I am analyzing the following question and answer.
The question is:
Let $X$ and $Y$ be two independent random variables with pdfs $f_X(x) = \frac{3}{4}(1-x^2)$ for $-1 \leq x \leq 1$ ($0$ otherwise) and $f_Y(y) = \frac{1}{3}$ if $0\leq y \leq 1$ and $f_Y(y)=\frac{2}{3}$ if $2\leq y \leq 3$ ($0$ otherwise).
Let $Z= X + Y$. Determine $f_Z(z)$.
And the answer is given by the convolution integral:
$f_Z(z) = \int_{-\infty}^{\infty}f_X(t)f_Y(z-t)dt$.
Now, I understand the idea behind the convolution of two random variables. But I am struggling with a systematic way to figure out the solutions to these types of problems.
In particular, I have no problem determining the maximum and minimum of $Z$, but I am having trouble breaking the range of $Z$ into subcases that can then be integrated over. I am going to list the subcases given in the solution. If someone could help/advise in how to figure out the upper and lower limits of integration in each subcase, and how to methodically ensure that subcases have been exhausted. That would be much appreciated.
*NOTE: I am not going to write the evaluated integrals in terms of $z$, I just need assistance figuring out a methodical way to actually set these integrals up and get each subcase in turn. I can compute them from there.
Here are the cases given in the solution:
For $z \in [-1,0]$:
$$f_Z(z) = \int_{-1}^z \frac{1}{3}\cdot\frac{3}{4}(1-t^2)dt$$
For $z \in [0,1]$:
$$f_Z(z) = \int_{z-1}^z \frac{1}{3}\cdot\frac{3}{4}(1-t^2)dt$$
For $z \in [1,2]$:
$$f_Z(z) = \int_{z-1}^1 \frac{1}{3}\cdot\frac{3}{4}(1-t^2)dt + \int_{-1}^{z-2} \frac{2}{3}\cdot\frac{3}{4}(1-t^2)dt$$
For $z \in [2,3]$:
$$f_Z(z) = \int_{z-3}^{z-2} \frac{2}{3}\cdot\frac{3}{4}(1-t^2)dt$$
For $z \in [3,4]$:
$$f_Z(z) = \int_{z-3}^{1} \frac{2}{3}\cdot\frac{3}{4}(1-t^2)dt$$.
I guess I am also wondering if a typical problem such as this is difficult, or more so tedious. I think right now I am finding problems of this type difficult, because determining the bounds and exhausting cases seems to be very tricky.