TITLE: Entropy of convolution of measures
QUESTION [6 upvotes]: Let $G$ be a countable, discrete group, and let $\mu_1,\mu_2$ be probability measures on the group $G$. We define the entropy of $\mu_i$ as
$H(\mu_i)=\sum\limits_{g \in G}-\mu_i(g)\log(\mu_i(g))$ (with the convention that $0 \cdot \log0=0$).
Recall that convolution of $\mu_1$ and $\mu_2$ is defined as:
$(\mu_1 \star \mu_2)(g)=\sum\limits_{h \in G}\mu_1(gh^{-1})\mu_2(h)$.
Now, it's a fact that $H(\mu_1 \star \mu_2) \leq H(\mu_1)+H(\mu_2)$. The usual proof goes by showing that entropy decreases under factor maps. 
My question is: Is there a completely hands on proof of this fact- preferably using the concavity of $\log x$ or convexity of $-x\log x$?

REPLY [3 votes]: Knowing that the convolution of densities correspond to the sum of independent variables, letting $Z=X+Y$ , and using the chain rule: $H(Z, Y ) = H( Z) + H(Y|Z)=H(Y)+H(Z|Y) $ we have
$$
\begin{array}{rclllll}
H(Z) &=& H(Z|Y)&+&H(Y)&-&H(Y|Z)  \\ 
&=&H(X)&+& H(Y) &-&H(Y|Z)\\ 
&\le&H(X)&+& H(Y) &&\\
\end{array}
$$
For the second row we've used $H(Z|Y)=H(X+Y|Y)=H(X|Y)=H(X)$ where the last equality follows from the independence of $X,Y$, and the previous from $H(X+a)=H(X)$
(Notice that here I'm using a different notation than in the question, in $H(X)$ $X$ is the random variable, not the probability measure; that's why $H(\mu_1 \star \mu_2)$ in your notation corresponds to $H(X+Y)=H(Z)$ here.)
BTW: This also says that the equality holds only if $H(Y|Z)=0$, that is, when the sum is one-to-one, when knowing the  sum you can know the summands.