Probability Expectation proof

Question

I dont know what to do, I have to proof this $$ E(X^n)=n\displaystyle\int_{0}^{\infty}(1-F_{X}(x))x^{n-1}dx-n \displaystyle\int_{-\infty}^{0}F_{X}(x)x^{n-1}dx$$ but i dont have any idea. Thanks.

Answer 1

$n\int_0^{\infty} x^{n-1} (1-F_X(x))dx=n\int_0^{\infty} x^{n-1} \int_{(x,\infty)} d\mu_X(y) dx$ where $\mu_X(E)=P(X^{-1}(E))$. By Fubini/Tonelli Theorem this becomes $\int_0^{\infty} \int_0^{y} nx^{n-1} dx d\mu_X(y) =\int_0^{\infty} y^{n}d\mu_X(y) $. Similarly, the second term on RHS becomes $\int_{-\infty}^{0} y^{n}d\mu_X(y) $. Adding these we get $\int_{\mathbb R} y^{n}d\mu(y)$ which is $EX^{n}$.