Complimentary Cumulative Distribution Function

Question

Let $X$ be a nonnegative r.v. with cdf $F$ and $E[X]<\infty$. Consider the indicator function. Show that

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

There is a hint provided that says: First claim that $\int_{0}^{\infty} 1_{{X>x}} dx = X$. Use that result to show that $$E[X^m] = m \int_{0}^{\infty}x^{m-1}F^c(x)dx, m \geq 1$$

Despite being already closed, this comment is to link this post as one of the (abstract) duplicates to the current choice of mother post. — Lee David Chung Lin, Nov 13 '18 at 12:23

score 2 · Answer 1 · answered Oct 16 '15 at 21:54

First, note that the result of the hint with $m=1$ proves the original question, so we just need to show this more general result.

So, \begin{align*} m\int_0^\infty x^{m-1} F^c(x) \mathop{dx} &= m \int_0^\infty x^{m-1} E[\mathbf{1}_{X>x}]\mathop{dx} \\ &= E\left[m \int_0^\infty x^{m-1} \mathbf{1}_{X>x} \mathop{dx}\right] & \text{Fubini (switch expectation and integration)} \\ &= E\left[\int_0^X mx^{m-1} \mathop{dx}\right] \\ &= E[X^m]. \end{align*}

Complimentary Cumulative Distribution Function

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