Question

Let Z be a standard normal random variable,and, for a fixed x, set

$X=\{\begin{array}{ll}Z& \text{if}Z>x\\ 0& \text{otherwise}\end{array}$

$\text{Show that}E\left[X\right]=\frac{1}{\sqrt{2\pi }}{e}^{-x^2/2}\text{.}$

Short answer

$E\left(X\right)={\int }_{z>x}z\cdot d{P}_Z={\int }_{z>x}z{f}_Z\left(z\right)dz={\int }_x^{\mathrm{\infty }}z\frac{1}{\sqrt{2\pi }}{e}^{-\frac{z^2}{2}}dz$

Let's solve this integral using the substitution $u=Z^2/2$which implies$du=zdz.$

$\frac{1}{\sqrt{2\pi }}{\int }_{\frac{x^2}{2}}^{\mathrm{\infty }}{e}^{-u}du=\frac{1}{\sqrt{2\pi }}{e}^{-\frac{x^2}{2}}$

Step-by-step solution

Given Information

Given in the question that

Let Z be a standard normal random variable

$X=\{\begin{array}{ll}Z& \text{if}Z>x\\ 0& \text{otherwise}\end{array}$

$E\left[X\right]=\frac{1}{\sqrt{2\pi }}{e}^{-x^2/2}\text{.}$

Explanation

Observe that $X$can be written as $X=Z·I_{(Z>x)}$. So, the mean of$X$is

$E\left(X\right)={\int }_{z>x}z\cdot d{P}_Z={\int }_{z>x}z{f}_Z\left(z\right)dz={\int }_x^{\mathrm{\infty }}z\frac{1}{\sqrt{2\pi }}{e}^{-\frac{z^2}{2}}dz$

Let's solve this integral using the substitution $u=z^2/2$which implies $du=zdz$. So, the integral above is equal to

$\frac{1}{\sqrt{2\pi }}{\int }_{\frac{u^2}{2}}^{\mathrm{\infty }}{e}^{-u}du=\frac{1}{\sqrt{2\pi }}{e}^{-\frac{x^2}{2}}$

So, we have proved that

$E\left[X\right]=\frac{1}{\sqrt{2\pi }}{e}^{-x^2/2}\text{.}$

Final Answer

$E\left(X\right)={\int }_{z>x}z\cdot d{P}_Z={\int }_{z>x}z{f}_Z\left(z\right)dz={\int }_x^{\mathrm{\infty }}z\frac{1}{\sqrt{2\pi }}{e}^{-\frac{z^2}{2}}dz$

Let's solve this integral using the substitution $u=Z^2/2$which implies $du=zdz$

$\frac{1}{\sqrt{2\pi }}{\int }_{\frac{x^2}{2}}^{\mathrm{\infty }}{e}^{-u}du=\frac{1}{\sqrt{2\pi }}{e}^{-\frac{x^2}{2}}$