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Let $X$ be a non-negative random variable with density function $f$. Show that $$ E(X^r) = \int_0^\infty r x^{r-1} P(X > r)\,dx. $$

I tried using integration by parts to obtain \begin{align} \int_0^\infty r x^{r-1} P(X > r)\,dx &= \int_0^\infty r x^{r-1} (1 - F(x))\,dx, \\ &= [x^r]_0^{\infty} + \int_0^\infty x^r f(x)\,dx. \end{align}

Where is the mistake in the above calculation and how do I get rid of the first term on the right to obtain the required result? Any hints would be much appreciated.

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The first term should be $[x^r(1-F(x)]_0^{\infty}=0-0$.

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  • $\begingroup$ Doh! I should be more careful. $\endgroup$ – Raj Apr 25 '16 at 19:11
  • $\begingroup$ You need to show that $\lim_{x\to\infty}( x^r(1-F(x))=0$, which does require some care. $\endgroup$ – John Dawkins Apr 26 '16 at 0:32
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An alternative solution

$$ \begin{align} E(X^r) & = \int_0^{\infty} r x^{r-1} P(X > x) \:dx \\ &= \int_0^{\infty} r x^{r-1} \left[ \int_{y=x}^{\infty} f(y) \:dy \right] \:dx \\ &= \int_{y=0}^{\infty} f(y) \left[ \int_{x=0}^{y}r x^{r-1} \,dx \right]\:dy \\ &= \int_0^{\infty} y^r f(y)\,dy \end{align} $$

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  • $\begingroup$ Thank you for the alternative method. $\endgroup$ – Raj Apr 25 '16 at 19:50

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