I have to find $$H(A\mid C)-H(A\mid B)$$ Where I will denote:
$$H(A\mid C) = -\sum_{k>0}p(c_k)\sum_{i>0}p(a_i\mid c_k)\log(p(a_i\mid c_k)) \\ H(A\mid B) = -\sum_{j>0}p(b_j)\sum_{i>0}p(a_i\mid b_j)\log(p(a_i\mid b_j))$$ with $p(A=a_i) = p(a_i), p(B=b_j) = p(b_j), p(C=c_k) = p(c_k)$
When I subtract it, I will obtain: $$\sum_{k>0} p(c_k)\sum_{i>0} p(a_i\mid c_k) \log\left(\frac{1}{(p(a_i\mid c_k))}\right) - \sum_{j>0} p(b_j)\sum_{i>0} p(a_i\mid b_j) \log\left(\frac{1}{(p(a_i\mid b_j))}\right)$$
But the answer needs to be $$\sum_{k>0}\sum_{j>0}p(b_j,c_k)\sum_{i>0}p(a_i\mid b_j)\left(\log \frac{1}{p(a_i\mid c_k)} - \log\frac{1}{p(a_i\mid b_j)} \right)$$ What do I have to do? I tried everything (like the Bayes function), but I can't go further. Can someone help me with the next step(s)?
\begin{align} &\mathsf{E}[\ln\mathsf{P}(A\mid C)-\ln\mathsf{P}(A\mid B)] \\ &\qquad =\sum_{i,j,k}p_{A,B,C}(a_i,b_j,c_k)[\ln p_{A\mid C}(a_i\mid c_k)-\ln p_{A\mid B}(a_i\mid b_j)] \\ &\qquad =\sum_{j,k}p_{B,C}(b_j,c_k)\sum_i p_{A\mid B,C}(a_i\mid b_j,c_k)[\ln p_{A\mid C}(a_i\mid c_k)-\ln p_{A\mid B}(a_i\mid b_j)] \\ &\qquad =\sum_k p_C(c_k)\sum_{i}p_{A\mid C}(a_i\mid c_k)\ln p_{A\mid C}(a_i\mid c_k) \\ &\qquad\quad- \sum_j p_B(b_j)\sum_{i}p_{A\mid B}(a_i\mid b_j)\ln p_{A\mid B}(a_i\mid b_j) \\ &\qquad =H(A\mid B)-H(A\mid C). \end{align}