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?

A better way to prove it would be to use Fubini's theorem to change the order of integration. This also gives you a condition when the result you have is true.
Consider $I = \displaystyle \int_0^{\infty}n x^{n-1} (1-F(x))dx$.
Using the fact that $\displaystyle \int_x^{\infty} f(y)dy = 1 - F(x)$, we get $I = \displaystyle \int_0^{\infty}n x^{n-1} \int_x^{\infty} f(y) dy dx$.
Now we first integrate with respect to $y$ (the inner integral) and $y$ goes from $x$ to $\infty$ and then integrate with respect to $x$ (the outer integral), $x$ goes from $0$ to $\infty$.
Change the order of integration. i.e. integrate with respect to $x$ first and then with respect to $y$.
Note that this can be done provided the integral $I < \infty$ (See Fubini's theorem). This is the condition svenkatr and trutheality get as well.
Changing the order of integration, we get
$I = \displaystyle \int_0^{\infty} \displaystyle \int_{0}^{y} nx^{n-1}f(y)dxdy$.
Note that now $x$ in the inner integral goes from $0$ to $y$ and $y$ goes from $0$ to $\infty$.
Now the inner integral with respect to $x$ can performed easily and now we get
$I = \displaystyle \int_0^{\infty} y^{n}f(y)dy = E[X^n]$.
Hence, we have $\displaystyle \int_0^{\infty}n x^{n-1} (1-F(x))dx = E[X^n]$.
