I am struggling with the following question:
Let $X \sim Beta(a,1)$ and $Y \sim Beta(a,1)$ and $Z=XY$. Assume $a>b>0$. What is the probability density function of the random variable Z?
Hint: $$A_z=\{ (x,y):xy\le z \}=\{ (x,y): 0 \le x \le z \} \cup \{ (x,y) : y \le \frac{z}{x} , z \le x \le 1\}$$
First, I am assuming that the hint is specifying the range of x and y, but I am not sure how this is useful.
Second, I was able to find a solution online, but it did not have any explanation and so it really wasn't much help in understanding how to solve this problem.
Third, the teaching assistant for the class said that I should try to find the CDF first. I have made the following attempt...
$$ P(Z \le z) = P(XY \le z) = P(0 \le x \le z) + P(Y \le \frac{z}{x}, z \le x \le 1) $$ Now, if we assume that these are independent, then we have that
$$ P(Y \le \frac{z}{x}, z \le x \le 1) = P(Y \le \frac{z}{x})*P( z \le x \le 1) $$
Here, we notice that $f_X(x)=ax^{a-1}$ and $f_Y(y)=by^{b-1}$. Thus, we have
$$ P(Z \le z) = \int_0^z ax^{a-1} dx + \biggl( \int_0^{\frac{z}{x}} by^{b-1} dy \biggl) \biggl( \int_z^1 ax^{a-1} dx \biggl) $$
At this point, I am not sure if my work is correct or if I'm even on the right track. Any help would be highly appreciated! Thanks in advance.