Given a continuous positive r.v. (I think this means $Z \geq 0$), with pdf $f_Z$ and CDF $F_Z$, how would I show that the following expression
$$\mathbb{E}(Z^p) = p \int_0^\infty (1-F_Z(x))x^{p-1} \, dx\text{?}$$
I don't know where to start, I tried maybe changing it to a by-parts question or something using $px^{p-1} = \frac{d}{dx}(x^p)$ but that's all I can think of.
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Did
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Twenty-six colours
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The existence of the PDF is not required and the result is direct using Fubini, as has been explained several times on the site. – Did May 21 '17 at 11:52
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Note that for $X \geq 0$ we have $\mathbb{E}(X) = \int_0^{\infty} \mathbb{P}(X \geq x) \, \mathrm{d}x$. Now let $Y= Z^p$ then $\mathbb{P}(Y < y) = \mathbb{P}(Z < y^{1/p})$ by monotonicity of $x^p$ on $[0,\infty)$. Hence we have $$\mathbb{E}(Z^p) = \int_0^{\infty} \mathbb{P}(Y > y) \, \mathrm{d}y = \int_0^{\infty} 1-F_Y(y) \, \mathrm{d}y = \int_0^{\infty}1-F_Z(y^{1/p}) \, \mathrm{d}y$$
Now we use the substitution $x^p = y$ to get $$\mathbb{E}(Z^p) = \int_0^{\infty}p(1-F_Z(x))x^{p-1} \, \mathrm{d}x.$$
Alternatively you can mimic the proof for $\mathbb{E}(Z) = \int_0^{\infty} \mathbb{P}(Z \geq z) \, \mathrm{d}z$ using Fubini for $\mathbb{E}(Z^p)$.
Zain Patel
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Cheers. I don't see how it goes to zero. Won't $x^p$ be able to dominate it so the non-integral part diverges? – Twenty-six colours May 21 '17 at 03:07
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They're positive integers so it says to prove it is true in general for $p = 1,2,3 \cdots,$ – Twenty-six colours May 21 '17 at 03:10
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Observation: If $x^{p-1}$ dominated over $1-F_Z(x)$ for large $x$ then the original integral for your expectation wouldn't exist. – Zain Patel May 21 '17 at 03:14
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Ah I see. So for the expectation to exist, then the integrand would've had to exist, which means $x^p$ wouldn't have dominated in the first place? – Twenty-six colours May 21 '17 at 03:17
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That $\displaystyle \lim_{x\to\infty} x^p (1-F_Z(x)) =0$ deserves more ink than it got here. $\qquad$ – Michael Hardy May 21 '17 at 03:29
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Wait I'm still a bit unsure. The integrand is $f$, not $F$, so why does $x^p$ not dominate $1-F$? – Twenty-six colours May 21 '17 at 04:01
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Sorry: correction to the other comment: If $x^{p-1}$ doesn't dominate $1-F_Z(x)$, this doesn't imply $x^p$ does not dominate $1-F_Z(x)$ does it? – Twenty-six colours May 21 '17 at 04:13
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