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 $\Phi( \frac{a-bX}{c} )$ 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.
Best Answer
Assume that the $\Phi$ function is for some other independent standard normal random variable Y, and simply rewrite the problem like so:
$$E\left(\Phi\left( \frac{a-bX}{c}\right)\right)=P(Y<\frac{a-bX}{c})=P\left( Y+\frac{bX}{c}<\frac{a}{c}\right)$$
Now, since $X$ ~ $N(0,1), \frac{bX}{c}$ ~ $N(0,\frac{b^2}{c^2})$, so $Y+\frac{bX}{c}$ ~ $N(0,1+\frac{b^2}{c^2})$
So,
$$ P\left( Y+\frac{bX}{c}<\frac{a}{c}\right) = P\left( \frac{Y+\frac{bX}{c}}{\sqrt{1+\frac{b^2}{c^2}}}<\frac{a}{c\sqrt{1+\frac{b^2}{c^2}}}\right)=\Phi\left( \frac{a}{c\sqrt{1+\frac{b^2}{c^2}}}\right) $$
Therefor: $E\left(\Phi\left( \frac{a-bX}{c}\right)\right)=\Phi\left( \frac{a}{c\sqrt{1+\frac{b^2}{c^2}}}\right)$