next question. This is getting exciting, we are getting to the heart of the matter...

[Heinera]

Do you believe it is possible to produce a 4xN spreadsheet of any size, from any source, introducing as much error as one likes, which produces a CHSH value above 2 ("statistically") by even 0.000000000001.

Jesus Christ. The answer is no. Understood by all but Joy Christian.

[minkwe]

Do you believe it is possible to produce a 4xN spreadsheet of any size, from any source, introducing as much error as one likes, which produces a CHSH value above 2 ("statistically") by even 0.000000000001.

Jesus Christ. The answer is no.

You are aware that Richard argues that it can be violated by statistical error don't you. Richard argues that the CHSH is just a theoretical bound which can be violated a little due to statistical/experimental error. I'm happy you now see the issue. Nothing can violate the CHSH, not even non-locality or QM.

[Heinera]

Do you believe it is possible to produce a 4xN spreadsheet of any size, from any source, introducing as much error as one likes, which produces a CHSH value above 2 ("statistically") by even 0.000000000001.

Jesus Christ. The answer is no.

You are aware that Richard argues that it can be violated by statistical error don't you. Richard argues that the CHSH is just a theoretical bound which can be violated a little due to statistical/experimental error. I'm happy you now see the issue. Nothing can violate the CHSH, not even non-locality or QM.

No, he doesn't. A 4xN spreadsheet means that all correlations are measured on the same set of vectors. Richard is fully aware of the fact that the CHSH inequality is absolute in this case (he even gives a short proof in his paper. You should read it.).

What he says, is that if each correlation is computed on four different subsets, then the CHSH inequality can be violated. But then there is no 4xN spreadsheet.

He furthermore says: If the four different subsets are selected by a random process, then the violation can only be due to random errors. By increasing the size of the sets, we can get these random errors as close to zero as we want.

[minkwe]

What he says, is that if each correlation is computed on four different subsets, then the CHSH inequality can be violated. But then there is no 4xN spreadsheet.

Funny, so you agree that Richard has been arguing that the upper bound of the CHSH inequality derived for a single set, applies to averages from 4 different sets, but only statistically?

[Heinera]

What he says, is that if each correlation is computed on four different subsets, then the CHSH inequality can be violated. But then there is no 4xN spreadsheet.

Funny, so you agree that Richard has been arguing that the upper bound of the CHSH inequality derived for a single set, applies to averages from 4 different sets, but only statistically?

The upper bound of the CHSH inequality derived for a single set, also applies to correlations from 4 different sets selected randomly, but only statistically.

[minkwe]

What he says, is that if each correlation is computed on four different subsets, then the CHSH inequality can be violated. But then there is no 4xN spreadsheet.

Funny, so you agree that Richard has been arguing that the upper bound of the CHSH inequality derived for a single set, applies to averages from 4 different sets, but only statistically?

The upper bound of the CHSH inequality derived for a single set, also applies to correlations from 4 different sets selected randomly, but only statistically.

Not only is this false, it demostrates that you too do not understand what upper bound means. Richard has since stopped talking about upper bounds because he realizes the silliness of suggesting that an upper bound is violated. Didn't you just confirm to me that the CHSH upper bound is never violated, even statistically, when applied to the system it was derived for. Upper bounds are mathematical tautologies, they do not care where you get your data. They can never be violated by anything.

If you want to prove that averages from different sets have a different upper bound, do that. Start from different sets and derive the upper bound from that. Why start with a single set to begin with like the CHSH. It is a mathematical crime. It is sad that you do not see it.

[Heinera]

You need to realize that the two words "upper bound" can be used in two different ways: Absolute upper bound, and upper bound for expectations. Read up on probability theory. When we are talking about an upper bound for expectations, the bound will usually be violated half of the time.

[minkwe]

You need to realize that the two words "upper bound" can be used in two different ways: Absolute upper bound, and upper bound for expectations. Read up on probability theory.

You need to read up on probability. I just asked you if it was possible for the CHSH to be violated by a 4xN spreadsheet from any source, including as much error as you like, even by 0.0000001 and you said no. Now you are telling me that there is such a thing as an upper bound for expectations which can be violated?

Let us see: We have an expression:

E(U) + E(V) + E(X) + E(Y)

Those are expectation values, each has a maximum value of 1. Please tell me what you think the upper bound of that expression is. Can it be violated by experimental error.

Now we have another expression:

E(X) + E(-X) + E(Y) + E(-Y)

Please tell me what you think the upper bound for this expression is. Can it be violated by anything whatsoever?

[Heinera]

E(X) + E(-X) + E(Y) + E(-Y)

Please tell me what you think the upper bound for this expression is. Can it be violated by anything whatsoever?

No. But that's not the point. The point is:

E(X) + E(-U) + E(Y) + E(-V),
and
E(X) = E(U),
E(Y) = E(V).

Now, can E(X) + E(-U) + E(Y) + E(-V) be violated by anything whatsoever?

[minkwe]

No. But that's not the point.

You can't even bring yourself to fully answer the simple questions

What is the upper bound for
E(U) + E(V) + E(X) + E(Y)

and what is the upper bound for
E(X) + E(-X) + E(Y) + E(-Y)

Can anything violate them whatsoever, even by experimental error?

First tell me what the upper bounds are, then tell me whether they can ever be violated by anything, including experimental/experimental error. Note those are expectation values.

Remember we addressing your claim that:

You need to realize that the two words "upper bound" can be used in two different ways: Absolute upper bound, and upper bound for expectations. Read up on probability theory. When we are talking about an upper bound for expectations, the bound will usually be violated half of the time.

Once we establish that, then we can go on to discuss specific possibilities of what X, Y, U, V, may be.