$E[\Pr(X<Y)]$ where $X$ and $Y$ are normally distributed? 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 $X: F(z) = \Pr(X &lt; z)$. 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 $\mathbb E[F(Y)] = 1/2$.
Use this expression for  expectation, $\mathbb{E}(X)= \int_{\mathbb{R}} xdF(x)$   $$\mathbb{E}(F(Y))=\int_{\mathbb{R}} F(y) dF(y) = \int_0^1 x dx = \frac{1}{2}$$