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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.

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    $\begingroup$ It is true that $$\mathbf{E}[\Phi(X)]=\mathbf{P}(Z\leq X)$$ where $Z$ is a standard normal random variable independent of $X$. This is because we can write $\Phi(X)=\mathbf{P}(Z\leq X\mid X)$. $\endgroup$ Commented May 4, 2022 at 18:09
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    $\begingroup$ $$\underset{X}{\mathop{\mathbb{E}}}\,\left[ \Pr \left[ Z<X\left| X \right. \right] \right]=\underset{X}{\mathop{\mathbb{E}}}\,\left[ \underset{Z}{\mathop{\mathbb{E}}}\,\left[ {{I}_{\left\{ Z<X \right\}}}\left| Z \right. \right] \right]=\mathbb{E}\left[ {{I}_{\left\{ Z<X \right\}}} \right]=\Pr \left[ Z<X \right].$$ Taken from math.stackexchange.com/a/4140516/415432 and math.stackexchange.com/a/1125935/415432 $\endgroup$ Commented May 4, 2022 at 19:19

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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 $$ \mathbb E[Y]=\int_0^1\Phi\Big(\sigma\,\Phi^{-1}(u)+\mu\Big)\,du\,. $$

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