Bell nonlocality and conditional independence











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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.










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    up vote
    4
    down vote

    favorite
    1












    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.










    share|improve this question


























      up vote
      4
      down vote

      favorite
      1









      up vote
      4
      down vote

      favorite
      1






      1





      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.










      share|improve this question















      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.







      mathematics quantum-correlation






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      edited Nov 15 at 6:22

























      asked Nov 14 at 6:44









      ahelwer

      1,008111




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






          share|improve this answer





















          • 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
            Nov 14 at 19:30










          • 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 $|psirangle = |1rangle otimes 1 rangle$, where exactly the above square Venn diagram is obtained, but not in an entangled state, such as the Bell state
            – David Bar Moshe
            Nov 15 at 10:02












          • 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
            6 hours ago


















          up vote
          2
          down vote













          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.






          share|improve this answer




























            up vote
            2
            down vote













            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
            $$
            P(e_1text{ and }e_2)=P(e_1)P(e_2)
            $$

            where $e_1$ and $e_2$ are events, and practically, you can see what it means through Bayes' theorem
            $$
            P(e_1|e_2)=frac{P(e_1text{ and }e_2)}{P(e_2)}=P(e_1)
            $$

            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.






            share|improve this answer





















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              3 Answers
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              active

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              3 Answers
              3






              active

              oldest

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              active

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              active

              oldest

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              up vote
              4
              down vote



              accepted










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






              share|improve this answer





















              • 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
                Nov 14 at 19:30










              • 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 $|psirangle = |1rangle otimes 1 rangle$, where exactly the above square Venn diagram is obtained, but not in an entangled state, such as the Bell state
                – David Bar Moshe
                Nov 15 at 10:02












              • 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
                6 hours ago















              up vote
              4
              down vote



              accepted










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






              share|improve this answer





















              • 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
                Nov 14 at 19:30










              • 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 $|psirangle = |1rangle otimes 1 rangle$, where exactly the above square Venn diagram is obtained, but not in an entangled state, such as the Bell state
                – David Bar Moshe
                Nov 15 at 10:02












              • 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
                6 hours ago













              up vote
              4
              down vote



              accepted







              up vote
              4
              down vote



              accepted






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






              share|improve this answer












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







              share|improve this answer












              share|improve this answer



              share|improve this answer










              answered Nov 14 at 8:38









              David Bar Moshe

              6445




              6445












              • 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
                Nov 14 at 19:30










              • 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 $|psirangle = |1rangle otimes 1 rangle$, where exactly the above square Venn diagram is obtained, but not in an entangled state, such as the Bell state
                – David Bar Moshe
                Nov 15 at 10:02












              • 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
                6 hours ago


















              • 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
                Nov 14 at 19:30










              • 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 $|psirangle = |1rangle otimes 1 rangle$, where exactly the above square Venn diagram is obtained, but not in an entangled state, such as the Bell state
                – David Bar Moshe
                Nov 15 at 10:02












              • 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
                6 hours ago
















              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
              Nov 14 at 19:30




              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
              Nov 14 at 19:30












              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 $|psirangle = |1rangle otimes 1 rangle$, where exactly the above square Venn diagram is obtained, but not in an entangled state, such as the Bell state
              – David Bar Moshe
              Nov 15 at 10:02






              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 $|psirangle = |1rangle otimes 1 rangle$, where exactly the above square Venn diagram is obtained, but not in an entangled state, such as the Bell state
              – David Bar Moshe
              Nov 15 at 10:02














              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
              6 hours ago




              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
              6 hours ago












              up vote
              2
              down vote













              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.






              share|improve this answer

























                up vote
                2
                down vote













                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.






                share|improve this answer























                  up vote
                  2
                  down vote










                  up vote
                  2
                  down vote









                  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.






                  share|improve this answer












                  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.







                  share|improve this answer












                  share|improve this answer



                  share|improve this answer










                  answered Nov 14 at 8:11









                  James Wootton

                  5,7551942




                  5,7551942






















                      up vote
                      2
                      down vote













                      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
                      $$
                      P(e_1text{ and }e_2)=P(e_1)P(e_2)
                      $$

                      where $e_1$ and $e_2$ are events, and practically, you can see what it means through Bayes' theorem
                      $$
                      P(e_1|e_2)=frac{P(e_1text{ and }e_2)}{P(e_2)}=P(e_1)
                      $$

                      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.






                      share|improve this answer

























                        up vote
                        2
                        down vote













                        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
                        $$
                        P(e_1text{ and }e_2)=P(e_1)P(e_2)
                        $$

                        where $e_1$ and $e_2$ are events, and practically, you can see what it means through Bayes' theorem
                        $$
                        P(e_1|e_2)=frac{P(e_1text{ and }e_2)}{P(e_2)}=P(e_1)
                        $$

                        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.






                        share|improve this answer























                          up vote
                          2
                          down vote










                          up vote
                          2
                          down vote









                          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
                          $$
                          P(e_1text{ and }e_2)=P(e_1)P(e_2)
                          $$

                          where $e_1$ and $e_2$ are events, and practically, you can see what it means through Bayes' theorem
                          $$
                          P(e_1|e_2)=frac{P(e_1text{ and }e_2)}{P(e_2)}=P(e_1)
                          $$

                          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.






                          share|improve this answer












                          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
                          $$
                          P(e_1text{ and }e_2)=P(e_1)P(e_2)
                          $$

                          where $e_1$ and $e_2$ are events, and practically, you can see what it means through Bayes' theorem
                          $$
                          P(e_1|e_2)=frac{P(e_1text{ and }e_2)}{P(e_2)}=P(e_1)
                          $$

                          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.







                          share|improve this answer












                          share|improve this answer



                          share|improve this answer










                          answered Nov 14 at 8:30









                          DaftWullie

                          10.5k1534




                          10.5k1534






























                               

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