# Integral of a CDF

I'm solving a problem where I've this 'expectation': $$\int_{0}^y x\cdot f(x) dx$$ where $f(x)$ is a PDF with support on $[0, z]$, with $z>y$. Is there a way to rewrite it without the integral and as a function of the CDF? I've tried by parts, but without great success: $$\int_{0}^y x\cdot f(x) dx = y\cdot F(y) -\int_0^y F(x) dx$$ because I don't a way of solving for the second part.

• There's no further simplification possible. Are you sure those are all the details given in the problem? – Alex R. Mar 14 '18 at 18:40

For cdfs $F$ of distributions with supports on $(0,a)$, $a$ being possibly $+\infty$, a useful representation of the expectation is $$\mathbb{E}_F[X]=\int_0^a x \text{d}F(x)=\int_0^a \{1-F(x)\}\text{d}x$$ since by an integration by part \begin{align*}\int_0^a x \text{d}F(x)&=-\int_0^a x \text{d}(1-F)(x)\\&=\left[x(1-F(x))\right]_0^a+\int_0^a \{1-F(x)\}\text{d}x\\&=a(1-F(a))-0(1-F(0))+\int_0^a \{1-F(x)\}\text{d}x\end{align*} In the current case, one can turn the integral into an expectation as $$\int_0^y x\text{d}F(x)=F(y)\int_0^y x\frac{\text{d}F(x)}{F(y)}=\mathbb{E}_{\tilde{F}}[X]$$with $$\tilde{F}(x)=F(x)\big/F(y)\mathbb{I}_{(0,y)}(x)$$Thus $$\int_0^y x\text{d}F(x)=F(y)\int_0^y \{1-F(x)\big/F(y)\}\text{d}x$$ which is the representation that you found.
• Why can you write \begin{align*}\int_0^a x \text{d}F(x)&=-\int_0^a x \text{d}(1-F)(x)? is it because the derivative of $(1-x) = -dx$ so the derivative of $-(1-x) = dx$? – user106860 Nov 7 '18 at 4:22
• The equation in proper format is$$\int_0^a x \text{d}F(x)=-\int_0^a x \text{d}(1-F)(x)$$and there is no other derivative than $$\frac{\text{d}}{\text{d}x}F(x)=-\frac{\text{d}}{\text{d}x}(1-F)(x)$$ – Xi'an Nov 7 '18 at 19:51