This is a follow up of sorts relating to the question posted here:
Determine the PDF of Z = XY when the joint pdf of X and Y is given.
I want to check to make sure I have some kind of understanding or that my reasoning is correct concerning this problem.
Given $f_{X,Y}(x,y) = 2(1-x)$ for $x,y \in [0,1]$ and $f_{X,Y}(x,y)=0$ otherwise. Find the pdf of $Z=XY$.
So, I choose the method of finding the cdf of $Z$ and then taking its derivative to get the pdf of $Z$.
So, $F_Z(z)=0$ for $z \leq 0$ because in this case $F_Z(z)=\mathrm{Pr}(Z \leq z) = \mathrm{Pr}(Z \leq z \leq 0) = \mathrm{Pr}(Z \leq 0) = \mathrm{Pr}(XY \leq 0)$ $ = \mathrm{Pr}(XY = 0) + \mathrm{Pr}(XY < 0) = 0 + 0 = 0.$
Then for $z \geq 1$, $F_Z(z)=1$ since if $z \geq 1$ then $F_Z(z) = \mathrm{Pr}(Z \leq z) = \mathrm{Pr}(Z \leq 1 \leq z) = \mathrm{Pr}(Z \leq 1) + \mathrm{Pr}(1 \leq Z \leq z)$ $= \mathrm{Pr}(XY \leq 1) + 0 = 1$ since $\mathrm{Pr}(XY \leq 1)$ covers $[0,1] \times [0,1]$ which is all the area of $\mathbb{R}^2$ that has non-zero probability.
From here I pick a $z \in (0,1)$ then go through with the integration over the region. I do get after this process that $F_Z(z) = z^2-2z\mathrm{ln}(z)$ as was shown to be the CDF in the original question.
I understand this may be quite grotesque what I have written. Does this make sense what I have written?
Thank you and sorry if it doesn't make sense.