Question on integrating probability

Question

Let $X,Y$ be two independent random variables with distribution functions $F_X,F_Y$. I need to show that $$P(Y\leq y+X)=\int F_Y(y+x)\mathrm{d}F_X(x)$$

I have no idea where to start.

Answer 1

If the CDF of $\left(X,Y\right)$ is $F_{X,Y}$ then: $$P\left(Y\leq y+X\right)=\iint f\left(u,v\right)dF_{X,Y}\left(u,v\right)$$ where $f$ sends $\left(u,v\right)$ to $1$ if $v\leq y+u$ and to $0$ otherwise.

By independence of $X$ and $Y$ we have $F_{X,Y}\left(u,v\right)=F_{X}\left(u\right)F_{Y}\left(v\right)$ resulting in: $$\iint f\left(u,v\right)dF_{X,Y}\left(u,v\right)=\iint f\left(u,v\right)dF_{Y}\left(v\right)dF_{X}\left(u\right)=\int F_{Y}\left(y+u\right)dF_{X}\left(u\right)$$

I deliberately used $u$ and $v$ instead of $x$ and $y$ to avoid confusion that can rise because in your question $y$ is already used as a constant.