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I am going through a blog post written by Thomas Vidick. It states following three assumptions by Bell.

Measurement independence (“free will”): the state $\lambda$ is independent of ${x,y}$ (since Alice and Bob can make their choice of measurement settings independently and freely), $p(\lambda|x,y)=p(\lambda)$.

No-signaling (“no violation of special relativity”): since Alice and Bob perform their measurements in locations that are arbitrarily far apart, the outcome of Alice’s experiment cannot contain any information about Bob’s own choice of measurement. Formally, for any choice $x$ at Alice’s side and any two choices $y,y'$ at Bob’s side, the marginal distribution on outcomes that Alice observes must be the same whether Bob decides to make measurement $y$ or $y'$: $$\displaystyle \forall a,x,y,y',\lambda,\qquad \sum_b p(a,b|x,y,\lambda) \,=\, \sum_b p(a,b|x,y',\lambda)$$.

Outcome independence (“local realism”): since the experiments are performed locally, the outcome $a$ (resp. $b$) can be directly computed as a function of the state of the system, $\lambda$, and the choice of measurement $x$ (resp. $y$). Formally, $$\displaystyle \forall a,b,x,y,\lambda,\qquad p(a,b|x,y,\lambda) \,=\, p(a|x,\lambda)p(b|y,\lambda)$$.

Based on these assumptions I want to prove Exercise 1 in the article.

Exercise 1. Show that any family of distributions satisfying assumptions 1.–3. can be expressed as

$$\displaystyle p(a,b|x,y) = \int_\lambda p(a|x,y,\lambda)p(b|x,y,\lambda) \text{d}\lambda$$

I consider this trivial as both events on the right hand side are independent hence the probability is their product. We evaluate the probability for each given $\lambda$ hence the result is the sum of the probabilities for each $\lambda$.

Is this intuition correct?

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    $\begingroup$ Outcome independence is $p(a|b, x,y,\lambda) = p(a| x,y,\lambda)$. So you have a little more work to obtain the factorization. $\endgroup$
    – Trimok
    Commented Oct 10, 2013 at 17:53

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The formula is a simple corollary of assumption 1,3 adding total probability formula $$p(a,b|x,y)=\int p(a,b|x,y,\lambda)p(\lambda)d\lambda=\int p(a|x,y,\lambda)p(b|x,y,\lambda)p(\lambda|x,y)d\lambda$$

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  • $\begingroup$ Yes, now I understand. $\endgroup$ Commented Sep 14, 2013 at 12:19

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