I dont know what to do, I have to proof this $$ E(X^n)=n\displaystyle\int_{0}^{\infty}(1-F_{X}(x))x^{n-1}dx-n \displaystyle\int_{-\infty}^{0}F_{X}(x)x^{n-1}dx$$ but i dont have any idea. Thanks.
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I think you mean $E(X^{n})$? – Ekesh Kumar Nov 29 '19 at 23:05
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Show that $\mathbb E[X] = \int_0^\infty (1-F_X(x))\ \mathsf dx - \int_{-\infty}^0 F_X(x)\ \mathsf dx$ using integration by parts, then use induction on $n$ to prove the general formula. Recall that $\mathbb E[X] = \int_{-\infty}^\infty x\ \mathsf dF_X(x)$ - let $u=x$ and $\mathsf dv = \mathsf dF_X(x)$, and go from there. – Math1000 Nov 29 '19 at 23:10
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The proof for $n=1$ is shown here which is easily generalised. – StubbornAtom Nov 30 '19 at 11:01
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$n\int_0^{\infty} x^{n-1} (1-F_X(x))dx=n\int_0^{\infty} x^{n-1} \int_{(x,\infty)} d\mu_X(y) dx$ where $\mu_X(E)=P(X^{-1}(E))$. By Fubini/Tonelli Theorem this becomes $\int_0^{\infty} \int_0^{y} nx^{n-1} dx d\mu_X(y) =\int_0^{\infty} y^{n}d\mu_X(y) $. Similarly, the second term on RHS becomes $\int_{-\infty}^{0} y^{n}d\mu_X(y) $. Adding these we get $\int_{\mathbb R} y^{n}d\mu(y)$ which is $EX^{n}$.
Kavi Rama Murthy
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