Dimension of joint probability space in the Bell Scenario In the Bell(CHSH) scenario with the normal two qubits+two measurement settings, how does the dimension of the joint probability space $ P(ab|xy)$ drops from 16 to 8?
In general, for two d-dimensional quantum states and m measurement settings, the dimension is $m^2(d-1)^2+2m(d-1)$, how is this obtained?
 A: Trying to reconstruct the reduction, I obtained slightly different results. I haven’t worked a lot on Bell inequalities, so I am not 100% sure of the quantitative results, but I’m pretty sure the approach I give is the good one,
but I am less sure for the normalization conditions.
CHSH Scenario
In the CHSH scenario, the naive initial dimension is $2^4=16$.
The no signalling condition (forbiding Alice to get information on $x$ and Bob to get information on $y$) imposes, for each $a$ and $x$, $∑_{b}P(ab|x0)=∑_{b}P(ab|x1)$ and, for each $b$ an $y$, $∑_{a}P(ab|0y)=∑_{a}P(ab|1y)$. That amounts to 8 constraints, reducing the dimension to $16-8=8$.
I think the four normalization conditions $\sum_{ab}P(ab|xy)=1$ are then
reduced to $3$ independent condition, reducing the final dimension to $5$
Generic scenario
More generally, in the $d$-dimensional setting with $m$ measurements we initially have a dimension of $m^2d^2$. Then, 
the no-signalling condition , gives the following $2m(m-1)d$ constraints:
\begin{align}
  P(a|x)&=\sum_{b=0}^{d-1}P(ab|xy), &&∀a∈\{0,…,d-1\} ∀x,y∈\{0,…,m-1\} \\
  P(b|y)&=\sum_{a=0}^{d-1}P(ab|xy), &&∀b∈\{0,…,d-1\} ∀x,y∈\{0,…,m-1\}. 
\end{align}
The Normalization condition then reduces to the $2m$ constraints $∑_{a}P(a|x)=1$ and $∑_{b}P(b|x)=1$, with only $2m-1$ independent ones? 
We have then
\begin{align}
  m^2d^2 -2m(m-1)d - (2m -1) &=m^2(d^2-2d) + 2m(d-1) +1
\\
    &=m^2(d-1)^2 - m^2 + 2m(d-1) +1
\end{align}
Which looks a bit like the formula you gave, but is not identical. I hope this helps (and I’d like to know where my mistakes are !)
