Show that $E(Y^p)=\int_{0}^{\infty}px^{p-1}P(Y\geq x)dx$

Studying some probability theory and I came across this question. Show that if $$Y$$ is a non-negative random variable, and $$p>0$$, $$E(Y^p)=\int_{0}^{\infty}px^{p-1}P(Y\geq x)dx$$ I'm a bit stuck on this question: I know that by Markov's inequality, $$\frac{E(Y^p)}{y^p}\geq P(Y>y)$$ since $$Y=\lvert Y\rvert$$ here. Also, $$y^p=\int_{0}^y px^{p-1}dx$$ and I believe these two facts can be used to solve the problem, but I'm not sure how exactly.

Could anyone give me some help, point me in the right direction?

If you are willing to use Tonelli's theorem (as opposed to something more elementary) you have $$\int_0^\infty px^{p-1} P(Y \ge x) \, dx = \int_0^\infty px^{p-1} \int_\Omega \mathbb{1}_{\{Y \ge x\}} \, dP dx = \int_\Omega \int_0^\infty px^{p-1} \mathbb{1}_{\{Y \ge x\}} \, dxdP.$$ The inner integral evaluates as $$\int_0^\infty px^{p-1} \mathbb{1}_{\{Y \ge x\}} \, dx = \int_0^Y px^{p-1} \, dx = Y^p.$$

• Would Fubini's work here as well? Also, how is the first equality in the 2nd line derived? – Mog Oct 29 '18 at 23:20
• Yes, but Fubini requires the hypothesis that the expression inside the double integral is actually integrable on the product measure space. Tonelli's theorem works only for nonnegative functions and does not require integrability, making the proof a bit shorter. – Umberto P. Oct 29 '18 at 23:23
• On any measure space $\mu(E) = \int \mathbb{1}_E \, d\mu.$ – Umberto P. Oct 29 '18 at 23:24
• I believe I can apply Fubini's by stating that $px^{p-1}\mathbb 1_{Y\geq x}$ is non-negative, am I correct? (I'm not equipped with Tonelli's but am with Fubini's.) – Mog Oct 29 '18 at 23:38

If $$Y$$ has density $$f(y)$$, the trick is to (1) put $$y^p=\int_0^y px^{p-1}\,dx$$ in the formula for $$E(Y^p)$$, obtaining a double integral, then (2) interchange the order of integration: \begin{align} E(Y^p) &= \int_{y=0}^\infty y^pf(y)\,dy\\ &\stackrel{(1)}=\int_{y=0}^\infty\left(\int_{x=0}^y px^{p-1}\right)f(y)\,dy\\ &=\int_{y=0}^\infty\left(\int_{x=0}^y px^{p-1}f(y)\right)\,dy\\ &\stackrel{(2)}=\int_{x=0}^\infty\left(\int_{y=x}^\infty px^{p-1}f(y)\right)\,dx\\ &=\int_{x=0}^\infty px^{p-1}\left(\int_{y=x}^\infty f(y)\right)\,dx\\ &=\int_{x=0}^\infty px^{p-1} P(Y\ge x)\,dx \end{align}

• Thank you for your answer; my worry was that as the problem does not explicitly state that $Y$ has a density, wasn't sure if I could do it this way. A question though, if I took this route, order of integration doesn't have to come from Fubini's or Tonelli's rather just from properties of integrals on the real line right? – Mog Oct 30 '18 at 0:09
• @pilotmath Actually Fubini is the justification for swapping the order of integration here. We often forget the reason why we are allowed to manipulate integrals on the real line the way we do. – grand_chat Oct 30 '18 at 0:28

\begin{align*} E [ Y^p] & = E [\int_0^{Y^p} ds ]\\ & = E [ \int_0^\infty {\bf 1}_{[0,Y^p)}(s) ds ] \\ & = \int_0^\infty P(s< Y^p) ds \\ & = \int_0^\infty P(Y > s^{1/p} ds \\ & \underset{x=s^{1/p}}{=} \int_0^\infty P(Y>x) p x^{p-1} dx. \end{align*}