# The $I_{3322}$ Inequality

Bell inequalities are equivalent if you can map one to another by

1. Relabelling the parties (in this case A and B)
2. Relabelling the inputs (in this case 1,2,3)
3. Relabelling the outputs (in this case -1 and 1 for Śliwa version, 0 and 1 for the Pál and Vértesi version, and undefined for the other two versions)
4. Scaling by a nonzero constant.

These equivalences are explored here, for example. The last two conditions are not relevant for this question, as the notations used here, correlators and Collins-Gisin, do not represent them.

We start by showing that the Śliwa version is equivalent to the Collins and Gisin's version. For that, we pass the latter from Collins-Gisin notation to correlator notation via the formulas $$4p_{ij} = E(A_iB_j) + E(A_i) + E(B_j) +1$$ $$2p_i^A = E(A_i)+1$$ $$2p_j^B = E(B_j)+1$$ Scaling the inequality by 4, for convenience, we end up with $$E(A_1B_1)+E(A_1B_2)+E(A_1B_3)+E(A_2B_1)+E(A_2B_2)-E(A_2B_3)+E(A_3B_1)-E(A_3B_2)+E(A_1)+E(A_2)-E(B_1)-E(B_2) \leq 4,$$ where we see that the asymmetry in the coefficients became an asymmetry in the signs. We can take care of that by relabelling Alice's outputs, mapping -1 to +1 and vice-versa. This changes the expectation values as $$E(A_iB_j) \mapsto -E(A_iB_j)$$ $$E(A_i) \mapsto -E(A_i)$$ $$E(B_j) \mapsto E(B_j)$$ And the inequality becomes $$-E(A_1B_1)-E(A_1B_2)-E(A_1B_3)-E(A_2B_1)-E(A_2B_2)+E(A_2B_3)-E(A_3B_1)+E(A_3B_2)-E(A_1)-E(A_2)-E(B_1)-E(B_2) \leq 4,$$ which is Śliwa's version.

Now, we can obtain Brunner et al.'s version either from Śliwa's version by changing back to Collins-Gisin notation, or directly from Collins and Gisin's version by relabelling Alice's output. It's more interesting to do the latter.

The idea of the Collins-Gisin notation is to show the probabilities only of $$n-1$$ outcomes out of $$n$$; the other probability is implicit from the normalisation condition. In this case $$n=2$$, so all the probabilities refer to obtaining the outcome $$0$$, and the probabilities of outcome $$1$$ are implicit (here the labels are arbitrary). Thus after relabelling Alice's outcome we have to rewrite the probabilities as functions of the outcome $$0$$. This is done via the identities $$p(01) = p^A(0)-p(00)$$ $$p(10) = p^B(0)-p(00)$$ $$p(11) = 1+p(00)-p^A(0)-p^B(0)$$ $$p^A(1) = 1-p^A(0)$$ $$p^B(1) = 1-p^B(0)$$ So if we relabel Alice's outcome from 0 to 1, this changes the probabilities in the inequality as $$p_{ij} \mapsto p_j^B-p_{ij}$$ $$p^A_{i} \mapsto 1-p^A_i$$ $$p^B_j \mapsto p^B_j$$ Applying this transformation to the Collins and Gisin's version, we obtain $$-p_{11}-p_{12}-p_{13}-p_{21}-p_{22}+p_{23}-p_{31}+p_{32}+p_1^A+p_1^B\leq 1.$$ Which is not equal to Brunner et al.'s version. They made a sign mistake.

This is a partial answer, that I will be updating as I get more insight.

First, in Pal and Vertesi's article they use correlator notation, but the inequality and bounds they give to it only hold if the operators $A_i$ and $B_j$ are actually projectors, so what they really want to mean is $E(A_iB_j)=p_{ij}$ and $E(P_i)=p^P_i$. This can be checked not only when computing the bounds they give for $I_{3322}$, but is also the case for the CHSH inequality they give in Eqs. (1) and (2).

Then, the inequality is the same as that given by Collins and Gisin when changing the labels of the measurements $1\leftrightarrow 2$ in both parties.

Second, if one transforms Sliwa's version to probability notation (given that the measurements are binary, we can substitute $E(P_i) = 2p^P_i - 1$ and $E(A_iB_j)=4p_{ij}-2(p^A_i+p^B_j)+1$) one obtains the form that appears in Brunner's article.

So, we have grouped the four expressions in two families. Now what is left is to connect them both.