X,Y univariate random variable with $F_{X,Y}(x,y)=G_1(x)G_2(y)$: are they independent? Let $X:\Omega\to\mathbb{R}$ and $Y:\Omega\to\mathbb{R}$ be univariate random variables with CDF $F_{X,Y}(x,y)$ such that:
$$
F_{X,Y}(x,y)=G_1(x)G_2(y),\forall (x,y)\in\mathbb{R}\times\mathbb{R}
$$
where $G_1:\mathbb{R}\to\mathbb{R}$, $G_2:\mathbb{R}\to\mathbb{R}$ are known functions.
Question: Is it true that $X$ and $Y$ are independent RVs?
Can anyone give me some hints? 
I tried to:
$$
F_X(x)=\lim_{y\to\infty}F_{X,Y}(x,y)=\lim_{y\to\infty}G_1(x)G_2(y)=G_1(x)\cdot\lim_{y\to\infty}G_2(y)
$$
but I don't know why (or if) $\lim_{y\to\infty}G_2(y)=1$.
 A: Yes, it's true that these assumptions imply $X$ and $Y$ are independent.
Simplify the notation by writing $F = F_{X,Y}$.  By definition,
$$F(x,y) = \Pr(X \le x, Y \le y).$$
Therefore the limit of $F(x,y)$ as $y$ increases without bound exists and is the chance that $X$ does not exceed $x$:
$$F_X(x) = \Pr(X \le x) = \lim_{y\to\infty} F(x,y) = G_1(x) \lim_{y\to\infty} G_2(y).$$
Choosing any $x$ for which $F_X(x)\ne 0$ shows $G_2^\infty = \lim_{y\to\infty}G_2(y)$ is nonzero.  (Such an $x$ must exist by the law of total probability, which asserts $\lim_{x\to\infty}F_X(x)=1$.)  Thus
$$G_1(x) = \frac{F_X(x)}{G_2^\infty}$$
for all $x$.  Exchanging the roles of $X$ and $Y$ and using analogous notation,
$$G_2(y) = \frac{F_Y(y)}{G_1^\infty}$$
for all $y$.  Taking the joint limit as both $x$ and $y$ grow without bound shows
$$1 = \lim_{x,y\to\infty} F(x,y) = G_1^\infty G_2^\infty.$$
Therefore
$$F(x,y) = G_1(x)G_2(y) = \frac{F_X(x)F_Y(y)}{G_1^\infty G_2^\infty} = F_X(x)F_Y(y),$$
demonstrating $X$ and $Y$ are independent.
