CHSH Correlations and Joint-Distribution Extensions

The four pair distributions in a CHSH experiment need not be marginals of one probability distribution for four jointly defined classical variables. Fine’s theorem characterizes exactly when such an extension exists in the binary, two-setting scenario. This page develops that extension problem and constructs a nontrivial example. The canonical CHSH article gives the local-bound derivation, Tsirelson bound, and quantum realization.

Required background. CHSH Inequality defines the four correlations and their local bound; Measure Theory Minimum supplies marginalization of finite probability distributions.

Four observed pairs and one proposed extension

Let $x,y\in\{0,1\}$ and $a,b\in\{-1,1\}$. Given four normalized, nonnegative distributions $p_{xy}(a,b)$, ask for 16 numbers

$$q(a_0,a_1,b_0,b_1)\ge0,\qquad \sum q=1,$$

whose $(a_x,b_y)$ marginals are $p_{xy}$. This is an ordinary finite linear feasibility problem. The unobserved joint probabilities for $a_0,a_1$ or $b_0,b_1$ are unknowns; they are not measured quantum joint distributions of incompatible observables.

A necessary condition is consistent one-party marginals:

$$\begin{aligned} \sum_b p_{x0}(a,b)&=\sum_b p_{x1}(a,b),\\ \sum_a p_{0y}(a,b)&=\sum_a p_{1y}(a,b). \end{aligned}$$

These are the no-signaling equalities for the observed scenario. Without them, no extension exists even if one chosen CHSH expression happens to lie between $-2$ and $2$.

Fine’s theorem and its precise scope

For the four valid pair distributions with consistent marginals, the following are equivalent:

1. A nonnegative normalized extension $q$ exists.
2. The behavior has a measurement-independent Bell-local model.
3. All eight CHSH inequalities hold:

$$\sum_{x,y}s_{xy}E_{xy}\le2, \qquad s_{xy}\in\{-1,1\},\quad \prod_{x,y}s_{xy}=-1,$$

where $E_{xy}=\sum_{a,b}ab\,p_{xy}(a,b)$. The equivalence with the complete inequality family is Fine’s theorem; we use that result rather than claiming to derive its general sufficiency from a single correlation bound.

The first two statements have a direct proof. Given $q$, use the four-entry table as the hidden variable and read out $a_x,b_y$ deterministically. Conversely, a local model with responses $p_A,p_B$ gives

$$q(\boldsymbol a,\boldsymbol b) =\int\mu(d\lambda) \prod_{x=0}^1p_A(a_x\mid x,\lambda) \prod_{y=0}^1p_B(b_y\mid y,\lambda).$$

Positivity is immediate. Summing all entries gives one, and summing the unselected entries recovers each observed pair. The construction chooses conditionally independent auxiliary outcomes for different settings; it does not assert that incompatible quantum measurements can actually be performed together.

Necessity of every CHSH inequality follows by averaging the deterministic bound over $q$. General sufficiency is special to this scenario. With more settings or outcomes, a chosen family of CHSH tests can miss other Bell inequalities.

An explicit noisy-correlation extension

Consider unbiased marginals with

$$p_{xy}(a,b)=\frac14(1+abE_{xy}), \qquad (E_{00},E_{01},E_{10},E_{11})=(t,t,t,-t).$$

For $|t|\le1$ these are valid no-signaling pair distributions. Define

$$\begin{aligned} K&=a_0b_0+a_0b_1+a_1b_0-a_1b_1,\\ q_t&=\frac1{16}(1+tK). \end{aligned}$$

Every deterministic table has $K=\pm2$. Thus $q_t\ge0$ precisely when $|t|\le1/2$. Summing any single sign variable kills terms containing it; the four required marginals are exactly the $p_{xy}$. This proves existence constructively throughout the local interval. Outside that interval, $|E_{00}+E_{01}+E_{10}-E_{11}|=4|t|>2$, so no alternative positive extension can work.

The same algebraic formula still reproduces the marginals for larger $t$, but some entries become negative. Such a signed extension is not a classical hidden-variable probability distribution. It shows concretely where positivity, rather than normalization alone, enters the theorem.

Exercise: inspect the boundary distribution

At $t=1/2$, determine how many of the 16 entries of $q_t$ are nonzero and their values. Why can the observed pairs nevertheless all have positive probabilities?

Solution

Flipping both Alice signs changes $K$ to $-K$ and pairs the tables. Eight therefore have $K=2$ and eight have $K=-2$. Their weights are $1/8$ and zero respectively. A pair marginal sums four entries and equals either $3/8$ or $1/8$, so all observed pair probabilities are positive. Zeros in the unobserved four-variable extension do not require zeros in its pair marginals. The boundary of the local set can occur while every observed outcome remains possible.

References

• Clauser, John F., Michael A. Horne, Abner Shimony, and Richard A. Holt, “Proposed experiment to test local hidden-variable theories,” Physical Review Letters 23, 880–884 (1969), doi:10.1103/PhysRevLett.23.880 — the CHSH constraint.
• Fine, Arthur, “Hidden Variables, Joint Probability, and the Bell Inequalities,” Physical Review Letters 48, 291–295 (1982), doi:10.1103/PhysRevLett.48.291 — the joint-distribution equivalence used here.