Let $Z\sim N(0,1)$ be a random variable, then $E[\max\{Z-1,0\}]$ is ?
I know by letting $\varphi$ be the standard normal density function and $X=\max\{Z-1,0\}$, we can simplify the expected value as $$ \operatorname{E}(X) = 0 \cdot \Pr(X=0) + \int_?^? (z-1)\varphi(z)\,dz. $$ I am not just sure about the integral part that have written and its limits. Should it be the following?
$$ \operatorname{E}(X) = 0 \cdot \Pr(X=0) + \int_1^\infty (z-1)\varphi(z)\,dz. $$
Then how do we solve that?
Set $X=\max\{Z-1,0\}$ is positive a.s. and for $x>0$ we have
$$\mathbb P\{X>x\}=\mathbb P\{Z>x+1\}=\int_{x+1}^\infty e^{-\frac{x^2}{2}}\,\mathrm d x,$$
Then, $$\mathbb E[X]=0\mathbb P\{X=0\}+\int_0^\infty \mathbb P\{X> x\}\,\mathrm d x=...$$