I am trying to calculate the expected value of a Normal CDF, but I have gotten stuck. I want to find the expected value of $\Phi\left( \frac{a-bX}{c} \right)$ where $X$ is distributed as $\mathcal{N}(0,1)$ and $\Phi$ is the standard normal CDF.
I know I can transform $\frac{a-bX}{c}$ to be a normal random variable $\mathcal{N}\left(\frac{a}{c},\frac{b^2}{c^2}\right)$ where $\frac{b^2}{c^2}$ is the variance of the normal random variable. I'm not sure if this helps though.
I think that the expected value of a CDF is $0.5$ but since $\Phi$ is the CDF of a standard normal CDF and $\frac{a-bX}{c}$ is not standard normal I do not think the expected value should be $0.5$. I tried integrating the CDF, but I do not believe I did it correctly.
When $a = -2.3338$, $b = 0.32164$, $c = 0.94686$, I believe the correct answer is approximately $0.009803$. I found this through simulation.
I would appreciate any help or suggestions.
Let $c$ in your question be absorbed into $a$ and $b$, so that you are concerned with $\Phi(a-bX)$. Let $X$ and $Y$ be independent standard normals, and consider their joint pdf $\phi(x,y) = \phi(x)\,\phi(y)$ as a surface covering the $X,\!Y$ plane, where $\phi(\cdot)$ is the standard normal pdf.
For any given $x$, $\Phi(a-bx)\,\phi(x)\,\text{d}x$ is the volume under the joint pdf in the thin slice at $x$ with $Y < a-bx$. Then the conditional expected value of $\Phi(a-bX)$, given that $X$ is in some interval, is just the volume under the pdf in the $X$-interval and below the line $Y = a - bX$, divided by all the volume under the pdf in the $X$-interval.
I know of no closed-form expression for the numerator in the general case where the X-interval is of the form $(t,\infty)$.