Mixture Density and Relation to Function of R.V.

Question

Let us consider a probability density given by
$$
f(x) = a f_1(x) + (1-a) f_2(x),
$$
where $x \in \mathbb{R}$ and $f_1, f_2$ are known probability densities and $0 \leq a \leq 1$ is given constant. Additionally, we know
$$
\mu_1 := \int_{-\infty}^{\infty} x f_1(x)\,dx, \quad \mu_2 := \int_{-\infty}^{\infty} x f_2(x)\,dx,\\
\sigma_1^2 := \int_{-\infty}^{\infty} (x-\mu_1)^2 f_1(x)\,dx, \quad \sigma_2^2 := \int_{-\infty}^{\infty} (x-\mu_2)^2 f_2(x)\,dx.
$$
Then, if I'm not mistaken, the expectation and the variance of of the mixture pdf should be
$$
E[X] = \mu_2 + a (\mu_1 – \mu_2), \quad \mathrm{Var}[X] = \sigma_{2}^2 + a(\sigma_{1}^2 – \sigma_{2}^2) + a(1-a)(\mu_1 – \mu_2)^2.
$$
If $X_1$ and $X_2$ are two random variables distributed according to the pdf $f_1$ and $f_2$, respectively. Can we relate the above mixture density to a random variable $X$, which is a function of $X_1$ and $X_2$? Let us e.g. consider
$$
Z = a X_1 + (1-a) X_2.
$$
The expectation and variance of $Z$ are then
$$
E[Z] = a E[X_1] + (1-a) E[X_2] = a \mu_1 + (1-a) \mu_2
$$
and
$$
\mathrm{Var}[Z] = \mathrm{Var}[a X_1 + (1-a) X_2] = a^2 \mathrm{Var}[X_1] + (1-a)^2 \mathrm{Var}[X_2] + 2 a (1-a) \mathrm{Cov}[X_1, X_2].
$$
While the expectation of $Z$ is the same as the expectation of $X$, the relation between the variance of $Z$ and the variance of $X$ is rather unclear to me.

Answer 1

Figured I’d turn the comments into answer:

Let $B$ be a Bernoulli RV with parameter $a\in[0,1]$ so $B=1$ with chance $a$ and $B=0$ with chance $1-a$, let $X_i$ be an RV with PDF $f_i(x)$ for $i=1,2$. Assume $B$ is independent of $X_i$ for all $i$.

Then we claim the RV defined by $$Z=BX_1 + (1-B)X_2$$ has the mixture density $$f(x)=af_1(x)+(1-a)f_2(x).$$

That $Z$ has the appropriate expected value is easily verified. The variance can be computed and verified via law of total variance, comment for further details if you have trouble, clarifications, or if you spot any mistakes I missed.