Expected value via Darth Vader Rule on fuction

Question

$$E(Max(0,X - d )) = \int_{d}^{\infty}1-F(X)dx $$ How can i prove this? $$E(Max(0,X - d )) =E(g(X))= \int_{0}^{\infty}(P(g(X)>x) dx = \int_{0}^{d}(P(0>x) dx +\int_{d}^{\infty}(P(X - d > x) dx $$ But it is clearly wrong and i dont know why.

Answer 1

$Y:=\max(0,X-d)$ is a positive random variable and for every positive random variable we have:$$\mathbb EY=\int_0^{\infty}P(Y>y)dy$$

Note that $\{Y>y\}=\{X>d+y\}$ for every $y>0$ so that we can substitute: $$P(Y>y)=P(X>d+y)=1-F_X(d+y)$$leading to:$$\mathbb EY=\int_0^{\infty}1-F_X(d+y)dy=\int_d^{\infty}1-F_X(x)dx$$