probability theory - Show that $E(Z^p) = p int_0^infty (1-F_Z(x))x^{p-1} , dx$ for every $p>0$ and nonnegative random variable $Z$

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.

Answer

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 $\mathbb{E}(Z^p)$.