Consider a random variable $X$ with distribution function $F(x)$. Calculate the $r$th moment of $X$, $\mathbb E X^r$. I read that the desired moment can be calculated as follows.
$$ \begin{align} \mathbb E X^r &= \int_0^\infty x^r dF(x) - \int_0^\infty (-x)^r dF(-x) \\[5pt] &=r \int_0^\infty x^{r-1} \left[ 1-F(x)+(-1)^rF(-x) \right] dx. \end{align} $$
Could anyone explain to me why this is true, please? In addition, when do we need to find the moments in this way? Thank you!
Here's a sketch for the continuous case:
For any nonnegative continuous real random variable $X$ and any integer $r\ge 1$,
$$X^r = \int_0^X rx^{r-1}dx = \int_0^\infty r x^{r-1}[X>x]dx
$$
where
$$[X>x] =
\begin{cases}
1 & \text{if }X>x \\
0 & \text{if }X\le x.
\end{cases}$$
Therefore, using Tonelli's theorem and the fact that $E[X>x] = P(X>x)$,
$$E (X^r) = r \int_0^\infty x^{r-1}P(X>x)dx.
$$
Now, for any continuous random variable $X$ (not necessarily nonnegative), we have $X = Y - Z$, where
$Y=X^+$ and $Z=X^-$ are the positive and negative parts of $X$.
Then, since $YZ = 0$, the Binomial Theorem gives
$$X^r = (Y - Z)^r = Y^r +(-Z)^r
$$
and because both $Y$ and $Z$ are nonnegative random variables,
$$\begin{align}
E(X^r) &= E(Y^r) + (-1)^rE(Z^r)\\
&=r\int_0^\infty y^{r-1}P(Y>y)dy + (-1)^r r\int_0^\infty z^{r-1}P(Z>z)dz\\
&=r\int_0^\infty y^{r-1}P(X>y)dy + (-1)^r r\int_0^\infty z^{r-1}P(X<-z)dz\\
&=r\int_0^\infty x^{r-1}\big(P(X>x) + (-1)^r P(X<-x)\big)dx\\
&=r\int_0^\infty x^{r-1}\big(1-F(x) + (-1)^r F(-x)\big)dx.\\
\end{align}
$$
As to your question "why" this is necessary, many times it is easy to approximate an empirical distribution with some other theoretical cdf. Then this gives a shortcut to investigating the moments of such a distribution if you don't have the time/patience to derive the pdf and then derive the moments from there. Also, it gives us general albeit significant insight into the relationship between the moments of a distribution and the cdf, where we are used to only seeing it in terms of the expected value with the pdf.
