I came across this question. Essentially, the answer says that if you can write a joint pdf in the form $p(x,y)=p_1(x)p_2(y)$, then the random variables are independent of each other. My question is this:
Given a pdf: $$p(x,y)=4e^{-2(x+y)}$$ How can we find $p_1$ and $p_2$? There are multiple ways of factoring this into the form $p_1(x)p_2(y)$. How can we know that $p_1(x)=2e^{-2x}$ for instance?
I suppose one way of checking whether or not we have the correct $p_1$ and $p_2$ is to check whether or not are normalizable, but that still seems to be a trial and error way of going about it.
In this particular case, you can see that x and y are perfectly exchangeable in the equation $\forall (x,y), \ p(x,y) = p(y,x)$
Therefore they must have the same pdf. Therefore $p(x,x)=p(x)^2$.