For random variables $X$ and $Y$, $F_{X,Y}(x,y)=F_X(x)F_Y(y)$ if and only if $f_{X,Y}(x,y)=f_X(x)f_Y(y)$

Question

I have seen either statement used as a definition of independent random variables. I was trying to prove their equivalence for discrete random variables. I am able to prove that if the joint density function (or mass function as some books call it for the discrete case) factors into individual densities then the joint distribution function factors as well, by writing the events $X \leq x$ and $Y \leq y$ as (potentially infinite) disjoint unions of events and using the basic axioms of probability. However, I am not able to see how to prove the converse? Can someone give a hint for that?

Answer 1

I'll try to get you started in the case of continuous random variables $X,Y$. Recall that (assuming things are nice) $$\frac{\partial^2{F_{X,Y}}}{\partial x \partial y} = f_{X,Y}$$ And if we assume the distribution splits we have: $$\frac{\partial^2 F_{X,Y}}{\partial x \partial y} = \frac{\partial^2}{\partial x \partial y} \left(F_X(x)F_Y(y)\right)$$

Try continuing from there (note that $F_X(x)$ is independent of y).