Expected Value of a function of the standard normal distribution

Question

Any help is appreciated. Thanks

Answer 1

You've got a good start. Now, notice that $$ (z-a)^{+}=\begin{cases}z-a & \text{if }z\geq a\\0 & \text{else}\end{cases}, $$ so that $$ \mathbb{E}[(Z-a)^+]=\int_{-\infty}^{\infty}(z-a)^{+}\frac{1}{\sqrt{2\pi}}e^{-\frac{z^2}{2}}\,dz=\int_{-\infty}^{a}0\,dz+\int_a^{\infty}(z-a)\frac{1}{\sqrt{2\pi}}e^{-\frac{z^2}{2}}\,dz. $$ The first integral just evaluates to $0$, and disappears; what should you do with the rest?

Try breaking it up over subtraction, and see if you can figure out where each part of the given formula is coming from.