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The 3-page essay at http://vixra.org/abs/1909.0216 is titled:"Einstein vs Bell? Bell’s inequality refuted, Bell’s error corrected."

Bell's inequality (BI) is refuted in 8 lines of elementary mathematics. Bell's error is defined and corrected in 11 more.

No knowledge of QM, Joy's Clifford Algebras, nor Richard's probability theory is required.

Reason:

(i) Bell (1964b) -- http://cds.cern.ch/record/111654/files/ ... 00_001.pdf -- prepares us for the derivation of BI with a valid (and elementary) physical boundary condition; his eqn (1).

(ii) He then breaches that condition; see his move from the valid first line to the invalid second line on p.198 of Bell (1964b).

To facilitate discussion, I suggest we number the three unnumbered relations atop p.198 as (14a) to (14c).

I look forward to critical comments from both sides of this debate; and from those in the middle or on the side-lines.

Reinforcing the view that it is all rather elementary; and with best regards: Gordon Watson http://vixra.org/abs/1909.0216

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The 3-page essay at http://vixra.org/abs/1909.0216 is titled:"Einstein vs Bell? Bell’s inequality refuted, Bell’s error corrected."

Bell's inequality (BI) is refuted in 8 lines of elementary mathematics. Bell's error is defined and corrected in 11 more.

No knowledge of QM, Joy's Clifford Algebras, nor Richard's probability theory is required.

Reason:

(i) Bell (1964b) -- http://cds.cern.ch/record/111654/files/ ... 00_001.pdf -- prepares us for the derivation of BI with a valid (and elementary) physical boundary condition; his eqn (1).

(ii) He then breaches that condition; see his move from the valid first line to the invalid second line on p.198 of Bell (1964b).

To facilitate discussion, I suggest we number the three unnumbered relations atop p.198 as (14a) to (14c).

I look forward to critical comments from both sides of this debate; and from those in the middle or on the side-lines.

Reinforcing the view that it is all rather elementary; and with best regards: Gordon Watson http://vixra.org/abs/1909.0216

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- Gordon Watson
**Posts:**255**Joined:**Wed Apr 30, 2014 4:39 am

Gordon Watson wrote:.

The 3-page essay at http://vixra.org/abs/1909.0216 is titled:"Einstein vs Bell? Bell’s inequality refuted, Bell’s error corrected."

Bell's inequality (BI) is refuted in 8 lines of elementary mathematics. Bell's error is defined and corrected in 11 more.

No knowledge of QM, Joy's Clifford Algebras, nor Richard's probability theory is required.

Reason:

(i) Bell (1964b) -- http://cds.cern.ch/record/111654/files/ ... 00_001.pdf -- prepares us for the derivation of BI with a valid (and elementary) physical boundary condition; his eqn (1).

(ii) He then breaches that condition; see his move from the valid first line to the invalid second line on p.198 of Bell (1964b).

To facilitate discussion, I suggest we number the three unnumbered relations atop p.198 as (14a) to (14c).

I look forward to critical comments from both sides of this debate; and from those in the middle or on the side-lines.

Reinforcing the view that it is all rather elementary; and with best regards: Gordon Watson http://vixra.org/abs/1909.0216

.

Your logic is faulty, Gordon. I found your paper extremely difficult to read. You succeed in making simple things very difficult, and this enables you to make incorrect logical deductions.

It is indeed all extremely elementary. No knowledge of QM is needed to follow Bell's arguments, though of course sooner or later, one should verify the classical QM derivation of the EPR-B correlations, by matrix algebra calculations (over the complex numbers) involving the 2x2 Pauli matrices and various 4x4 matrices built up from them. Clifford algebra and Geometric Algebra are sophisticated and beautiful mathematical structures which one can use to elegantly reproduce classical QM calculations. Joy Christian uses them for other purposes.

I don't understand what your problem is with "Richard's probability theory". Bell does an elementary probability calculation using the notations and terminology of 60's physicists since he was a physicist writing in the 60's for physicists.

But to get to the heart of the matter, you compare a new inequality of your own, labelled (5) in your new paper, to Bell's original three correlations inequality, reproduced and labelled (6) in your new paper. You note that a particular set of three correlations E(a, b) = -1/2, E(b, c) = -1/2, E(a, c) = +1/2 violates Bell's inequality but satisfies yours.

