In Bell nonlocality, why does $P(ab|xy)\neq P(a|x)P(b|y)$ mean the variables are not statistically independent?



I've been working through the paper Bell nonlocality by Brunner et al. after seeing it in user glS' answer here. Early on in the paper, the standard Bell experimental setup is defined:

Bell experiment

Where $x, y \in \{0,1\}$, $a, b \in \{-1, 1\}$, and the two people (Alice & Bob) measure a shared quantum system generated by $S$ according to their indepedent inputs $x$ and $y$, outputting the results as $a$ and $b$.

The paper then has the following equation:

$P(ab|xy) \ne P(a|x)P(b|y)$

And claims the fact this is an inequality means the two sides are not statistically independent. It's been a long time since I took probability & statistics in university, so I'm interested in this equation, what it means, and why it is a test for statistical independence. Why is this equation used, and what is the intuitive meaning of each side? I have basic knowledge of conditional probability and Bayes' theorem.


Posted 2018-11-14T06:44:34.750

Reputation: 3 448

@ Danylo Y found a major error in my answer, can you please remove your acceptance, so I can delete it – David Bar Moshe – 2020-12-01T14:42:14.903



Statistical independence of random variables means that their joint probability is the product of their marginal probabilities: $$P(A∩B) = P(A) P(B)$$ The following Venn diagram, in which the event $A$ occupies the left half of the big square, while the event $B$ occupies the lower half, exemplifies the above relation. ($A^c$ and $B^c$ denote the complements).

enter image description here

The above relation extends to conditional probabilities:

$$P(A∩B|X) = P(A|X) P(B|X)$$ Since we can imagine the above diagram inside a subset in which the event $X$ happened. So, in the question's notation: $$P(ab|xy) = P(a|xy) P(b|xy)$$ Now, the locality of the measurements (explicitly illustrated in the black boxes) implies that the variable $a$ is independent of $y$ and similarly the variable $b$ is independent of $x$. Thus $$P(a|xy) = P(a|x)$$ $$P(b|xy) = P(b|y)$$ Thus, in our case, independence would imply that: $$P(ab|xy) = P(a|x) P(b|y)$$

David Bar Moshe

Posted 2018-11-14T06:44:34.750

Reputation: 1 859

@DavidBarMoshe Your explanation is misleading. What's the point of introducing $X$ if you assume that $P(X)=1$? Otherwise $P(A∩B|X) = P(A|X) P(B|X)$ is wrong in general. Conditioning doesn't preserve independence. – Danylo Y – 2020-12-01T14:06:37.337

@ Danylo Y @ahelwer Indeed, this answer is wrong, I'll delete it – David Bar Moshe – 2020-12-01T14:35:09.443

Ah, okay! So really the important inequalities here are $P(a|xy) \ne P(a|x)$ and $P(b|xy) \ne P(b|y)$, which stated together is $P(ab|xy) \ne P(a|x)P(b|y)$? – ahelwer – 2018-11-14T19:30:10.167

Yes, the principle is that conditioning of quantum probabilities does not preserve independence, even if the observables belong to a commutative subalgebra of the system's operator algebra. In our case the observables are $I \otimes \sigma_z$ and $ \sigma_z \otimes I$ which commute. The conditioning preserves the independence in a separable state, for example $|\psi\rangle = |1\rangle \otimes 1 \rangle$, where exactly the above square Venn diagram is obtained, but not in an entangled state, such as the Bell state – David Bar Moshe – 2018-11-15T10:02:34.667

Just to be 100% sure here - in the equation's notation, $P(ab|xy)$ means $P(a \cap b | x \cap y)$ and not $P(a \cdot b | x \cdot y)$? Since multiplication would be defined over the values of $a,b,x,y$... – ahelwer – 2018-11-18T23:59:22.993

Yes, this is correct. – David Bar Moshe – 2018-11-19T06:58:00.693


It perhaps helps to express $P(ab|xy)$ in words:

the probability that Alice gets answer A and Bob gets answer B given that choices x and y were made

Now independence in classical probability holds if and only if $$ P(e_1\text{ and }e_2)=P(e_1)P(e_2) $$ where $e_1$ and $e_2$ are events, and practically, you can see what it means through Bayes' theorem $$ P(e_1|e_2)=\frac{P(e_1\text{ and }e_2)}{P(e_2)}=P(e_1) $$ i.e. for independent events, the condition probability is independent of the conditioning.

Now, we could rewrite $$ P(ab|xy)=P(a|bxy)P(b|xy) $$ you lose a bit of symmetry by doing it like this, but only briefly. Now, the idea of independence in the current context is that Alice's result should not depend on anything that happens on Bob's side, so $P(a|bxy)=P(a|x)$ and Bob's result should be independent of anything that happens on Alice's side, so $P(b|xy)=P(b|x)$. Hence, independence between Alice and Bob would imply $$ P(ab|xy)=P(a|x)P(b|y). $$ So, if this condition holds for all $a,b,x,y$, then the probability distribution is independent, otherwise, the results are not statistically independent.


Posted 2018-11-14T06:44:34.750

Reputation: 35 722


The equation

$P(ab|xy) = P(a|x)P(b|y)$

would imply that any dependence that the output $ab$ has on the inputs $xy$ (expressed by the lhs) is solely due to $a$ depending on $x$ alone, and $b$ depending on $y$ alone. This is expressed by the rhs by treating the value of $a$ and its dependence on $x$ as an independent event from the value of $b$ and its dependence on $y$, and hence the probability of a particular $ab$ is the product of these independent probabilities.

In Bell's equalities, we see that this is not the case. Correlations between $a$ and $b$ depend explicitly on which pair of inputs $xy$ is chosen. Specifically, for the CHSH inequality, whether or not $a$ agrees with $b$ depends on both $x$ and $y$, and hence $a$ cannot be said to be independent of $y$ (nor $b$ of $x$). This behaviour means that there will be at least some values of $a$, $b$, $x$ and $y$ where the above equality does not hold.

James Wootton

Posted 2018-11-14T06:44:34.750

Reputation: 9 708