I would add that it's a bit weird though that $J(\rho) = 1$ for both the maximally mixed state and the maximally entangled state. Surely, a measure of correlations ought to give you 0, at least for the maximally mixed state?

@user1936752 well, not really. For the max mixed state, you have a purely classical situation, with $X$ and $Y$ being fully correlated. The correlation is maximal because knowing $Y$ gives you full information about $X$, and thus $J(\rho)=I(X:Y)=H(\rho_A)=H(\vec p)$. The fact that you get the same value for the maximally entangled state is because you are not actually exploiting the entanglement of the state, and therefore you cannot see the "quantumness" at all. This is like $|0\rangle+|1\rangle\sim |0\rangle\!\langle0|+|1\rangle\langle1|$ if you only measure in the computational basis

@user1936752 sorry, I don't follow. For the maximally mixed state, the reduced state is also maximally mixed, thus measuring in the computational basis $J(\rho)=S(\rho_A)$, which can be as high as $\log N$ with $N$ the dimension of the embedding space. You do have correlations, as measuring something on $B$ determines what will be measured at $A$. What you might mean is that for the maximally mixed state the correlations are all of this kind, as you do not have non-classical correlations

According to the defintion of $J$, what you do is take a bipartite state. So here we have $I_{AB} = I_A\otimes I_B$. $S(\rho_A) = 1$ since the reduced state is also maximally mixed. According to the defintion, what you do next is measure system $B$ in the computational basis assuming the outcome is $i$, check $S(\rho_A^i)$. This is still maximally mixed since measuring in $B$ can't change the $A$ part. So for the maximally mixed state, $J(\rho) = 0$. If Alice and Bob hold a maximally mixed state, Bob can do whatever he wants with his state and there can be no effect on Alice's side.

@user1936752 I don't know where you are taking these numbers from. $S(\rho_A)$ is not $1$ in general. If $\rho_A=\sum_i p_i |i\rangle$ as in my example, then $S(\rho_A)$ coincides with the Shannon entropy of the vector $\vec p$ of probabilities. This can go as high as $\log N$. Also, measuring in $B$ definitely does change the $A$ part of the system: measuring $i_0$ in the computational basis at $B$ "collapses" the state of $A$ at $|i_0\rangle$. This is a "classical" kind of collapse, at it really just describes how the description of $A$ should change knowing the results of $B$.

For the maximally mixed state, I'm pretty sure you have any correlations. (You can replace 1 in my argument with log(d), that's just me being lazy). But the first point is important - measuring something in B does not do anything to A if the global state is maximally mixed.

I gotta go now, but we can chat later if it's still not clear

@user1936752 sorry but that's nonsense. You might be thinking of product states, for which that holds, but it's definitely not true for maximally mixed states

A maximally mixed state can always be written as a product state. $I_{AB} = I_A\otimes I_B$. The statement that you can observe correlations if two parties share a maximally mixed state is simply not true

If you still think a max mixed state can generate correlations, please provide a reference.