The reason they violate Bell's inequality is actually because it is impossible to find functions A(a, lambda), B(b, lambda), and rho(lambda) satisfying the usual constraints *and* such that E(a, b) = int A(a, lambda)B(b, lambda) rho(lambda) d lambda, etc. If you had wanted to disprove Bell by a counterexample featuring these values of the three correlations, you would have had to exhibit functions A, B and rho which reproduce your three correlations E while at the same time satisfying the standard requirements (A and B take values +/-1; rho is nonnegative and integrates to 1). But you don't do this! So your paper is empty of any content! It just exhibits muddled thinking!

I suggest you do try to find a concrete example of functions A, B and rho which do the job you want them to do. (Joy Christian thinks he did manage to do that).

I can tell you in advance that you will not succeed, since Bell's inequality is a true and trivial result in probability theory, known already by George Boole in the 1850s, and since then independently discovered again and again by many others, as Karl Hess points out (yet again) in his most recent paper.

If you do succeed, then please program the resulting model and let's do my computer challenge. 10 thousand Euro's says you'll fail.

- gill1109
- Mathematical Statistician
**Posts:**1653**Joined:**Tue Feb 04, 2014 10:39 pm**Location:**Leiden

gill1109 wrote:Gordon Watson wrote:.

The 3-page essay at http://vixra.org/abs/1909.0216 is titled:"Einstein vs Bell? Bell’s inequality refuted, Bell’s error corrected."

Bell's inequality (BI) is refuted in 8 lines of elementary mathematics. Bell's error is defined and corrected in 11 more.

No knowledge of QM, Joy's Clifford Algebras, nor Richard's probability theory is required.

Reason:

(i) Bell (1964b) -- http://cds.cern.ch/record/111654/files/ ... 00_001.pdf -- prepares us for the derivation of BI with a valid (and elementary) physical boundary condition; his eqn (1).

(ii) He then breaches that condition; see his move from the valid first line to the invalid second line on p.198 of Bell (1964b).

To facilitate discussion, I suggest we number the three unnumbered relations atop p.198 as (14a) to (14c).

I look forward to critical comments from both sides of this debate; and from those in the middle or on the side-lines.

Reinforcing the view that it is all rather elementary; and with best regards: Gordon Watson http://vixra.org/abs/1909.0216

.

Your logic is faulty, Gordon. I found your paper extremely difficult to read. You succeed in making simple things very difficult, and this enables you to make incorrect logical deductions.

It is indeed all extremely elementary. No knowledge of QM is needed to follow Bell's arguments, though of course sooner or later, one should verify the classical QM derivation of the EPR-B correlations, by matrix algebra calculations (over the complex numbers) involving the 2x2 Pauli matrices and various 4x4 matrices built up from them. Clifford algebra and Geometric Algebra are sophisticated and beautiful mathematical structures which one can use to elegantly reproduce classical QM calculations. Joy Christian uses them for other purposes.

I don't understand what your problem is with "Richard's probability theory". Bell does an elementary probability calculation using the notations and terminology of 60's physicists since he was a physicist writing in the 60's for physicists.

But to get to the heart of the matter, you compare a new inequality of your own, labelled (5) in your new paper, to Bell's original three correlations inequality, reproduced and labelled (6) in your new paper. You note that a particular set of three correlations E(a, b) = -1/2, E(b, c) = -1/2, E(a, c) = +1/2 violates Bell's inequality but satisfies yours.

The reason they violate Bell's inequality is actually because it is impossible to find functions A(a, lambda), B(b, lambda), and rho(lambda) satisfying the usual constraints *and* such that E(a, b) = int A(a, lambda)B(b, lambda) rho(lambda) d lambda, etc. If you had wanted to disprove Bell by a counterexample featuring these values of the three correlations, you would have had to exhibit functions A, B and rho which reproduce your three correlations E while at the same time satisfying the standard requirements (A and B take values +/-1; rho is nonnegative and integrates to 1). But you don't do this! So your paper is empty of any content! It just exhibits muddled thinking!

I suggest you do try to find a concrete example of functions A, B and rho which do the job you want them to do. (Joy Christian thinks he did manage to do that).

I can tell you in advance that you will not succeed, since Bell's inequality is a true and trivial result in probability theory, known already by George Boole in the 1850s, and since then independently discovered again and again by many others, as Karl Hess points out (yet again) in his most recent paper.

If you do succeed, then please program the resulting model and let's do my computer challenge. 10 thousand Euro's says you'll fail.

