I have a textbook example as follows:
The joint distribution of $X$ and $Y$ is given by $$f_{XY}(x,y)=\begin{cases} \frac{3}{11}(5x+y) \hspace{0.2in} x,y>0, \hspace{0.2in}x+2y<2 \\ 0 \hspace{0.4in} \text{elsewhere}\end{cases}$$
Find the probability density of $Z=X+Y$.
I am currently on the section that deals with the sum of two continuous, independent random variables, and I am having trouble understanding how the solution finds the bounds of integration when using the formula: $$f_{X+Y}(a)=\int_{-\infty}^\infty f_X(a-y)f_Y(y)dy=\int_{-\infty}^\infty f_Y(a-y)f_Xdx$$ Where $f_X(x)$ and $f_Y(y)$ are the pdfs of $X$ and $Y$ respectively. The solution is as follows:
Note first that the region of integration is the interior of the triangle at $(0,0)$, $(0,1)$, and $(2,0)$. [this part I understand] If $0 \leq a <1$, then $$F_Z(a)=\text{Pr}(Z\leq a)=\int_0^a\int_0^{a-y} \frac{3}{11}(5x+y)dxdy=\frac{3}{11}a^3$$
If $1 \leq a < 2$ then the two lines $x+y=a$ and $x+2y=2$ intersect at $(2a-2, 2-a)$. In this case, $$F_Z(a)=\text{Pr}(Z \leq a)=\int_0^{2-a}\int_0^{a-y} \frac{3}{11}(5x+y)dxdy+\int_{2-a}^1\int_0^{2-2y} \frac{3}{11}(5x+y)dxdy=\frac{3}{11}\left(-\frac{7}{3}a^3+9a^2-8a+\frac{7}{3}\right)$$
My issue is understanding how the $a$ variable comes in, and how the bounds on above integrals are found. Is there a general, easy-to-understand process for this? I am familiar with double integration, but the composition of two variables here is confusing me. Thank you in advance!
Best Answer
$\hskip 1.7 in$
The shadowed region corresponds to all points $(x,y)$ s.t. $f_{X+Y}(x,y)>0$. So, when $0\le a\le 1$ e.g. $a_1$ the region of integration is bounded by the orange line "$a_1-a_1$" i.e. $$\{y>0,x>0,y\le a_1-x\}$$
On the other hand, when $a\ge 1$ e.g. $a_2$ the region of integration is the intersection of the region bounded by the line "$a_2-a_2$" and the shadowed one i.e.
$$\{x>0,y>0,y\le a_2-x,y<1-x/2\}$$