Expectetion of $Y^{\alpha}$ with $\alpha >0$

Question

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:

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

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

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

Answer 1

$E(Y^\alpha)=\int_0^\infty t^\alpha f_y(t)dt$. 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 $E(Y^\alpha)=-t^\alpha G_y(t)\rbrack_0^\infty +\alpha \int_0^\infty t^{\alpha-1}G_y(t)dt={\alpha \int_0^\infty t^{\alpha-1}P(Y\gt t)dt}$.

Answer 2

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.