# Calculate $\mathbb{E}[F(Y)]$

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 :)

-
I might be missing something, but don't quite get why are you considering $X$ as a random variable. $\phi(\cdot)$ is just a function like any other. If you want to compute $E(g(Y))$,that's just $\int g(y) f_Y(y) dy$ – leonbloy Jan 11 '13 at 15:01

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 $\Phi(\cdot)$?
Thanks a lot ! $X-Y\sim \mathbb{N}(-m,1+\sigma^{2})$ because $X$ and $Y$ are independent. Then, $\mathbb{P}(X-Y\le 0) = \Phi\left(\frac{m}{\sqrt{1+\sigma^{2}}}\right)$. Your result is correct because I check it for simples values : If $m=0$ and $\sigma=1$, then, $\Phi(Y)\sim \mathcal{U}(0,1)$, $\mathbb{E}[\Phi(Y)]=\Phi(0)$. – Gauss Jan 11 '13 at 16:23