Question
Let $X\sim \mathcal{N}(\mu,\sigma^2)$, $\Phi$ denotes the standard Gaussian CDF. What is $\mathbb{E}[\Phi(X)]$?
Solution
By definition $\Phi(X) = P[Z \le X]$ where $Z$ is a standard Gaussian variable. Then
$$ \Phi(X) = P[Z \le X] = P[Z - X \le 0] $$
where $Z - X \sim \mathcal{N}(-\mu, 1+\sigma^2)$. Therefore
$$ \begin{aligned} P[Z - X \le 0] &= P\!\left( \frac{Z - X + \mu}{\sqrt{1+\sigma^2}} \le \frac{\mu}{\sqrt{1+\sigma^2}} \right) \\ &= P\!\left( \tilde{Z} \le \frac{\mu}{\sqrt{1+\sigma^2}} \right) \\ &= \Phi\!\left( \frac{\mu}{\sqrt{1+\sigma^2}} \right). \end{aligned} $$
A special case is when $X$ has zero mean, in which case $\mathbb{E}[\Phi(X)] = 1/2$.