renormalization of sum of continuous random variables

Question

I want to sum two random variables.

So $Z = X+Y$ and $f_{X+Y}(z)=\int_{-\infty}^{\infty}f_{xy}(x,z-x)dx $

So I wanted to test this out a bit. If $$f_{xy}(x,y) = 6∙10^{-4}(x^2+y^2)$$ when $-5\leq x \leq5$ and $-5\leq y \leq5$ and 0 elsewhere.

The joint distribution $$\int^5_{-5} \int^5_{-5} f_{xy}(x,y)dxdy = 1$$

But when I try to calculate the pdf of the sum of X and Y

$$f_{X+Y}(z)=\int_{-5}^{5}6∙10^{-4}(x^2+(z-x)^2)dx = \frac{1}{500}(3z^2+50) $$

If I integrate this function from -10 to 10 with respect to z, I get 6.0. Not 1.0. $$\int_{-10}^{10}\frac{1}{500}(3z^2+50)=\frac{1}{500}(z^3+50z)|^{10}_{-10}=\frac{1}{500}(2000+(50)(20))=6$$

What am I doing wrong? $f_{X+Y}(z)$ is a probability distribution so it should integrate to 1. And I shouldn't have to re-normalize it.

Answer 1

Check your limits of integration. Remember the joint density $f(x,y)$ is zero outside of $|x|\le 5$ and $|y|\le 5$, so the convolution requires the constraints $|x|\le 5$ and $|z-x|\le 5$.

Unfortunately this means you have to evaluate $f_{X+Y}(z)$ by cases, depending on the value of $z$. If you plot the region defined by the constraints, you'll have two cases: (1) $0\le z\le 10$, where you integrate $x$ from $z-5$ to $5$, and (2) $-10\le z\le 0$, where you integrate $x$ from $-5$ to $z+5$. Outside of these two cases the value of $f_{X+Y}(z)$ is zero.

Answer 2

There's no $z-x$ in the upper limit of the integral for $f_{X+Y}(z)$, it's just $z$.

Anyway, you have some setup problems. $f_{X+Y}(z)=\int_{-\infty}^\infty f_{X,Y}(x,z-x) dx$ is true. But you have to be more careful about where that integrand is zero: it's zero whenever $x$ or $z-x$ is not in $[-5,5]$. This immediately restricts the range of integration to $[-5,5]$ so the first argument can't be too big, and it restricts the range of integration to $z-x \in [-5,5] \Leftrightarrow x \in [-5+z,5+z]$ so that the second argument can't be too big. Then you have to take the intersection, so you wind up integrating on $[\max \{ -5,-5+z \},\min \{ 5,5+z \}]$.

This naturally creates a piecewise character in the PDF as you pass through $z=0$, because for $z<0$ you integrate over $[-5,5+z]$ and for $z>0$ you integrate over $[-5+z,5]$. This piecewise effect is quite common.