Sunday, 23 September 2018

probability - Expectetion of Yalpha with alpha>0





Let Y be a positive random variable. For α>0 show that



E(Yα)=α0tα1P(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)=1Fy(t). Therefore Gy(t)=fy(t). Integrate E(Yα) by parts and get E(Yα)=tαGy(t)]0+α0tα1Gy(t)dt=α0tα1P(Y>t)dt.


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