Let Y be a positive random variable. For α>0 show that
E(Yα)=α∫∞0tα−1P(Y>t)dt.
My ideas:
E(Yα)=∫∞−∞tαfY(t)dt
=∫∞0tαfY(t)dt
=∫∞0(∫tα0dy)fY(t)dt
Answer
E(Yα)=∫∞0tαfy(t)dt. Let Gy(t)=P(Y>t)=1−Fy(t). Therefore G′y(t)=−fy(t). Integrate E(Yα) by parts and get E(Yα)=−tαGy(t)]∞0+α∫∞0tα−1Gy(t)dt=α∫∞0tα−1P(Y>t)dt.
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