Prove $E[X]=\int_{0}^{\infty}(1-F(x))\,dx-\int_{-\infty}^{0}F(x)\,dx$.

Question

$F(x)$ is the distribution function of $\mathbb X$, and $f(x)$ is the derivation of $F(x)$, Prove that $$\int_{0}^{\infty}(1-F(x))dx-\int_{-\infty}^{0}F(x)dx=E(X)$$ 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?

Answer 1

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:

$$\begin{align}\int_{0}^{\infty}(1-F(x))\,dx - \int_{-\infty}^{0}F(x)\,dx &= \int_{0}^{\infty}xf(x)\,dx + \int_{-\infty}^{0}xf(x)\,dx \\ &= \int_{-\infty}^{\infty}xf(x)\,dx = E(X).\end{align}$$

In the above proof, we use the facts that $\lim_{x\to +\infty} x(1-F(x))=0$ and $\lim_{x\to-\infty} xF(x)=0$. A proof of the first fact is given in this question and answer.