Calculate $\mathbb{E}[N(X)]$, where $N(·)$ is the cdf of the standard normal distribution, and $X$ is a standard normal random variable.

Question

I'm stuck with this problem: Calculate $\mathbb{E}[N(X)]$, where N(·) is the cdf of the standard normal distribution, and X is a standard normal random variable.

Here's where I'm stuck: We know that:

$$\mathbb{E}[g(X)]=\int_{-\infty }^{\infty}g(x)f_{X}(x)dx$$

And we are given that:

$$ g(x) = N(x) = \frac{1}{\sqrt{2 \pi}} \int_{-\infty}^{x}e^{-\frac{u^2}{2}}du $$

$$ f_{X}(x)=\frac{1}{\sqrt{2 \pi}} e^{-\frac{x^2}{2}} $$

So, the solution should be to develop this:

$$\begin{split} \mathbb{E}[N(X)] &= \int_{-\infty }^{\infty} \left( \frac{1}{\sqrt{2 \pi}} \int_{-\infty}^{x}e^{-\frac{u^2}{2}}du \right) \left( \frac{1}{\sqrt{2 \pi}} e^{-\frac{x^2}{2}} \right) dx\\ &= \frac{1}{2 \pi} \int_{-\infty }^{\infty} \left( \int_{-\infty}^{x}e^{-\frac{u^2}{2}}du \right) e^{-\frac{x^2}{2}} dx\\ &= \frac{1}{2 \pi} \int_{-\infty }^{\infty} \int_{-\infty}^{x} e^{-\frac{u^2+x^2}{2}}dudx \end{split}$$

My questions are:

1. Can we combine the exponents moving the term with x inside, the last step above?
2. If yes, this looks like a good opportunity to change variables to polar. If that's the case, how to change the limits of the integrals from $x$ and $u$ to $r$ and $\theta$?

Any help would be extremely appreciated! Thanks!

Answer 1

@Golden_Ratio's answer is the most general as it holds for an arbitrary random variable. However, the problem at hand can be easily determined by denoting $\Phi$ and $\phi$ as the standard normal cdf/pdf and then writing $$ \mathsf E\Phi(X)=\int_{-\infty}^\infty\Phi(x)\phi(x)\,\mathrm dx=\int_{-\infty}^\infty\Phi(x)\,\mathrm d\Phi(x)=\frac{\Phi^2(x)}{2}\bigg|_{-\infty}^\infty=\frac{1}{2}. $$

Answer 2

Let $X$ be a random variable having invertible CDF $F$. Then the random variable $F(X)$ is uniformly distributed on $(0,1)$ since

$$ P(F(X)\leq y)=P(X\leq F^{-1}(y)) =F(F^{-1}(y))=y.$$

So it immediately follows that $E[F(X)]=1/2.$

Answer 3

$A(x)=\int\limits_{-\infty}^xe^{-\frac{u^2}{2}}du$ and $dA=e^{-\frac{x^2}{2}}dx$ so integral becomes $\frac{1}{2\pi}\int\limits_0^{\sqrt{2\pi}}AdA=\frac{1}{2}$.