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

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


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.

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    It is $\int_0^\infty \Pr(X^r \gt x)\,dx$, which is $\int_0^\infty \Pr(X \gt x^{1/r})\,dx$. Make the change of variable $u^r=x$. – 2012-08-04
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    @Justin: I'm quite curious about the reference of this exercise. Could you tell me in which book you encountered it? – 2012-10-15
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    Even though this post is slightly different, I’d like to link it to the current choice of [mother post](https://math.stackexchange.com/questions/172841). Also see the meta post for [(abstract) duplicates](https://math.meta.stackexchange.com/a/29382/356647). – 2018-11-13

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