Thanks Richard,

You'll see the case made more clearly in v2. I'm still time-poor as I write.

Re the functions that you refer to. As implied in ¶2.3: TLR derives the same expectations as QM. As I state in ¶1.1(v), the results require no knowledge of QM. In this particular matter, Bell gives us the QM expectation. And TLR independently agrees with it.

However, didn't Bell make the claim (or imply) that such functions did not affect his result? That he need only work with the expectations to prove his point?

My comment re "Richard's probability theory" (PT) -- one side of the middle-way -- was reference to your use of PT to justify Bell's work. (AND NOT the good work you hopefully do in Court.)

Your repeat my concern above:

"Bell does an elementary probability calculation using the notations and terminology of 60's physicists since he was a physicist writing in the 60's for physicists."

But he gets a wrong answer: as I show.

For Bell's inequality produces a lower upper bound than logic, QM, experiments and PT allow. He also expected his dilemma re AAD to be resolved!

Since I pin-point the location of BE1 (Bell's first error) in his move from (what I call) his (14a) to his (14b): I would welcome your comments re the physics and your PT-based analysis thereof.

As for your computer challenge: Can I first program the two computers to move toward each other under the influence of their masses? Might you see such a challenge (for starters) as equivalent to your Bell-based challenge? And worth the prize-money?

Thanks again; Gordon

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- Gordon Watson
**Posts:**255**Joined:**Wed Apr 30, 2014 4:39 am

Gordon Watson wrote:However, didn't Bell make the claim (or imply) that such functions did not affect his result? That he need only work with the expectations to prove his point?

No, he did not make any such claim. His logic is very transparent, I think. The question is whether or not such functions could exist. Assume they do exist, and assume QM predictions are at least approximately true, leads to a contradiction. Therefore they do not exist.

Gordon Watson wrote:Since I pin-point the location of BE1 (Bell's first error) in his move from (what I call) his (14a) to his (14b): I would welcome your comments re the physics and your PT-based analysis thereof.

Sure, I hope to come back to that.

Gordon Watson wrote:As for your computer challenge: Can I first program the two computers to move toward each other under the influence of their masses? Might you see such a challenge (for starters) as equivalent to your Bell-based challenge? And worth the prize-money?

You are allowed to program just whatever you like. I am not going to look at your programs. You can even program the whole thing on one computer. My requirements are only that your computer code will also run on other people's computers, that it includes facilities for choosing the sample size, for fixing the seed to the random number generators, and for external submission of lists of settings. This makes any experiments we do completely transparent and reproducible. No one is obliged to study your code or argue about the ideas which you put into it. I will merely make some random checks to verify things such as: given the results of the first nine trials, Alice's tenth setting does not influence Bob's tenth outcome. Your task is to show that you can systematically violate Bell inequalities in the kind of rigorous framework used in the "loophole-free Bell experiments" starting in 2015 in Delft.

- gill1109
- Mathematical Statistician
**Posts:**1653**Joined:**Tue Feb 04, 2014 10:39 pm**Location:**Leiden

gill1109 wrote:Gordon Watson wrote:However, didn't Bell make the claim (or imply) that such functions did not affect his result? That he need only work with the expectations to prove his point?

No, he did not make any such claim. His logic is very transparent, I think. The question is whether or not such functions could exist. Assume they do exist, and assume QM predictions are at least approximately true, leads to a contradiction. Therefore they do not exist.

OK; thanks. So my logic is totally transparent, I think: I say such functions exist, assume QM predictions are at least approximately true (deriving QM predictions independently, but let's discuss this later), and find no contradiction.

But I do find a Bellian error; see below.

gill1109 wrote:Gordon Watson wrote:Since I pin-point the location of BE1 (Bell's first error) in his move from (what I call) his (14a) to his (14b): I would welcome your comments re the physics and your PT-based analysis thereof.

Sure, I hope to come back to that.

That will be very much appreciated; especially in the light of my thoughts above. Thanks.

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- Gordon Watson
**Posts:**255**Joined:**Wed Apr 30, 2014 4:39 am

Gordon Watson wrote:Since I pin-point the location of BE1 (Bell's first error) in his move from (what I call) his (14a) to his (14b): I would welcome your comments re the physics and your PT-based analysis thereof.

