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This is a made-up example, just to understand a concept. If changing the probability values aids your explanation, that's fine by me.

Say you have a physical quantity $E$ that can take values 1, 2, 3 with probabilities 0.4, 0.25, 0.35 respectively (working in a quantum framework). You have a positive operator valued measure, $E_1, E_2, E_3$, with $E_i$ corresponding to your measurement resulting in value $i$. If $\rho$ is the density operator representing the current state, then you have:

Tr($\rho E_1$) = 0.4, Tr($\rho E_2$) = 0.25, Tr($\rho E_3$) = 0.35

Given just these probability values, is it possible to construct $E_1, E_2, E_3$ "backwards"?

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Yes. Not only is it possible, the solution is not unique and there exists a solution that is independent of $\rho$.

So for $p_k$ being probabilities with $\sum_k p_k = 1$, you can say $E_k = p_k \mathbf{1}$, where $\mathbf{1} = \sum_j |j\rangle \langle j|$ (ie the identity matrix). Then $Tr[ E_k \rho] = p_k Tr[\rho] = p_k$. This works for all $\rho$.

There are also nontrivial solutions. If say your state is a pure state $|\psi\rangle$ in some $d$ dimensional Hilbert space then you only need to find the basis $|k\rangle$ for which $|\psi\rangle = \sum_k \sqrt{p_k} e^{i \phi_k} |k\rangle$.

Also notice something with the probability equations, in $Tr[\rho E_1]$ if you don't assume that $\rho$ is the state it could well be $E_1$ is the state and $\rho$ is the measurement (although unlike $\rho$, $E_k$ need not be normalized). So the question is the same as asking how to construct three states for which measurement of $\rho$ gives the said probabilities.

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if the construction is not unique, for me it's impossible to know which operators has given the measurement, maybe it's only a matter of interpretation of the question. – Ikiperu May 23 '13 at 8:39
That's right. You need to do what's called measurement tomography, which is the same thing as state tomography but instead of engineering the right measurements you engineer the right states to measure. The basic idea is have a set of states that form a complete basis in the Hilbert-Schmidt space of operators, measure them repeatedly and use the resulting expectations to reconstruct the matrix elements. See for example – SMeznaric May 23 '13 at 10:22

If you count the number of variables that you want to find you get 18 (=9+9, 2 hermitian operators, the third is fixed by the others), but you impose only four conditions on these variables (=2x2, the trace may be complex, the third trace is already counted), therefore I think it's impossible.

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