Confused about the expression relating the CDF and expected value of a random variable

Question

Let X be a random variable that takes on nonnegative values and has distribution function $F(x)$. $E(X) = \int_{0}^{\infty}1-F(x)dx$

The proof my book gives is: This statement was included at the end with respect to the last inequality:

Since both terms are non-negative, the only way this can happen is for the inequality to be an equality and the limit to be 0.

Based on the integration by parts formula, looks like $u = x$ and $dv = f(x)dx = dF(x)$ since $E(X) = \int_{0}^{\infty}xf(x)dx$.

1. It appears that $v = F(x)-1$; why is it not $F(x)$? Given that $dF(x)/dx = f(x)$ so I thought the antiderivative of $f(x)$ would be $F(x)$?

2. How is $a(1-F(a))$ is less or equal to $\int_{a}^{\infty} xf(x)dx$ (in the middle part)?

3. I don't really understand the purpose of the last equation; why do we have $E(X)$ on both sides of the inequality?

Answer 1

1. The antiderivative of $f(x)$ is $F(x)+c \;\; \forall c \in R\;$ and in this case $c=-1$ in order to get the proof.
2. You have to integrate $xf(x)dx$ from $a$ to infinity so $x\geq a$, then $\int_a^{+\infty} xf(x)dx \geq \int_a^{+\infty}af(x)dx=a(1-F(a))$
3. $\int_0^{+\infty} xf(x)dx \geq \int_0^{a}af(x)dx+a(1-F(a))\;\;\; \forall a \geq 0$, so letting a tend to infinity we have $\lim_{a\to+\infty}\int_0^{+\infty} xf(x)dx\geq \lim_{a\to+\infty} (\int_0^{a}xf(x)dx+a(1-F(a)))$ $E(X)\geq \int_0^{+\infty} xf(x)dx+\lim_{a\to+\infty}a(1-F(a))$ and $E(X)\geq E(X)+\lim_{a\to+\infty}a(1-F(a))$, since by definition $E(X)=\int_0^{+\infty} xf(x)dx$