I encountered a knowledge gap while attempting to solve a problem I posed to myself. I have presented the problem along with my reasoning below.
Let $G$, $Q$ and $\Theta$ be 3 random variables such that $g= f(\theta,q)$ where $f$ is a real-valued function. If $Q \sim \mathcal{N}(\mu;\sigma^2)$ and $\Theta \sim \mathcal{B}(1,p)$, we can write that: $$ N(q) = \frac{1}{\sigma\sqrt{2\pi}} exp\bigl\{-\frac{(q-\mu)^2}{2\sigma^2}\bigr\} $$
where $N(q)$ is the density function of $Q$.
And, $$ B(\theta) = p\delta(\theta-1) + (1-p) \delta(\theta) $$ where $B(\theta)$ is the density function of $\Theta$.
Thus, the density function of $G$ can be expressed as: $$ f_G (g) = \int_{\mathbb{R}}\int_{\mathbb{R}} \delta(g-f(\theta,q))N(q)B(\theta) dq d\theta $$ Integrating first over $\theta$ gives: \begin{equation*} f_G (g) = \int_{\mathbb{R}} p \delta(g-f(1,q))N(q) dq + \int_{\mathbb{R}}(1-p) \delta(g-f(0,q))N(q) dq \end{equation*}
Finally, integrating over $q$, using the sifting property of the Dirac delta function, gives: $$ f_G (g) = p \sum_i \frac{N(q_i)}{\Bigl\lvert \left. \frac{\partial f(\theta, q)}{\partial q} \right|_{\theta=1, q_i}\Bigr\rvert} + (1-p) \sum_i \frac{N(q_i)}{\Bigl\lvert \left. \frac{\partial f(\theta, q)}{\partial q} \right|_{\theta=0, q_i}\Bigr\rvert} $$
where the $q_i$'s corresponds are solutions to $g-f(\theta,q) = 0$ for $\theta = 1$ in the first summation and for $\theta = 0$ in the second summation. Obviously, $f_G$ is a density function therefore: $$ \int_{\mathbb{R}} f_G(g)dg = 1 $$
Albeit the use of the word "Obviously", I fail to properly prove the previous assumption. Moreover, I would like to be able to express the CDF of this random variable.
Please feel free to correct my wordings as it may not be rigorous enough. Looking forward to any insights, one can provide.
Much appreciated!
By paying more attention to the equation, I was able to come up with solution.
Indeed, from the equation obtained previously by integrating over $\theta$, we have that:
$$ \int_{\mathbb{R}} f_G(g)dg = p \int_{\mathbb{R}}\int_{\mathbb{R}} \delta(g-f(1,q))N(q) dq dg + (1-p) \int_{\mathbb{R}}\int_{\mathbb{R}} \delta(g-f(0,q))N(q) dq dg $$
Since $q$ is fixed when integrating over $g$, it simplifies to : $$ \int_{\mathbb{R}} f_G(g)dg = p \int_{\mathbb{R}}N(q) dq + (1-p) \int_{\mathbb{R}} N(q) dq $$ The end of the proof of the normalization condition follows naturally from the normalization condition of the density functions of $Q$ and $\Theta$.
From this simple calculation, we can provide the CDF of the above PDF.
Let $F_G(g)$ being the CDF of $G$, i.e. $F_G(g) = \int_{-\infty}^{g} f_G(g_0)dg_0$.
In a similar fashion to the proof of the normalization condition, we have that: $$ F_G(g) = p \int_{\mathbb{R}}\Bigl\{\int_{-\infty}^{g} \delta(g_0-f(1,q))dg_0\Bigr\}N(q) dq + (1-p) \int_{\mathbb{R}}\Bigl\{\int_{-\infty}^{g} \delta(g_0-f(0,q)) dg_0\Bigr\}N(q) dq $$
Using the fact that: $$ \int_{-\infty}^{t} \delta(x-a) dx = \mathcal{H}(t-a) $$
where $\mathcal{H}(x)$ is the Heaviside function defined as: $$ \mathcal{H}(x) = \begin{cases} 0 & \text{if } x < 0 \\ 1 & \text{if } x \geq 0 \end{cases} $$
Thus, we get: $$ F_G(g) = p \int_{\mathbb{R}}\mathcal{H}(g-f(1,q))N(q) dq + (1-p) \int_{\mathbb{R}}\mathcal{H}(g-f(0,q))N(q) dq $$
Therefore: $$ F_G(g) = p \int_{\{q \mid g-f(1,q) \geq 0\}}\mathcal N(q) dq + (1-p) \int_{\{q \mid g-f(0,q) \geq 0\}} N(q) dq $$
Hoping that my naïve question would be of assistance to someone else pondering the same issue as I did.