TITLE: Calculate $\mathbb{E}[F(Y)]$
QUESTION [5 upvotes]: I try to resolve this problem, but I have some difficulties to get a clear result.
The problem :
Let X be a normal random variable with mean 0 and variance 1 (ie. $X\sim \mathcal{N}(0,1)$).
Let Y be a normal random variable with mean $m$ and variance $\sigma^{2}$ (ie. $Y\sim \mathcal{N}(m,\sigma^{2})$).
X and Y are independent random variables.
What I want is to compute $I=\mathbb{E}[\Phi(Y)]$ where $\Phi$ is the the cumulative distribution function (CDF) of $X$.
*What I done is wrong *
Sorry for my english :)

REPLY [6 votes]: As leonbloy says, you just need to use
$$E[\Phi(Y)] = \int_{-\infty}^\infty \Phi(y)\frac{1}{\sigma \sqrt{2\pi}}\exp\left(-\frac{(y-\mu)^2}{2\sigma^2}\right)\,\mathrm dy.$$
But, rather than trying to evaluate the integral directly, consider that we can
write it as 
$$\begin{align*}
E[\Phi(Y)] &= \int_{-\infty}^\infty \Phi(y)\frac{1}{\sigma \sqrt{2\pi}}\exp\left(-\frac{(y-\mu)^2}{2\sigma^2}\right)\,\mathrm dy\\
&= \int_{-\infty}^\infty P\{X \leq y\}f_Y(y)\,\mathrm dy\\
&= \int_{-\infty}^\infty P\{X \leq y \mid Y = y\}f_Y(y)\,\mathrm dy\\
&= P\{X \leq Y\}\\
&= P\{X-Y\leq 0\}.
\end{align*}$$
What kind of random variable is $X-Y$? Can you find its mean and variance
without doing any integrations? Can you write an expression for this probability
in terms of $\Phi(\cdot)$?