It is known fact for random variables $(X,Y) \sim p(x,y)=p(x)p(y|x)$ the mutual information is concave function of $p(x)$ for fixed $p(y|x)$.
I have two confusions in interpreting the above fact:
1) when it says "....concave function of $p(x)$", does that mean function of the probability of one particular realization of $X$ i-e $x$ ?
2) If yes, let us suppose $p(x)$ is dependent on certain parameter $\theta$ i-e $p(x;\theta)$. When $\theta$ increases probability of one particular realization $x$ gets decreased i-e $p(X=x)$ decreases as $\theta$ increases. How the concavity of mutual information between $X$ and $Y$ relates to $\theta$?
No.
A concave function has the property $f(\frac{a+b}{2}) \ge \frac{f(a)+f(b)}{2}$ [*]
In the case of the mutual information, the variables ($a,b$ above) are the probabilities densities themselves. So that would translate as
$$ I_{p_m(x)} \ge \frac{I_{p_1(x)}+I_{p_2(x)}}{2} $$
where $p_m(x) =\frac{p_1(x)+p_2(x)}{2}$. In words: if we construct a (weighted) average of two density functions, then the mutual information of the averaged density is greater than the (weighted) average of the original mutual informations.
[*] Replace for strict inequality if strictly concave. Further, notice that the inequality must also be true for other linear combinations, $\alpha a + (1-\alpha) b$, I'm using a particular case for the sake of illustration)