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$F(x)$ is the distribution function of $\mathbb X$, and $f(x)$ is the derivation of $F(x)$, Prove that $\int_{0}^{\infty}(1-F(x))dx-\int_{-\infty}^{0}F(x)=E(X)$. Note that $E(X)=\int_{-\infty}^{\infty}xf(x)dx$. I integrated it by parts, but I got something like $xF(x)|_{-\infty}^{\infty}-\int_{0}^{\infty}F(x)$, and I have to show that it equals to $\int_{0}^{\infty}(1-F(x)dx$. I have no idea how to proceed since the limit seems uncertain and I don't how to use that fact that $E(|x|)<\infty$, can anyone help me out?

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Integration by parts is the right idea, but let's start with the left side of $\displaystyle\int_{0}^{\infty}(1-F(x))\,dx - \int_{-\infty}^{0}F(x)\,dx = E(X)$ instead of the right side.

We have $\displaystyle\int_{0}^{\infty}(1-F(x))\,dx = \underbrace{\left[x(1-F(x))\right]_{0}^{\infty}}_{0}-\int_{0}^{\infty}-xF'(x)\,dx = \int_{0}^{\infty}xf(x)\,dx$,

and $\displaystyle\int_{-\infty}^{0}F(x)\,dx = \underbrace{\left[xF(x)\right]_{-\infty}^{0}}_{0} - \int_{-\infty}^{0}xF'(x)\,dx = -\int_{-\infty}^{0}xf(x)\,dx$.

Combining those two yields:

$\displaystyle\int_{0}^{\infty}(1-F(x))\,dx - \int_{-\infty}^{0}F(x)\,dx = \int_{0}^{\infty}xf(x)\,dx + \int_{-\infty}^{0}xf(x)\,dx = \int_{-\infty}^{\infty}xf(x)\,dx = E(X)$

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  • $\begingroup$ Thanks! But can you justify on why $x(1-F(x))\to 0$ as $x\to\infty?$, since $(1-F(x))$ goes to 0 but $x$ goes to infinity. $\endgroup$ – jack Jul 5 '15 at 20:43

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