Proof: $X\ge 0, r>0\Rightarrow E(X^r)=r\int_0^{\infty}x^{r-1}P(X>x)dx$

Question

As the title states, the problem at hand is proving the following:

$X\ge 0, r>0\Rightarrow E(X^r)=r\int_0^{\infty}x^{r-1}P(X>x)dx$

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Attempt/thoughts on a solution

I am guessing this is an application of Fubini's Theorem, but wouldn't that require writing $P(X>x)$ as an expectation? If so, how is this accomplished?

Thoughts and help are appreciated.

Answer 1

Proof: Consider the expectation of the identity $$ X^r=r\int_0^{X}x^{r-1}\,\mathrm dx=r\int_0^{+\infty}x^{r-1}\mathbf 1_{X>x}\,\mathrm dx. $$