Sunday, 29 June 2014

probability - Moments and non-negative random variables?



I want to prove that for non-negative random variables with distribution F:
E(Xn)=0nxn1P({Xx})dx





Is the following proof correct?



R.H.S=0nxn1P({Xx})dx=0nxn1(1F(x))dx




using integration by parts:
R.H.S=[xn(1F(x))]0+0xnf(x)dx=0+0xnf(x)dx=E(Xn)





If not correct, then how to prove it?


Answer



Here's another way. (As the others point out, the statement is true if E[Xn] actually exists.)



Let Y=Xn. Y is non-negative if X is.



We know
E[Y]=0P(Yt)dt,


so
E[Xn]=0P(Xnt)dt.


Then, perform the change of variables t=xn. This immediately yields
E[Xn]=0nxn1P(Xnxn)dx=0nxn1P(Xx)dx.


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