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Let $Y$ be a positive random variable. For $\alpha>0$ show that

$E(Y^{\alpha})=\alpha \int_{0}^{\infty}t^{\alpha -1}P(Y>t)dt$.

My ideas:

$E(Y^{\alpha})= \int_{-\infty}^{\infty}t^{\alpha}f_{Y}(t)dt$

=$\int_{0}^{\infty}t^{\alpha}f_{Y}(t)dt$

=$\int_{0}^{\infty}(\int_{0}^{t^{\alpha}}dy)f_{Y}(t)dt$

irchans
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VarúAnselmo Sui
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2 Answers2

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$E(Y^\alpha)=\int_0^\infty t^\alpha f_y(t)dt$. Let $G_y(t)=P(Y\gt t)=1-F_y(t)$. Therefore $G'_y(t)=-f_y(t)$. Integrate $E(Y^\alpha)$ by parts and get $E(Y^\alpha)=-t^\alpha G_y(t)\rbrack_0^\infty +\alpha \int_0^\infty t^{\alpha-1}G_y(t)dt={\alpha \int_0^\infty t^{\alpha-1}P(Y\gt t)dt}$.

herb steinberg
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By Tonelli's theorem (i.e. switching the order of integration), \begin{align*} \int_{[0,\infty)}\alpha t^{\alpha-1}P[Y>t]\,dt &= \int_{[0,\infty)}\alpha t^{\alpha-1}\int_\Omega 1_{\{\omega':Y(\omega')>t\}}(\omega)\,P(d\omega)\,dt \\ &= \int_\Omega\int_{[0,\infty)} \alpha t^{\alpha-1}1_{\{\omega':Y(\omega')>t\}}(\omega)\,dt\,P(d\omega) \\ &= \int_\Omega\int_{[0,Y(\omega))}\alpha t^{\alpha-1}\,dt\,P(d\omega) \\ &= \int_\Omega Y(\omega)^\alpha\,P(d\omega) \qquad\text{fundamental theorem of calculus} \\ &= E(Y^\alpha), \end{align*} as desired. This method also does not require that the distribution function be continuous, merely integrable.

Alex Ortiz
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  • I don't understand the $3_{rd}$ '=' sign. Could you explain that to me? – Jasper Mar 16 '19 at 16:56
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    @Jasper: In the integral $\int_{[0,\infty)}\alpha t^{\alpha-1}1_{{\omega':Y(\omega')>t}}(\omega),dt$, $\omega$ is fixed, so if $t\ge Y(\omega)$, then $1_{{\omega':Y(\omega')>t}}(\omega) = 0$. Hence we only integrate over those $t$ such that $0 \le t < Y(\omega)$, i.e., where $1_{{\omega':Y(\omega')>t}}(\omega) = 1$. Hope that helps. – Alex Ortiz Mar 16 '19 at 17:18
  • Ahh yes, thank you! – Jasper Mar 16 '19 at 18:15