I see no error in the move from what you call Bell's (14a) to his (14b). Notice how the leading sign changes from a minus to none. So going from 14a to 14b we not only, under the integral sign, expand a product and replace a square of +/-1 by +1, we also reorder the two terms. You could do just one step at a time giving us (14a), (14aa), (14ab), (14b). Maybe that will help you understand the argument?

For others - Gordon Watson and I are talking about the first two (unnumbered) lines of displayed formula on page 198 of J.S. Bell (1964) "On the Einstein Podolsky Rosen paradox" https://cds.cern.ch/record/111654/files/vol1p195-200_001.pdf. Physics vol 1, nr.3, pp. 195–290.

I see no point in rewriting this proof in the language of modern probability theory for Gordon, just now. He can read past papers by me, written with an audience of mathematicians, probabilists and statisticians in mind, if he is interested. Let me just say: it is a nice exercise from Probability 101.

Boole already had it as an exercise for the reader! He did not bother to publish solutions, or an instructor's manual. He expected his readers to figure it out for themselves. Are you, Gordon, saying that he was wrong, too? And so are all those many researchers who disparage Bell by saying that Bell's inequality is already in Boole's work, and even some who suggest that Bell *stole* his inequality from George Boole? Come off it ...!

- gill1109
- Mathematical Statistician
**Posts:**1653**Joined:**Tue Feb 04, 2014 10:39 pm**Location:**Leiden

There is of course nothing wrong with the equivalence between (14a) and (14b). Perhaps it will be clearer to Gordon if we take it step by step in the opposite direction, from (14b) to (14a):

We start with (14b):

Simply multiply out to get

Now since must equal 1 we simplify the above to

.

Swap the two product terms:

and put the minus outside the expression:

We start with (14b):

Simply multiply out to get

Now since must equal 1 we simplify the above to

.

Swap the two product terms:

and put the minus outside the expression:

- Heinera
**Posts:**696**Joined:**Thu Feb 06, 2014 1:50 am

Yeah, I don't know why Gordon keeps harping about some error in the derivation of the inequality. There is no error in the inequality itself as it is mathematically proven. The error arises when you say a, b and c can happen all at the same time. They can't. It is that simple.

.

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- FrediFizzx
- Independent Physics Researcher
**Posts:**1715**Joined:**Tue Mar 19, 2013 7:12 pm**Location:**N. California, USA

FrediFizzx wrote:Yeah, I don't know why Gordon keeps harping about some error in the derivation of the inequality. There is no error in the inequality itself as it is mathematically proven. The error arises when you say a, b and c can happen all at the same time. They can't. It is that simple.

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I think Gordon is saying exactly the same thing as you are saying here, just in different terms.

- minkwe
**Posts:**1128**Joined:**Sat Feb 08, 2014 10:22 am

Heinera wrote:There is of course nothing wrong with the equivalence between (14a) and (14b).

Mathematically, there is nothing wrong. Physically, it is a whole other question.

When the A(a)A(c) pair of values is obtained from a different particle pair than the A'(a)A'(b) pair of values, the factorization within the integral can't proceed, since essentially A'(a) is a different random variable from A(a). Any assumptions within the integral apply to individual instances, thus, factorization within the integral implies equality between random variables that should not necessarily be equal unless there is a massive conspiracy. Probability considerations do not apply within the integral.

Perhaps it will be clearer to Gordon if we take it step by step in the opposite direction, from (14b) to (14a):

It is difficult for mathematicians without physical intuition to understand the point being made. It looks like an innocent mathematical operation but this is not about arithmetic.

We start with (14b):

Simply multiply out to get

Now since must equal 1 we simplify the above to

.

Swap the two product terms:

and put the minus outside the expression:

Exactly, now that the arithmetic operations are clear, let us discuss what they mean for the physics of the EPR experiment. Since we are working within the integral, let us focus on just one instance which MUST involve ALL the terms that came into play within the integral.

The three random variables/terms are equivalent to 3 coin tosses. But in an EPR experiment, you have 4 coin tosses. By applying your derivation to the EPR experiment you have made an assumption that the first and third coin tosses are so correlated that they will always be the same value. No amount of local realism can justify such a silly assumption. This is the point at which the spooky business is introduced into Bell.

Note that the fact that this assumption is made *within* the integral means you can't explain your way out of it with statistics/probability.

