So I am given
$$
f_{X,Y}(x,y) =\begin{cases}\frac{1}{\pi}&\mathrm{\ if \ } x^2+y^2\le1\\ 0&\mathrm{\ otherwise\ }\end{cases}
$$
And am asked to find joint probability density function for $X+Y$.
I'm assuming that I must use
$$
f_{X+Y}(z)=\int f_X(x)f_Y(z−x)\,\mathrm{d}x
$$
however, I'm not sure where to go from there. I have seen examples of solving for $X+Y$ but only for examples when $X$ and $Y$ are independent and I can use the marginal densities. In this case would I solve for the marginal density of $X$ normally but for $Y$ solve in terms of $(z-x)$ rather than $y$?
Best Answer
Hint:
$\displaystyle f_{X+Y}^{\,}(z)=\int f_{X,Y}^{\,}(x,z−x)\,\mathrm{d}x$ is a more basic assumption.
You know that inside the disk the joint density is a uniform $\frac{1}{\pi}$, and that you are inside the disk, given $z$, when $x^2+(z-x)^2 \le 1$; this is a quadratic inequality which will give you an interval when $-\sqrt{2} \le z \le \sqrt{2}$, and so the integral is not much harder than solving quadratic equation.