probability - Expectation of a continuous random variable explained in terms of the CDF

Problem:

Let $F_X(x)$ be the CDF of a continuous random variable $X$. Show that:

$$E[X]= \int_0^\infty(1-F_X(x)) \, dx -\int_{-\infty}^0F_X(x) \, dx.$$

Attempt:

A comprehensible explanation of the intuition regarding the expectation $E[X]$ and CDF for a non-negative random is found here: Intuition behind using complementary CDF to compute expectation for nonnegative random variables.

However, I am still at a loss about how to show the general case when $-\infty < x < \infty$.

I am again solving this as an exercise in my probability course and any help is greatly appreciated!

Answer

You can also see it by interchanging the order of integrals. We have

$$\begin{eqnarray*} \int_{0}^{\infty}(1-F_{X}(x))dx=\int_{0}^{\infty}P(X>x)dx & = & \int_{0}^{\infty}\int_{x}^{\infty}dF_{X}(t)dx\\ & = & \int_{0}^{\infty}\int_{0}^{t}dF_{X}(t)dx\\ & = & \int_{0}^{\infty}t\;dF_{X}(t) \end{eqnarray*}$$
and,
$$\begin{eqnarray*} \int_{-\infty}^{0}F_{X}(x)dx=\int_{-\infty}^{0}P(X\leq x)dx & = & \int_{-\infty}^{0}\int_{-\infty}^{x}dF_{X}(t)dx\\ & = & \int_{-\infty}^{0}\int_{t}^{0}dF_{X}(t)dx\\ & = & \int_{-\infty}^{0}-t\;dF_{X}(t) \end{eqnarray*}$$
Since,

$$E[X]=\int_{-\infty}^{0}t\;dF_{X}(t)+\int_{0}^{\infty}t\;dF_{X}(t)$$
the result follows.