Standard Normal Expectation

Question

Let $Φ$ be the CDF of $Z\sim N(0,1)$. Find $E[ZΦ(Z)]$ and $E[Z^2\cdot Φ(Z)]$

Expectation for a continuous RV is the integral from $-\infty$ to $\infty$ of xf(x)dx. How do I apply that here?

Note: $\Phi$ is the CDF, not PDF.

Answer 1

NOTE: $\phi(x)$ represents Gaussian cdf , and $\psi(x)$ represents Gaussian pdf

$E[Z\phi(Z)]$

$=\int_{-\infty}^{\infty}\int_{-\infty}^{z}{z\cdot\frac{1}{\sqrt{2\pi}}\exp(-\frac{x^2}{2})\cdot\frac{1}{\sqrt{2\pi}}\exp(-\frac{z^2}{2})dxdz}$

(Change of order of integration)

$=\int_{-\infty}^{\infty}\int_{-\infty}^{x}{z\cdot\frac{1}{\sqrt{2\pi}}\exp(-\frac{x^2}{2})\cdot\frac{1}{\sqrt{2\pi}}\exp(-\frac{z^2}{2})dzdx}$

$=\int_{-\infty}^{\infty}{\frac{1}{{2\pi}}\exp(-\frac{x^2}{2})\cdot(\int_{-\infty}^{x}z{\cdot}\exp(-\frac{z^2}{2})dz)dx}$

$=\int_{-\infty}^{\infty}{\frac{1}{{2\pi}}\exp(-\frac{x^2}{2})\cdot(-\exp(-\frac{x^2}{2}))dx}$

$=\int_{-\infty}^{\infty}{\frac{1}{{2\pi}}\exp(-x^2)dx}$

$=\frac{1}{2\sqrt{\pi}}$

$E[Z^2\phi(Z)]$

$=\int_{-\infty}^{\infty}\int_{-\infty}^{z}{z^2\cdot\frac{1}{\sqrt{2\pi}}\exp(-\frac{x^2}{2})\cdot\frac{1}{\sqrt{2\pi}}\exp(-\frac{z^2}{2})dxdz}$ $=\int_{-\infty}^{\infty}\int_{-\infty}^{x}{z^2\cdot\frac{1}{\sqrt{2\pi}}\exp(-\frac{x^2}{2})\cdot\frac{1}{\sqrt{2\pi}}\exp(-\frac{z^2}{2})dzdx}$

(Change of order of integration)

$=\int_{-\infty}^{\infty}{\frac{1}{{2\pi}}\exp(-\frac{x^2}{2})\cdot( \int_{-\infty}^{x}z^2{\cdot}\exp(-\frac{z^2}{2} )dz)dx}$

(Integrate by parts)

$=\int_{-\infty}^{\infty}{\frac{1}{{2\pi}}\exp(-\frac{x^2}{2})\cdot( z\cdot\int z{\cdot}\exp(-\frac{z^2}{2})dz-\int \frac{d}{dz}z{\cdot}(\int z\cdot\exp(-\frac{z^2}{2})dz)dz )dx}$

$=\int_{-\infty}^{\infty}{\frac{1}{{2\pi}}\exp(-\frac{x^2}{2})\cdot( x{\cdot}\exp(-\frac{x^2}{2})-\int_{-\infty}^{x} -\exp(-\frac{z^2}{2})dz) )dx}$

($\phi(x)=\int_{-\infty}^{x}\frac{1}{\sqrt{2\pi}}\exp(-\frac{z^2}{2})dz$)

$=\int_{-\infty}^{\infty}{\frac{1}{{2\pi}}\exp(-\frac{x^2}{2})\cdot( x{\cdot}\exp(-\frac{x^2}{2})+\sqrt{2\pi}\phi(x)) )dx}$

$=\int_{-\infty}^{\infty}{\frac{1}{{2\pi}} x{\cdot}\exp(-x^2)dx+\sqrt{2\pi}\int_{-\infty}^{\infty}\phi(x)\cdot\frac{1}{{2\pi}}\exp(-\frac{x^2}{2}) dx}$

(Odd function integrates to zero)

$=0+\int_{-\infty}^{\infty}\phi(x)\cdot\frac{1}{\sqrt{2\pi}}\exp(-\frac{x^2}{2}) dx$

Let $I=\int\phi(x)\cdot\frac{1}{\sqrt{2\pi}}\exp(-\frac{x^2}{2}) dx$

(Integrate by parts)

$I=\phi(x)\cdot\int\frac{1}{\sqrt{2\pi}}\exp(-\frac{x^2}{2}) dx-\int (\frac{d}{dx}\phi(x))(\int\frac{1}{\sqrt{2\pi}}\exp(-\frac{x^2}{2}) dx)dx$

($\phi(x)=\int_{-\infty}^{x}\frac{1}{\sqrt{2\pi}}\exp(-\frac{z^2}{2})dz$)

$=\phi(x)\cdot\phi(x)-\int (\frac{d}{dx}\phi(x))(\int\frac{1}{\sqrt{2\pi}}\exp(-\frac{x^2}{2}) dx)dx$

($\frac{d}{dx}\phi(x)=\frac{1}{\sqrt{2\pi}}\exp(-\frac{z^2}{2})dz$)

$=\phi^2(x)-\int\frac{1}{\sqrt{2\pi}}\exp(-\frac{x^2}{2}) \cdot\phi(x)dx$

$=\phi^2(x)-I$

So, $I=\frac{\phi^2(x)}{2}$

So, $E[Z^2\phi(Z)]=\frac{\phi^2(\infty)-\phi^2(-\infty)}{2}=\frac{1-0}{2}=\frac{1}{2}$