I was looking at the proof of the theorem stating that for two independent rv's $X,Y$ with density functions $f,g$ the density function of the rv $X+Y$ is the convolution of $f$ and $g$.

After having defined $A_z=\{(x,y) \in \mathbb R^2 : x+y \leq z\}$ for $z \in \mathbb R$ we get

$P(X+Y \leq z) = P((X,Y) \in A_z) = \int_{A_z}f(x)g(y)\,dxdy=\int_{-\infty}^{\infty}\, dx \,f(x) \int_{-\infty}^{z-x}\, dy \, g(y)\, = \, ...$

Can someone please explain the last equality $\int_{A_z}f(x)g(y)\,dxdy=\int_{-\infty}^{\infty}\, dx \,f(x) \int_{-\infty}^{z-x}\, dy \, g(y)$? I don't understand how the boundaries have been chosen and why we can just take the functions out of the integrals. Thank you.


For fixed $x$ the inequality $x+y \leq z$ is same as $y \leq z-x$ so $y$ ranges from $-\infty$ to $z-x$. Once the inequality $x+y \leq z$ has been taken care of, there is no further restriction on $x$ so $x$ ranges from $-\infty$ to $\infty$. (You can also do it the other way around: keep $y$ fixed, integrate w.r.t. $x$ from $-\infty$ to $z-y$ and then integrate w.r.t $y$).

  • $\begingroup$ Got it, thank you very much! And why am I allowed to take the functions out of the integral? $\endgroup$ – Tesla Nov 19 '18 at 10:22
  • 1
    $\begingroup$ When you are integrating w.r.t. $y$, $f(x)$ acts like a constant, so you can pull it out. $\endgroup$ – Kavi Rama Murthy Nov 19 '18 at 10:24

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