In probing that the entropy $H(p)$ is a concave function one usually construct three probability densities as follows:
Let $X_1$ be a random variable with distribution $p_1$, taking values in the set $A$. Let $X_2$ be another random variable with distribution $p_2$ on the same set. Let $\theta=1$ with probability $\alpha$ and $\theta=2$ with probability $1-\alpha$. Let $Z=X_\theta$. Then the distribution of $Z$ is $\lambda p_1+(1-\lambda)p_2$. Since $H(Z)\geq H(Z|\theta)$ the concavity of $H$ follows easiliy.
My stupid question is how to understand this construction of probabilities. Specially what does it mean $Z=X_\theta$, and why for instance $H(Z|\theta)=\lambda H(p_1)+(1-\lambda)H(p_2)$ and not (!) $H(Z|\theta)=\lambda H((\lambda p_1+(1-\lambda)p_2)/\lambda)+(1-\lambda)H((\lambda p_1+(1-\lambda)p_2)/(1-\lambda))$?
(Part of the answer is given here but I still think that one can explain more intuitively what is $Z$.)
It's a (not very standard) notation for a mixture, that is random variable that is defined as
$$Z=\begin{cases}X_1 & \text{if } \theta = 1\\ X_2 & \text{if } \theta = 2 \end{cases}$$
Then $p(Z|\theta = 1) =p(X_1)$ and $H(Z | \theta = 1) = H(X_1)$ and then, by definition of conditional entropy
$$H(Z | \theta) = p(\theta=1) H(Z | \theta = 1) + p(\theta=2) H(Z | \theta = 2)= \alpha H(X_1) + (1-\alpha) H(X_2)$$