I want to prove that for non-negative random variables with distribution F:
E(Xn)=∫∞0nxn−1P({X≥x})dx
Is the following proof correct?
R.H.S=∫∞0nxn−1P({X≥x})dx=∫∞0nxn−1(1−F(x))dx
using integration by parts:
R.H.S=[xn(1−F(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(Y≥t)dt,
so
E[Xn]=∫∞0P(Xn≥t)dt.
Then, perform the change of variables t=xn. This immediately yields
E[Xn]=∫∞0nxn−1P(Xn≥xn)dx=∫∞0nxn−1P(X≥x)dx.
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