Bell Interview in Omni Magazine wrote:Bell: Then in 1932 [mathematician] John von Neumann gave a “rigorous” mathematical proof stating that you couldn’t find a nonstatistical theory that would give the same predictions as quantum mechanics. That von Neumann proof in itself is one that must someday be the subject of a Ph.D. thesis for a history student. Its reception was quite remarkable. The literature is full of respectable references to “the brilliant proof of von Neumann;” but I do not believe it could have been read at that time by more than two or three people.

Omni: Why is that?

Bell: The physicists didn’t want to be bothered with the idea that maybe quantum theory is only provisional. A horn of plenty had been spilled before them, and every physicist could find something to apply quantum mechanics to. They were pleased to think that this great mathematician had shown it was so. Yet the Von Neumann proof, if you actually come to grips with it, falls apart in your hands! There is nothing to it. It’s not just flawed, it’s silly. If you look at the assumptions it made, it does not hold up for a moment. It’s the work of a mathematician, and he makes assumptions that have a mathematical symmetry to them. When you translate them into terms of physical disposition, they’re nonsense. You may quote me on that: the proof of von Neumann is not merely false but foolish.

The irony!

- minkwe
**Posts:**1128**Joined:**Sat Feb 08, 2014 10:22 am

minkwe wrote:When the A(a)A(c) pair of values is obtained from a different particle pair than the A'(a)A'(b) pair of values, the factorization within the integral can't proceed, since essentially A'(a) is a different random variable from A(a).

You left out the lambda. In Bell's model there are not two functions A and A', there is only one function A(a, lambda). And A is not a random variable; it is a deterministic function of two arguments. The only random variable in the model is lambda, so your objection is based on a misunderstanding.

- Heinera
**Posts:**696**Joined:**Thu Feb 06, 2014 1:50 am

Heinera wrote:minkwe wrote:When the A(a)A(c) pair of values is obtained from a different particle pair than the A'(a)A'(b) pair of values, the factorization within the integral can't proceed, since essentially A'(a) is a different random variable from A(a).

You left out the lambda. In Bell's model there are not two functions A and A', there is only one function A(a, lambda). And A is not a random variable; it is a deterministic function of two arguments. The only random variable in the model is lambda, so your objection is based on a misunderstanding.

If lambda is a random variable, then any function of lambda such as A(a, lambda) is also a random variable. I learned that in high-school. Bell's functions A(a, lambda) are random variables.

***

- Joy Christian
- Research Physicist
**Posts:**2161**Joined:**Wed Feb 05, 2014 4:49 am**Location:**Oxford, United Kingdom

Joy Christian wrote:Heinera wrote:minkwe wrote:When the A(a)A(c) pair of values is obtained from a different particle pair than the A'(a)A'(b) pair of values, the factorization within the integral can't proceed, since essentially A'(a) is a different random variable from A(a).

You left out the lambda. In Bell's model there are not two functions A and A', there is only one function A(a, lambda). And A is not a random variable; it is a deterministic function of two arguments. The only random variable in the model is lambda, so your objection is based on a misunderstanding.

If lambda is a random variable, then any function of lambda such as A(a, lambda) is also a random variable. I learned that in high-school. Bell's functions A(a, lambda) are random variables.

***

As I wrote, in Bell's model A is a completely deterministic function of two arguments. It is meaningless to discuss the model while simultaneously leaving out lambda from the notation, as minkwe did.

- Heinera
**Posts:**696**Joined:**Thu Feb 06, 2014 1:50 am

Heinera wrote:Joy Christian wrote:Heinera wrote:

You left out the lambda. In Bell's model there are not two functions A and A', there is only one function A(a, lambda). And A is not a random variable; it is a deterministic function of two arguments. The only random variable in the model is lambda, so your objection is based on a misunderstanding.

If lambda is a random variable, then any function of lambda such as A(a, lambda) is also a random variable. I learned that in high-school. Bell's functions A(a, lambda) are random variables.

***

As I wrote, in Bell's model A is a completely deterministic function of two arguments. It is meaningless to discuss the model while simultaneously leaving out lambda from the notation, as minkwe did.

No, you wrote that "A is not a random variable." You wrote something that is obviously wrong. Admit it.

***

- Joy Christian
- Research Physicist
**Posts:**2161**Joined:**Wed Feb 05, 2014 4:49 am**Location:**Oxford, United Kingdom

Joy Christian wrote:Heinera wrote:Joy Christian wrote:If lambda is a random variable, then any function of lambda such as A(a, lambda) is also a random variable. I learned that in high-school. Bell's functions A(a, lambda) are random variables.

