Measuring Pauli strings using generators

Here I am going to show why $langle Z_1 Z_3 rangle$ generally cannot be estimated from $langle Z_1 Z_2 rangle$ and $langle Z_2 Z_3 rangle$. Let's start with an arbitrary three-qubit state:

begin{align*} |psi rangle = c_{000} &|000rangle + c_{001} |001rangle + c_{010} |010rangle + c_{011} |011rangle + +c_{100} &|100rangle + c_{101} |101rangle + c_{110} |110rangle + c_{111} |111rangle end{align*}

And because of this answer the expectation value for $Z_1 Z_2$, $Z_2 Z_3$ and $Z_1 Z_3$ (I am using this convention for qubit indexes $|q_1 q_2 q_3 rangle$):

$$ langle Z_1 Z_2 rangle = |c_{000}|^2 + |c_{001}|^2 - |c_{010}|^2 - |c_{011}|^2 - |c_{100}|^2 - |c_{101}|^2 + |c_{110}|^2 + |c_{111}|^2 langle Z_2 Z_3 rangle = |c_{000}|^2 - |c_{001}|^2 - |c_{010}|^2 + |c_{011}|^2 + |c_{100}|^2 - |c_{101}|^2 - |c_{110}|^2 + |c_{111}|^2 langle Z_1 Z_3 rangle = |c_{000}|^2 - |c_{001}|^2 + |c_{010}|^2 - |c_{011}|^2 - |c_{100}|^2 + |c_{101}|^2 - |c_{110}|^2 + |c_{111}|^2 $$

From this expressions one can see that $langle Z_1 Z_3 rangle$ generally cannot be estimated from $langle Z_1 Z_2 rangle$ and $langle Z_1 Z_3 rangle$. For prove let's consider this conterexample when $|c_{000}| = |c_{010}|$, $|c_{001}| = |c_{011}|$:

begin{equation*} langle Z_1 Z_2 rangle = - |c_{100}|^2 - |c_{101}|^2 + |c_{110}|^2 + |c_{111}|^2 langle Z_2 Z_3 rangle = |c_{100}|^2 - |c_{101}|^2 - |c_{110}|^2 + |c_{111}|^2 langle Z_1 Z_3 rangle = 2|c_{000}|^2 - 2|c_{001}|^2 - |c_{100}|^2 + |c_{101}|^2 - |c_{110}|^2 + |c_{111}|^2 end{equation*}

So for the same $langle Z_1 Z_2 rangle$ and $langle Z_2 Z_3 rangle$ there can be many different possible values for $langle Z_1 Z_3rangle$ because by varying the values $|c_{000}| = |c_{010}|$ and $|c_{001}| = |c_{011}|$ the expectations $langle Z_1 Z_2 rangle$ and $langle Z_2 Z_3 rangle$ will not be changed but $langle Z_1 Z_3 rangle$ will be changed. So $langle Z_1 Z_3 rangle$ cannot be estimated from $langle Z_1 Z_2 rangle$ and $langle Z_2 Z_3 rangle$ in this conterexample.