TITLE: Find the expected value of $F(X+a)$ where $X\sim\mathcal{N}(0,1)$ with CDF $F$
QUESTION [5 upvotes]: Let $X\sim\mathcal{N}(0,1)$ and denote the standard normal CDF by $F$. It is easy to see that $F(X)\sim\mathcal{U}(0,1)$. Namely, for $u\in(0,1)$ we have
\begin{align*}
P(F(X)\leq u)=P(X\leq F^{-1}(u)) = F(F^{-1}(u))= u, 
\end{align*}
so the expected value is 
\begin{align*}
E(F(X))=0.5.
\end{align*} 
Now consider the random variable $F(X+a)$ where $a>0$ is a constant. How can we find the expected value 
\begin{align*}
E(F(X+a))=\;?
\end{align*} 
I suspect the answer to be $F(a)$. Additionally, can we find a simple form for the density or distribution function of the random variable $F(X+a)$?

REPLY [3 votes]: Your conjectured value for $E[F(X+a)]$ is close but not quite correct. 
$$
E[F(X+a)] = \int_{-\infty}^\infty F(x+a)f(x)\, \mathrm dx
= \int_{-\infty}^\infty \left[\int_{-\infty}^{x+a}f(y)\, \mathrm dy\right]f(x)\, \mathrm dx
$$
Now suppose that $Y\sim N(0,1)$ is independent of $X$. Then, the integral on the right can be interpreted as calculating 
$P(Y \leq X+a) = P(Y-X \leq a)$. But, $Y-X \sim N(0,2)$ and so $$E[F(X+a)] = F(a/\sqrt 2)$$
and not $F(a)$ as you have conjectured.  See also this answer of mine on stats.SE for a more 
general version of the argument used here.