Consider two continuous random variables X and Y with joint p.d.f.
$f_{X,Y}(x,y) = \frac{x+2y}{24}, 0<x<2, 0<y<3$
Find the probability distribution of $W=X+Y$.
All I want to understand are the bounds for my double integral to find the probability distribution as a p.d.f. I understand that $0<w<5$ but from a lot of the problems I've seen before the bounds are still $\int^w_0\int^{w-x}_0$.
I don't need a whole solution, just want to understand the bounds. Thank you!
Indeed the support is $0<W<5$, however we are after the CDF, $\mathsf P(W\leq w)$.
For any $0<w<5$ we require $0<W\leq w$, which is $0<X+Y<w$ , and we also require $0<X<2$ and $0<Y<3$ so the bounds should be:
$$\max\{0,w-3\}< X< \min\{2, w-0\} \cap 0<Y<\min\{3,w-X\}, $$
Thusly
$$\begin{align}\Pr(W\leq w) & = \Pr(X+Y\leq w) \\[2ex] & = \mathbf 1_{0<w<5}\cdot \int_{\max\{0,w-3\}}^{\min\{2,w\}}\int_{0}^{\min(3,w-x)} f_{X,Y}(x,y)\,\mathrm {d} y\,\mathrm{d} x ~+~ \mathbf 1_{5\leqslant w} \\[2ex] & = {{\mathbf 1_{0<w<2}\cdot\int_0^w \int_{0}^{w-x}f_{X,Y}(x,y)\,\mathrm {d} y\,\mathrm{d} x }\\+{\mathbf 1_{2\leqslant w<3}\cdot\int_0^2\int_0^{w-x}f_{X,Y}(x,y)\,\mathrm {d} y\,\mathrm{d} x}\\+{\mathbf 1_{3\leqslant w<5}\cdot\int_{w-3}^2\int_0^{3}f_{X,Y}(x,y)\,\mathrm {d} y\,\mathrm{d} x}\\+{\mathbf 1_{5\leqslant w}}}\end{align}$$
