Moments and non-negative random variables? I want to prove that for non-negative random variables with distribution F:
$$E(X^{n}) = \int_0^\infty n x^{n-1} P(\{X≥x\}) dx$$
Is the following proof correct?
$$R.H.S = \int_0^\infty n x^{n-1} P(\{X≥x\}) dx = \int_0^\infty n x^{n-1} (1-F(x)) dx$$
using integration by parts:
$$R.H.S = [x^{n}(1-F(x))]_0^\infty + \int_0^\infty x^{n} f(x) dx = 0 + \int_0^\infty x^{n} f(x) dx = E(X^{n})$$
If not correct, then how to prove it?

If ${\rm E}(X^n) < \infty$, then $\int_0^\infty  {x^n f(x){\rm d}x}  < \infty $, and in turn, $\int_M^\infty  {x^n f(x){\rm d}x}  \to 0$ as $M \to \infty$. Since $\int_M^\infty  {x^n f(x){\rm d}x} \geq \int_M^\infty  {M^n f(x){\rm d}x}  = M^n [1 - F(M)]$, we have that $M^n [1 - F(M)] \to 0$ as $M \to \infty$. Hence, your solution is correct (assuming that ${\rm E}(X^n) < \infty$).
