Convolution of distributions of two non negative random variables

Question

If $X$ and $Y$ are independent random variables that follow some distributions $F$ and $G$ resp., then the r.v. $X+Y$ follows $(F\star G)$, the convolution of the two distributions.

And if $\Bbb P(X\ge0)=\Bbb P(Y\ge 0)=1$ we can write $(F\star G)(x)=\int\limits_0^x G(x-u)dF(u)$. At least that's what is written in my notes.

Now I was familiar with the definition $(f\star g)(x)=\int\limits_{-\infty}^{\infty}f(x-u)g(u)du$ for the convolution from some real analysis courses and wikipedia.

So the former definition is confusing for me, especially $dF(u)$

Answer 1

By the chain rule, $dF(u)=f(u)du$ if the pdf $f=F^\prime$ exists. This makes the expressions coincide. The advantage of using $dF$ is it handles cases where the distribution isn't continuous.

Answer 2

If $X,Y$ are independent random variables a.s. nonnegative and $f:\mathbb{R}^{2}\to\mathbb{R}$ is a suitable function then: $$\mathbb{E}f\left(X,Y\right)=\int_{0}^{\infty}\int_{0}^{\infty}f\left(u,v\right)dG\left(v\right)dF\left(u\right)$$

For a fixed $x>0$ let function $f$ be prescribed by: $$\left(u,v\right)\mapsto\mathbf{1}_{\left(0,x\right]}\left(u+v\right)$$

Then:

$$P\left(X+Y\leq x\right)=\mathbb{E}f\left(X,Y\right)=\int_{0}^{\infty}\int_{0}^{\infty}\mathbf{1}_{\left(0,x\right]}\left(u+v\right)dG\left(v\right)dF\left(u\right)=\int_{0}^{x}\int_{0}^{x-u}dG\left(v\right)dF\left(u\right)=$$$$\int_{0}^{x}G\left(x-u\right)dF\left(u\right)$$