***

As I wrote, in Bell's model A is a completely deterministic function of two arguments. It is meaningless to discuss the model while simultaneously leaving out lambda from the notation, as minkwe did.

No, you wrote that "A is not a random variable." You wrote something that is obviously wrong. Admit it.

***

You need to learn to distinguish between A and A(a, lambda).

- Heinera
**Posts:**696**Joined:**Thu Feb 06, 2014 1:50 am

Heinera wrote:Joy Christian wrote:Heinera wrote:Joy Christian wrote:If lambda is a random variable, then any function of lambda such as A(a, lambda) is also a random variable. I learned that in high-school. Bell's functions A(a, lambda) are random variables.

***

As I wrote, in Bell's model A is a completely deterministic function of two arguments. It is meaningless to discuss the model while simultaneously leaving out lambda from the notation, as minkwe did.

No, you wrote that "A is not a random variable." You wrote something that is obviously wrong. Admit it.

***

You need to learn to distinguish between A and A(a, lambda).

That is just nonsense. You are trying to justify your earlier nonsense with more nonsense.

***

- Joy Christian
- Research Physicist
**Posts:**2161**Joined:**Wed Feb 05, 2014 4:49 am**Location:**Oxford, United Kingdom

Joy Christian wrote:Heinera wrote:You need to learn to distinguish between A and A(a, lambda).

That is just nonsense. You are trying to justify your earlier nonsense with more nonsense.

***

No, it is not nonsense. A is a function, without any reference to a random variable. On the other hand, A(a, lambda) is a function of a random variable, assuming lambda is a random variable.

But you have anyway completely missed the point here; it is the omission of lambda in minkwe's argument that makes it misguided.

Last edited by Heinera on Thu Sep 19, 2019 12:51 am, edited 1 time in total.

- Heinera
**Posts:**696**Joined:**Thu Feb 06, 2014 1:50 am

Heinera wrote:Joy Christian wrote:Heinera wrote:You need to learn to distinguish between A and A(a, lambda).

That is just nonsense. You are trying to justify your earlier nonsense with more nonsense.

***

No, it is not nonsense. A is a function, without any reference to a random variable. On the other hand, A(a, lambda) is a function of a random variable, assuming lambda is a random variable.

But you have anyway completely missed the point here; it is the omission of lambda in minkwe's argument that makes it misguided.

That is nonsense. And it speaks volumes. It is you who is missing minkwe's point, and then trying you justify your faulty understanding with more and more nonsense.

***

- Joy Christian
- Research Physicist
**Posts:**2161**Joined:**Wed Feb 05, 2014 4:49 am**Location:**Oxford, United Kingdom

Joy Christian wrote:That is nonsense. And it speaks volumes. It is you who is missing minkwe's point, and then trying you justify your faulty understanding with more and more nonsense.

***

No worries, I'm used to see not-very-rigorous-minds confuse the two.

As to minkwe's (and your) misunderstanding: A(a, lambda) must have the same result as A(a, lambda), for the same values of a and lambda. There is no such thing as A'(a, lambda), different from A(a, lambda).

Last edited by Heinera on Thu Sep 19, 2019 1:11 am, edited 1 time in total.

- Heinera
**Posts:**696**Joined:**Thu Feb 06, 2014 1:50 am

.

In my experience the term "random variable" (RV) is defined differently in different contexts. In the early days (as an engineer), I personally took λ to be a random variable in the context of Bell's work: without reference to probability and statistics. To minimise controversy, I avoid such use now.

It might help us all if you could each provide your definition of an RV. And in the context of Bell's work, if you think it there differs from the definition used in probability and statistics.

PS: I use λ in FUNCTIONS: eg, A(a,λ).

Thanks; Gordon

.

In my experience the term "random variable" (RV) is defined differently in different contexts. In the early days (as an engineer), I personally took λ to be a random variable in the context of Bell's work: without reference to probability and statistics. To minimise controversy, I avoid such use now.

It might help us all if you could each provide your definition of an RV. And in the context of Bell's work, if you think it there differs from the definition used in probability and statistics.

PS: I use λ in FUNCTIONS: eg, A(a,λ).

Thanks; Gordon

.

- Gordon Watson
**Posts:**255**Joined:**Wed Apr 30, 2014 4:39 am

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