Let $Z\sim N(0,1)$ be a random variable, then $E[\max\{Z,0\}]$ is?

Question

Let $Z\sim N(0,1)$ be a random variable, then $E[\max\{Z,0\}]$ is ?

1. $\frac{1}{\sqrt{\pi}}$
2. $\sqrt{\frac{2}{\pi}}$
3. $\frac{1}{\sqrt{2\pi}}$
4. $\frac{1}{\pi}$

I know that $E[\max\{Z,Y\}]=\iint \max(z,y)f(z,y) \, dz \, dy$. How to use this in this case, since one variable is zero. Any hint would be helpful. Thanks.

Answer 1

I think for $\max \{Z,0\}$ its easy if $Z$ will be $Z \ge 0$ hence we search for $\operatorname{E}[Z]$ that is: $$\int x \left(\frac{1}{\sqrt{2\pi}}e^{-\frac{x^2}{2}}\right)dx =\frac{1}{\sqrt{2\pi}}$$

since it is symmetric it will also hold on the other side of $Z$.

Answer 2

Let $\varphi$ be the standard normal density function. The expected value of $X=\max\{Z,0\}$ is $$ \operatorname{E}(X) = 0 \cdot \Pr(X=0) + \int_0^\infty x \varphi(x)\,dx. $$ This integral is readily evaluated via the substitution $u = x^2/2$ and $du= x\,dx.$

In measure-theoretic language, we can define a measure $\nu$ by $$ \nu(A)=\Pr(\max\{Z,0\} \in A) = \int_{A\,\setminus\,(-\infty,0]} \varphi(x)\,dx + \begin{cases} 1/2 & \text{if } 0\in A, \\ 0 & \text{otherwise.} \end{cases} $$ Then the expected value of $X=\max\{Z,0\}$ is $$ \operatorname{E}(X) = \int_{[0,\infty)} x \, \nu(dx) = \cdots\, (\text{as above}). $$