Sorry if this is badly worded, not sure of the best way to express this.
Suppose we have $X \sim N(0,1)$ and $Y \sim N(0,\sigma)$ and the CDF of $X: F(z) = \Pr(X < z)$. If we take the expectation of $F(X)$ we get 0.5, as discussed here:
Expectation of CDF of continuous random variable $X$, evaluated at $X$
But what if we take the expectation of $F(Y)$? In this case the PDF in question is no longer the derivative of the CDF, so the integral seems less obvious. Any help would be appreciated!
This actually works for any distributions that are symmetric about $0$, i.e. $X$ and $-X$ have the same distribution and $Y$ and $-Y$ have the same distribution, where the distribution of $X$ is continuous.
Then the CDF of $X$ satisfies $$F(y) = \mathbb P(X \le y) = \mathbb P(X \ge -y) = 1 - F(-y)$$ so $$ \mathbb E[F(Y)] = 1 - \mathbb E[F(-Y)] = 1 - \mathbb E[F(Y)]$$ and therefore $\mathbb E[F(Y)] = 1/2$.
To be clear, we’d better to distinguish two notations, $\Pr \left[ X<Y\left| Y \right. \right]$ and $\Pr \left[ X<Y \right]$. The former means the conditional probability of the event $X<Y$ given the value of $Y$. The latter means the probability of the same event where $Y$ is treated as a random variable. It is essential to understand the relation between the probability and expectation of an event $A$, that is,$\Pr \left[ A \right]=\mathbb{E}\left[ {{I}_{\left\{ A \right\}}} \right]$ where ${{I}_{\left\{ A \right\}}}$ is a indication function for an event A.
Saying that, if you mean $\mathbb{E}\left[ \Pr \left[ X<Y \right] \right]$ as the expectation of the latter notation $\mathbb{E}\left[ \Pr \left[ X<Y \right] \right]=\Pr \left[ X<Y \right]$ since $\Pr \left[ X<Y \right]$ is a non-random variable and the expectation of a non-random variable is itself. If you mean $\mathbb{E}\left[ \Pr \left[ X<Y \right] \right]$ as the expectation of the former notation, that is, $\underset{Y}{\mathop{\mathbb{E}}}\,\left[ \Pr \left[ X<Y\left| Y \right. \right] \right]$, we can calculate as follows.
$$\underset{Y}{\mathop{\mathbb{E}}}\,\left[ \Pr \left[ X<Y\left| Y \right. \right] \right]=\underset{Y}{\mathop{\mathbb{E}}}\,\left[ \underset{X}{\mathop{\mathbb{E}}}\,\left[ {{I}_{\left\{ X<Y \right\}}}\left| Y \right. \right] \right]=\mathbb{E}\left[ {{I}_{\left\{ X<Y \right\}}} \right]=\Pr \left[ X<Y \right].$$
After all, $\mathbb{E}\left[ \Pr \left[ X<Y \right] \right]=\Pr \left[ X<Y \right]$ in the both cases.
Since $X\sim N\left( {{\mu }_{X}},\sigma _{X}^{2} \right)$ and $Y\sim N\left( {{\mu }_{Y}},\sigma _{Y}^{2} \right)$,$$X-Y\sim N\left( {{\mu }_{X}}-{{\mu }_{Y}},\sigma _{Z}^{2} \right),where\,\sigma _{Z}^{2}:=\sigma _{X}^{2}+\sigma _{Y}^{2}-\operatorname{cov}\left( X,Y \right).$$
We get $$\Pr \left[ X<Y \right]=\Pr \left[ X-Y<0 \right]=\Pr \left[ {\left( X-Y-\left( {{\mu }_{X}}-{{\mu }_{Y}} \right) \right)}/{{{\sigma }_{Z}}}\;<{-\left( {{\mu }_{X}}-{{\mu }_{Y}} \right)}/{{{\sigma }_{Z}}}\; \right]=\Phi \left( {-\left( {{\mu }_{X}}-{{\mu }_{Y}} \right)}/{{{\sigma }_{Z}}}\; \right)$$ where $\Phi \left( {} \right)$ is the C.D.F of the standard normal distribution.
Finally we get $\mathbb{E}\left[ \Pr \left[ X<Y \right] \right]=\Phi \left( {-\left( {{\mu }_{X}}-{{\mu }_{Y}} \right)}/{{{\sigma }_{Z}}}\; \right)$. When ${{\mu }_{X}}={{\mu }_{Y}}=0$ as your case, $\mathbb{E}\left[ \Pr \left[ X<Y \right] \right]=\Phi \left( 0 \right)={1}/{2}\;.$
Furthermore, even if neither $X$ nor $Y$ are normal distributions, $\mathbb{E}\left[ \Pr \left[ X<Y \right] \right]={1}/{2}\;$ holds only if $X-Y$ is a symmetrical distribution with 0 mean.
Use this expression for expectation, $\mathbb{E}(X)= \int_{\mathbb{R}} xdF(x)$ $$\mathbb{E}(F(Y))=\int_{\mathbb{R}} F(y) dF(y) = \int_0^1 x dx = \frac{1}{2}$$
