CDF of a sum of continuous and discrete dependent random variables Let $\psi_1$ be a Normal random variable with mean $\mu_1$ and standard deviation  $\sigma_1$. Let $\xi$ be defined as
$$
\xi=c\,\mathbb{1}_{\left\{\psi_2+\psi_1\leq 0\right\}},
$$
where $\mathbb{1}$ is the indicator function, and $\psi_2$ a Normal random variable with mean $\mu_2$ and standard deviation  $\sigma_2$. Thus $\xi$ is a discrete random variable that can be either $c$ or $0$. The problem is to compute the following CDF:
$$
F\left(\alpha\right)=\mathbb{P}\left[\psi_2+\xi\leq \alpha\right].
$$
Since the variable $\xi$ is discrete I cannot use 
$$
\mathbb{P}\left[X+Y\leq \alpha\right] = \int_{-\infty}^{\infty}\int_{-\infty}^{v=\alpha-u}f_{X,Y}\left(u,v\right)\,du\,dv =  \int_{-\infty}^{\infty}\int_{-\infty}^{v=\alpha-u} f_{Y\mid X}\left(v\mid u\right)\,f_{X}\left(u\right)\,dv\,du
$$
 A: Recall that $\{ \psi_2 + \xi \leqslant \alpha \} = \{\psi_2 + c [ \psi_2+\psi_1 \leqslant 0] \leqslant \alpha \} = \{\psi_1 \leqslant -\psi_2, \psi_2 + c \leqslant \alpha \} \lor \{\psi_1 > -\psi_2, \psi_2 \leqslant \alpha \} $
and the latter two events are disjoint. Hence
$$ \begin{eqnarray}
  F(\alpha) &=& \Pr\left(\psi_2 + \xi \leqslant \alpha\right) = \Pr\left(\psi_1 \leqslant -\psi_2, \psi_2 + c \leqslant \alpha\right) + \Pr\left(\psi_1 > -\psi_2, \psi_2 \leqslant \alpha\right) \\
  &=& \Pr\left( \psi_2 \leqslant \min(\alpha-c, -\psi_1) \right) + \Pr\left(-\psi_1 < \psi_2 \leqslant \alpha\right) \\
  &=& \mathbb{E}\left( \Phi\left(\frac{\min(\alpha-c, -\psi_1)-\mu_2}{\sigma_2}\right) \right) + \mathbb{E}\left( \Phi\left( \frac{\alpha-\mu_2}{\sigma_2}\right) - \Phi\left( \frac{-\psi_1-\mu_2}{\sigma_2}  \right)\right) \\
  &=&  \Phi\left( \frac{\alpha-\mu_2}{\sigma_2}\right)  + \mathbb{E}\left( \Phi\left(\frac{\min(\alpha-c, -\psi_1)-\mu_2}{\sigma_2}\right) - \Phi\left( \frac{-\psi_1-\mu_2}{\sigma_2}  \right) \right) \\
 &=&  \Phi\left( \frac{\alpha-\mu_2}{\sigma_2}\right)  + \mathbb{E}\left( \Phi\left(\frac{\min(\alpha-c, -\psi_1)-\mu_2}{\sigma_2}\right) - \Phi\left( \frac{-\psi_1-\mu_2}{\sigma_2}  \right) \mid \alpha-c < -\psi_1 \right) \\
&=&  \Phi\left( \frac{\alpha-\mu_2}{\sigma_2}\right)  + \mathbb{E}\left( \Phi\left(\frac{\alpha-c-\mu_2}{\sigma_2}\right) - \Phi\left( \frac{-\psi_1-\mu_2}{\sigma_2}  \right) \mid \alpha-c < -\psi_1 \right) \\
 &=&  \Phi\left( \frac{\alpha-\mu_2}{\sigma_2}\right) + \Phi\left(\frac{\alpha-c-\mu_2}{\sigma_2}\right) \Pr\left(\psi_1 < c-\alpha\right)  - \mathbb{E}\left(\Phi\left( \frac{-\psi_1-\mu_2}{\sigma_2}  \right) \mid \alpha-c < -\psi_1 \right) \\
 &=&  \Phi\left( \frac{\alpha-\mu_2}{\sigma_2}\right) + \Phi\left(\frac{\alpha-c-\mu_2}{\sigma_2}\right) \Phi\left(\frac{ c-\alpha - \mu_1}{\sigma_1}\right)  - \mathbb{E}\left(\Phi\left( \frac{-\psi_1-\mu_2}{\sigma_2}  \right) \mid \alpha-c < -\psi_1 \right) 
\end{eqnarray}
$$
