Mutual information of two random variables, where one of them is a mixture Suppose X and Y are two random variables, where Y is a mixture distribution of two other random variables W and Z 
with weight $\alpha$ and $1-\alpha$ respectively.
(To be more precise, Y = W with probability $\alpha$; Y = Z with probability $1-\alpha$)
I know that in general, $$\alpha I(X;W) + (1-\alpha) I(X;Z) \neq I(X;Y) $$
, but what exactly causes this difference? 
What is the information loss (or gain) if I separate $I(X;Y)$ linearly?
 A: The mixing loses information. A simple example:
Suppose $X\in\{0,1\}$ is a fair coin (Bernoulli with $p=1/2$), hence $H(X)=1$ bit. 
Let $Z=X$ and $W=1-X$ ("negated"). Let $Y$ be a fair mixture ($\alpha=1/2$) of $W$ and $Z$.
It's clear that $I(X;Z)=I(X;W)=1$ bit: knowing either $Z$ or $W$ gives us total information about $X$.
But knowing their mixture $Y$ tells us nothing about $X$ (in fact, there are independent), and $I(X;Y)=0$

(Update) In general:
Let $E$ be the indicator variable of the mixture ($E=1$ with prob $\alpha$, $0$ otherwise; so that $Y=W$ if $E=1$, elsewhere $Y=Z$ ). Variables $E,X$ are independent.
Then $$
\begin{align}
I(X;Y  \mid  E)&=H(X  \mid  E) - H(X  \mid  Y,E) \\
&=H(X ) - P(E=0) H(X  \mid  Y,E=0) - P(E=1) P(E=0) H(X  \mid  Y,E=0) \\
&=H(X) - (1-\alpha) H(X  \mid  Z) - \alpha  H(X  \mid  W) \\
&=(1-\alpha) H(X)  + \alpha  H(X)  - (1-\alpha) H(X  \mid  Z) - \alpha  H(X  \mid  W) \\
&= \alpha I(X;W) + (1-\alpha)  H(X ; Z)
\end{align}$$
Hence, the mixture of the mutual informations equals $I(X;Y  \mid  E)$, not $I(X;Y)$.
Further, using the chain rule for mutual information:
$$
\begin{align}
I(X;Y,E) &= I(X;Y) + I(X;E  \mid Y) \\
&=I(X;E) + I(X;Y  \mid E) 
\end{align}$$
Because $I(X;E)=0$ and $I(X;E  \mid Y) \ge 0$, this implies 
$$\alpha I(X;W) + (1-\alpha)  H(X ; Z)=I(X;Y  \mid E) =I(X;Y) + I(X;E  \mid Y) \ge I(X;Y)$$
That is, the mutual information of the mixture is less or equan than the mixture of the mutual informations.
A particular case that gives equality is when the support of the mixed variables $Z,W$ don't overlap. In this case, to know $Y$ is to know $E$, then $I(X;E  \mid Y)=0$ and the mutual information of the mixture equals the mixture of the mutual informations.
