Question

For any real number $\mathrm{y}$, define ${\mathrm{y}}^{+}$by

${\mathrm{y}}^{+}=\begin{array}{ll}\mathrm{y},& \text{if}\mathrm{y}\ge 0\\ 0,& \text{if}\mathrm{y}<0\end{array}$

Let $\mathrm{c}$be a constant.

(a) Show that

$\mathrm{E}\left[(\mathrm{Z}-\mathrm{c}{)}^{+}\right]=\frac{1}{\sqrt{2\mathrm{\pi }}}{\mathrm{e}}^{-{\mathrm{c}}^2/2}-\mathrm{c}(1-\mathrm{\Phi }(\mathrm{c}\left)\right)$

when $\mathrm{Z}$is a standard normal random variable.

(b) Find $\mathrm{E}\left[(\mathrm{X}-\mathrm{c}{)}^{+}\right]$when $\mathrm{X}$is normal with mean $\mathrm{\mu }$and variance ${\mathrm{\sigma }}^2$.

Short answer

a. Using the concept of a random variable's expectation it is proved.

b. The value of $\mathrm{E}\left[(\mathrm{X}-\mathrm{c}{)}^{+}\right]$is$\frac{1}{\mathrm{\sigma }}\times \left[\frac{1}{\sqrt{2\mathrm{\pi }}}{\mathrm{e}}^{-\frac{{\left(\frac{\mathrm{c}-\mathrm{\mu }}{\mathrm{\sigma }}\right)}^2}{2}}-\frac{\mathrm{c}-\mathrm{\mu }}{\mathrm{\sigma }}\left(1-\mathrm{\Phi }\left(\frac{\mathrm{c}-\mathrm{\mu }}{\mathrm{\sigma }}\right)\right)\right]$.

Step-by-step solution

Random variable

Observe that the random variable $(\mathrm{Z}-\mathrm{c})+$is, in fact, a random variable.

$(\mathrm{Z}-\mathrm{c}{)}^{+}(\mathrm{\omega })$$=$$\left\{\begin{array}{l}\mathrm{Z}\left(\mathrm{\omega }\right)-\mathrm{c},\mathrm{Z}\left(\mathrm{\omega }\right)\ge \mathrm{c}\\ 0,\mathrm{Z}\left(\mathrm{\omega }\right)<\mathrm{c}\end{array}\right.$

for$\mathrm{\omega }\in \mathrm{\omega }$, where probability space is$(\mathrm{\omega },\mathrm{F},\mathrm{P})$.

Showing of standard normal random variable (part a)

a.

Intended consequence is,

$\mathrm{E}\left((\mathrm{Z}-\mathrm{c}{)}^{+}\right)={\int }_{\mathrm{\Omega }}(\mathrm{Z}-\mathrm{c}{)}^{+}(\mathrm{\omega }\left)\mathrm{dP}\right(\mathrm{\omega })$

$={\int }_{\text{mtight" style}="\text{margin-right:}0.07153\mathrm{em};">\mathrm{Z}\left(\mathrm{\omega }\right)\ge \mathrm{c}}\left(\mathrm{Z}\right(\mathrm{\omega })-\mathrm{c})\mathrm{dP}\left(\mathrm{\omega }\right)$

Density function,

$\mathrm{\phi }\left(\mathrm{z}\right)=\frac{1}{\sqrt{2\mathrm{\pi }}}{\mathrm{e}}^{-\frac{{\mathrm{z}}^2}{2}}$

So integral is,

${\int }_{\mathrm{Z}\left(\mathrm{\omega }\right)\ge \mathrm{c}}\left(\mathrm{Z}\right(\mathrm{\omega })-\mathrm{c})\mathrm{dP}\left(\mathrm{\omega }\right)={\int }_{\mathrm{z}\ge \mathrm{c}}(\mathrm{z}-\mathrm{c})\mathrm{\phi }\left(\mathrm{z}\right)\mathrm{dz}$

$={\int }_{\mathrm{c}}^{\infty }\mathrm{z\phi }\left(\mathrm{z}\right)\mathrm{dz}-\mathrm{c}{\int }_{\mathrm{c}}^{\infty }\mathrm{\phi }\left(\mathrm{z}\right)\mathrm{dz}$

First integral,

$\mathrm{u}=\frac{{\mathrm{z}}^2}{2}\Rightarrow \mathrm{du}=\mathrm{zdz}$

Substitute values we get,

${\int }_{\mathrm{c}}^{\infty }\mathrm{z\phi }\left(\mathrm{z}\right)\mathrm{dz}=\frac{1}{\sqrt{2\mathrm{\pi }}}{\int }_{\mathrm{c}}^{\infty }{\mathrm{ze}}^{-\frac{{\mathrm{z}}^2}{2}}\mathrm{dz}$

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

$=\frac{1}{\sqrt{2\mathrm{\pi }}}{\mathrm{e}}^{-\frac{{\mathrm{c}}^2}{2}}$

Second integral,

$\mathrm{E}\left((\mathrm{Z}-\mathrm{c}{)}^{+}\right)=\frac{1}{\sqrt{2\mathrm{\pi }}}{\mathrm{e}}^{-\frac{{\mathrm{c}}^2}{2}}-\mathrm{c}(1-\mathrm{\Phi }(\mathrm{c}\left)\right)$

Calculation of E(X-c)+ (part b)

b.

For,

$\mathrm{X}~\mathcal{N}\left(\mathrm{\mu },{\mathrm{\sigma }}^2\right)$.

Hence $\mathrm{Z}=\frac{\mathrm{X}-\mathrm{\mu }}{\mathrm{\sigma }}~\mathcal{N}(0,1)$.

So, $\mathrm{X}=\mathrm{\sigma Z}+\mathrm{\mu }$.

$(\mathrm{X}-\mathrm{c}{)}^{+}=(\mathrm{\sigma Z}+\mathrm{\mu }-\mathrm{c}{)}^{+}$

$=\frac{1}{\mathrm{\sigma }}{\left(\mathrm{Z}-\frac{\mathrm{c}-\mathrm{\mu }}{\mathrm{\sigma }}\right)}^{+}$

since $\mathrm{\sigma }>0$.

Using part (a) we get,

$\mathrm{E}\left((\mathrm{X}-\mathrm{c}{)}^{+}\right)=\frac{1}{\mathrm{\sigma }}\mathrm{E}\left[{\left(\mathrm{Z}-\frac{\mathrm{c}-\mathrm{\mu }}{\mathrm{\sigma }}\right)}^{+}\right]$

$=\frac{1}{\mathrm{\sigma }}\times \left[\frac{1}{\sqrt{2\mathrm{\pi }}}{\mathrm{e}}^{-\frac{{\left(\frac{\mathrm{c}-\mathrm{\mu }}{\mathrm{\sigma }}\right)}^2}{2}}-\frac{\mathrm{c}-\mathrm{\mu }}{\mathrm{\sigma }}\left(1-\mathrm{\Phi }\left(\frac{\mathrm{c}-\mathrm{\mu }}{\mathrm{\sigma }}\right)\right)\right]$