Integral with analytical solution with normal distribution I received very good answers a couple of days ago in a simpler related problem, see Integral with Normal Distributions, but I am struggling with this new question:
Let's define a function $F(\theta)=\int\limits_{-\infty}^{+\infty} \Phi  \left(\theta -a- b x\right) \phi\left(cx- \theta \right) \mathrm d x$
$\theta$ is a relevant parameter and $a$, $b$ and $c$ are all given scalars. $\Phi$ and $\phi$ respectively denote the CDF and PDF of the standard normal distribution
I have two questions:


*

*Is there a way to express analytically this integral as a function of $\theta$, $a$, $b$ and $c$? (The case with $a=0$, $b=c=1$ is the one solved in the previous question)

*I want to find conditions on $a$, $b$ and $c$ that make $F'(\theta)<1$.
 A: For any $a,b,\theta \in \mathbb{R}$ and $c > 0$, it holds
$$
F(\theta ) := \int_{ - \infty }^\infty  {\Phi (\theta  - a - bx)\phi (cx - \theta )\,{\rm d}x} = \frac{1}{c}\Phi \bigg(\frac{{(c - b)\theta  - ac}}{{\sqrt {c^2  + b^2 } }}\bigg).
$$
(I confirmed the result numerically.) 
Proof. First note that
$$
\frac{1}{c}\Phi \bigg(\frac{{(c - b)\theta  - ac}}{{\sqrt {c^2  + b^2 } }}\bigg) = \frac{1}{c}{\rm P}\bigg[Z \le \frac{{(c - b)\theta  - ac}}{{\sqrt {c^2  + b^2 } }}\bigg] = \frac{1}{c}{\rm P}\big[cX + bY \le (c - b)\theta  - ac\big],
$$
where $X$, $Y$, and $Z$ are independent ${\rm N}(0,1)$ random variables (note that $\sqrt {c^2  + b^2 } Z$ and $cX+bY$ are identically distributed). By the law of total probability, conditioning on $Y$, we thus get
$$
\frac{1}{c}\Phi \bigg(\frac{{(c - b)\theta  - ac}}{{\sqrt {c^2  + b^2 } }}\bigg) = \frac{1}{c}\int_{ - \infty }^\infty  {{\rm P}\big[cX + by \le (c - b)\theta  - ac\big]\phi (y)\,{\rm d}y}. 
$$
A change of variable $y=cx-\theta$ then gives
$$
\frac{1}{c}\Phi \bigg(\frac{{(c - b)\theta  - ac}}{{\sqrt {c^2  + b^2 } }}\bigg) = \frac{1}{c}\int_{ - \infty }^\infty  {{\rm P}\big[cX + b(cx - \theta ) \le (c - b)\theta  - ac\big]\phi (cx - \theta )c\,{\rm d}x} . 
$$
A little algebra shows that the expression on the right is equal to
$$
\int_{ - \infty }^\infty  {{\rm P}\big[X \le  \theta  - a -bx \big]\phi (cx - \theta )\,{\rm d}x},
$$
and hence 
$$
\frac{1}{c}\Phi \bigg(\frac{{(c - b)\theta  - ac}}{{\sqrt {c^2  + b^2 } }}\bigg) = \int_{ - \infty }^\infty  {\Phi (\theta  - a - bx)\phi (cx - \theta )\,{\rm d}x}. 
$$
