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.