can somebody please explain me how following equality works:
$E [|X|^pI(|X| \leq x) ] = \int_0^x y^pdP(|X| \leq x) = x^pP(|X| > x) + p\int_0^x y^{p-1}P(|X| > y) dy$.
First one should be just the definition of the expectation but i can't get the second one. I thought about partial integration but failed doing it.
On
This is basically integration by parts for Riemann Stieltjes integrals. See here for example: http://www.math.mcgill.ca/labute/courses/255w03/L3.pdf
So literarily, $\int_0^x udv =uv|_{0}^x-\int_0^x vdu$. If the CDF $P(|X|\leq x)$ is differentiable, then this is back to the usual Riemann integration by parts.
The identity doesn't look quite right. If $Y$ is a nonnegative RV with cumulative distribution function $F(y):=P(Y\le y)$ and tail probability function $R(y):=P(Y>y)=1-F(y)$, then $$ E[Y^pI(Y\le x)]=\int y^pI_{(0,x]}(y)dF(y)=\int_0^x y^pdF(y) =-\int_0^xy^pdR(y)\tag1$$ and using integration by parts the RHS of (1) equals $$ \left.-y^pR(y)\right\vert_0^x +\int_0^xR(y)py^{p-1}dy=-x^pP(Y>x)+\int_0^xpy^{p-1}P(Y> y)dy,\tag2 $$ but this isn't what your identity is saying. Expressed in terms of $F$ we can similarly obtain $$ E[Y^pI(Y\le x)]=x^pF(x)-\int_0^xF(y)py^{p-1}dy=x^pP(Y\le x)-\int_0^xpy^{p-1}P(Y\le y)dy,\tag3 $$ which also doesn't agree with your formula.