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:

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

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.

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 $u^r=x$. — André Nicolas, Aug 04 '12 at 21:06
@Justin: I'm quite curious about the reference of this exercise. Could you tell me in which book you encountered it? — , Oct 15 '12 at 22:59
Even though this post is slightly different, I’d like to link it to the current choice of mother post. Also see the meta post for (abstract) duplicates. — Lee David Chung Lin, Nov 13 '18 at 12:46

Did · Accepted Answer · 2012-08-06T14:27:11.977

2

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

edited Aug 06 '12 at 14:27

answered Aug 04 '12 at 21:05

Did

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

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