Here is a theorem on expectation of a random variable in terms of its distribution function
Theroem: Let $X$ be a (continuous or discrete) non-negative random variable with distribution function $F$. Then, $E(|X|) < \infty$ if and only if $\displaystyle \int_0^\infty 1-F(x)dx <\infty$, and in that case,
$$E(X) = \displaystyle \int_0^\infty1-F(x)dx$$.
Then, a corollary of the Theorem is given as:
Corollary: For any random variable $X$, $E(|X|) <\infty $ if and only if the integrals $\displaystyle \int_0^\infty 1-F(x)dx$ and $\displaystyle \int_{-\infty}^0 F(x)dx $ both converge, and in that case
$$E(X) = \displaystyle \int_0^\infty 1-F(x)dx – \displaystyle \int_{-\infty}^0 F(x)dx$$
I understand the Theorem, but I do not see how the Corollary follows from the Theorem. I understand the first claim of the Corollary, but I do not see why
$$E(X) = \displaystyle \int_0^\infty 1-F(x)dx – \displaystyle \int_{-\infty}^0 F(x)dx \tag{1}$$
holds in that case.
I have that:
$$E(|X|) = \displaystyle \int_0^\infty P\{|X| > x\}dx \\
= \displaystyle \int_0^\infty P\{X > x\} + \displaystyle \int_0^\infty P\{X < -x\}dx \\
= \displaystyle \int_0^\infty P\{X > x\} – \displaystyle \int_0^\infty P\{X < x\}dx \tag{2}$$,
but then I could not conclude (1) since the integrand in the second integral of the last line in (2) is $P\{X < x\}$, which is equal to $F(x)$ if X is a continuous random variable, but not equal to $F(x)$ if X is discrete variable.
What am I missing?
Best Answer
\begin{align} \mathsf{E}X&=\mathsf{E}X^+-\mathsf{E}X^-=\mathsf{E}[X\vee 0]-\mathsf{E}[-X\vee 0] \\ &=\int_0^\infty\mathsf{P}\{X>x\}dx-\int_{0}^\infty\mathsf{P}\{-X\ge x\}dx. \end{align}