Let, $X\sim N(\mu, \sigma^2)$ and let $\Phi(\cdot):\mathbb{R}\to[0,1]$ be the CDF of a standard normal distribution.
Then, what is the pdf of $Y=\Phi(X)$. Also, find $E[\Phi(X)]$.
Note:-
Here, $Y=\Phi(X)\ne P(Z\le X)= \text{ some constant }$. Rather, in this case, $Y=\Phi(X)$ is a non-degenerate random variable.
And, so $E[\Phi(X)]\ne E[P(Z\le X)]=P(Z\le X)=\Phi\left(\dfrac{-\mu}{\sqrt{1+\sigma^2}}\right)$
My attempt:-
$$E[Y]=E[\Phi(X)]=\int^{\infty}_{-\infty}\Phi(x)f_X(x)dx=\int^{\infty}_{-\infty}\left( \int^{x}_{-\infty} \dfrac{1}{\sqrt{2\pi}}\exp\left( \frac{-1}{2}t^2 \right)dt \right) \dfrac{1}{\sigma\sqrt{2\pi}}\exp\left( \frac{-1}{2}{\bigg[\dfrac{x-u}{\sigma}}\bigg]^2 \right)dx \\\ =\dfrac{1}{2\pi\sigma}\int^{\infty}_{-\infty}\int^{x}_{-\infty}\exp\left[\dfrac{-1}{2}\Bigg(t^2+\Big(\frac{x-\mu}{\sigma}\Big)^2\Bigg)\right]dt dx $$
This is where i am stuck. Any help would be appreciated.
The CDF of $X$ is $$ \mathbb P(X\le x)=\mathbb P\Bigg(\frac{X-\mu}{\sigma}\le \frac{x-\mu}{\sigma}\Bigg)=\Phi\Big(\frac{x-\mu}{\sigma}\Big). $$ Therefore $$ U:=\Phi\Big(\frac{X-\mu}{\sigma}\Big) $$ is uniform on $[0,1]$. From $Y=\Phi(X)$ we get $$ Y=\Phi\Big(\sigma\,\Phi^{-1}(U)+\mu\Big)\,,\quad U=\Phi\Bigg(\frac{\Phi^{-1}(Y)-\mu}{\sigma}\Bigg)\,. $$ The CDF of $Y$ is $$ \mathbb P(Y\le y)=\mathbb P\Bigg(U\le\Phi\Bigg(\frac{\Phi^{-1}(y)-\mu}{\sigma}\Bigg)\Bigg)=\Phi\Bigg(\frac{\Phi^{-1}(y)-\mu}{\sigma}\Bigg)\,. $$ The expectation of $Y=\Phi(X)$ is $$ \mathbb E[Y]=\int_0^1\Phi\Big(\sigma\,\Phi^{-1}(u)+\mu\Big)\,du\,. $$