concave in the vector of the ensemble probabilities $P(i)$. This is straightforward if we first note that the mutual information, because it is, in fact, symmetric between $i$ and $j$, can be written in the reverse form $$\begin{aligned}
  M(I\! :\! J) & = & H[P(j)] - \sum_j P(i) H[P(j|i)] ,
  \label{revMI}\end{aligned}$$ Since, for a fixed measurement, the mutual information is a function of the ensemble probabilities we will write it as $M(\{P(i)\})$. Denoting the pure states in the encoding ensemble as $\rho_i =
|\psi_i\rangle\langle\psi_i|$, and choosing the ensemble state $\rho = \sum_k P_k \sigma_k$, where the $\sigma_k$ are built from the encoding states so that $\sigma_k = \sum_i
P_{i|k} |\psi_i\rangle\langle\psi_i|$, then $$\begin{aligned}
  & & M(\{P(i)\}) \nonumber \\
                  & = & H[\sum_k P(k) P(j|k)] - \sum_i \sum_k P(i|k) P(k) H[P(j|i)] \nonumber \\
                & \ge & \sum_k P(k) H[P(j|k)] - \sum_k P(k) \sum_i P(i|k) H[P(j|i)] \nonumber \\
                  & = & \sum_k P(k) M(\{P(i|k)\}) ,
\label{cconc}\end{aligned}$$ being the desired concavity relation. The inequality in the third line is merely a result of the concavity of the Shannon entropy. Note that while we have written the measurement’s outcomes explicitly as being discrete in the about derivation, the result also follows if they are a 