Why is entropy additive? Although it seems simple, I can't get the derivation correct. Here is my reasoning:
$P(S)=P(A)P(B)$  
Where P is the probability and S, A, and B denote different systems. 
$S_A=-P(A)\ln P(A)$ and $S_B=-P(B)\ln P(B)$
Then $$S_{S}=-P(S)\ln P(S)=-P(S)\ln P(A)P(B)=-P(S)\ln P(A)-P(S)\ln P(B)$$
The problem is, since $P(A)\neq P(S)$ and $P(B)\neq P(S)$, how can entropy be additive?
 A: If you consider a system $C$ and two subsystems $A,B$ with associated probability distributions $p_C,p_A,p_B$ and want the entropy to add, you must assume the subsystems are independent in the sense that $p_C(X) = p_A(Y)p_B(Z)$ where $Y$ and $Z$ are a partitioning of the variables $X$ belonging to $C$. Then, the total entropy is
\begin{align} S_C & = \langle -\ln(p_C)\rangle\\& = -\int p_C(X)\ln(p_C(X))\mathrm{d}X \\
& = -\int p_A(Y)p_B(Z)(\ln(p_A(Y))+\ln(p_B(Z))\mathrm{d}Y\mathrm{d}Z \\
& = -\int p_A(Y)p_B(Z)\ln(p_A(Y))\mathrm{d}Y\mathrm{d}Z - \int p_A(Y)p_B(Z)\ln(p_B(Z)\mathrm{d}Y\mathrm{d}Z \\
& = S_A+ S_B
\end{align}
because $\int p_A(Y)\mathrm{d}Y = 1$ and likewise for $B,Z$.
The assumption of statistical independence is crucial. Entropy is, from the information-theoretic viewpoint, the expected amount of information encoded in a system. If two systems are not independent, obviously the information encoded in them together will be less than the sum of the information you can extract from each without knowing the other. That is, classical entropy is subadditive, but only additive if the systems you are adding together are statistically independent.
