8
$\begingroup$

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) $\lim_{y\to\infty}G_2(y)=1$.

$\endgroup$
  • 2
    $\begingroup$ Does the relationship $F_{X,Y}(x,y)=G_1(x)G_2(y)$ hold for all $x$ and $y$ or just at a specific $(x,y)$? $\endgroup$ – Dilip Sarwate Oct 14 '15 at 22:17
  • 2
    $\begingroup$ Also, is $F_{X,Y}(x,y)$ the CDF? $\endgroup$ – Vimal Oct 14 '15 at 22:23
  • 1
    $\begingroup$ Are you trying to ask whether knowing how to factor the distribution function of a bivariate random variable $(X,Y)$ into a product of functions of $x$ and $y$ separately suffices to conclude $X$ and $Y$ are independent? $\endgroup$ – whuber Oct 14 '15 at 22:30
  • $\begingroup$ Sorry for the confusion, I'll edit the question now. $F_{X,Y}(x,y)$ is the CDF and the property holds for all $x,y$. $\endgroup$ – Guilherme Salomé Oct 14 '15 at 22:32
  • 1
    $\begingroup$ $\lim_{y\to\infty}G_2(y)=1$ doesn't have to be true. Consider $H_1(x) = G_1(x) * 0.5$ and $H_2(y) = G_2(y) * 2$ and consider that $F_{X,Y}(x,y)=H_1(x)H_2(y)$ but both $G_1$ and $H_1$ can't have a limit of 1. $\endgroup$ – bsdfish Oct 14 '15 at 22:50
7
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ The curious thing is that $G_1(\cdot)$ and $G_2(\cdot)$ both can be be negative-valued functions with, say, $G_1^\infty = -2$ and $G_2^\infty = -\frac 12$ and it will all still work out OK. $\endgroup$ – Dilip Sarwate Oct 15 '15 at 1:59
  • 2
    $\begingroup$ @DilipSarwate: it is not very curious in that, if $(G_1,G_2)$ satisfies the relation, so does $(-G_1,-G_2)$, so you can safely assume both $G_1$ and $G_2$ are positive-valued. Similarly, if $(G_1,G_2)$ satisfies the relation, so does $(\alpha G_1,\alpha^{-1} G_2)$, for any $\alpha\in\mathbb{R}^*$. $\endgroup$ – Xi'an Oct 15 '15 at 6:44
  • $\begingroup$ @Xi'an I do understand perfectly well. I just wanted to emphasize (since the OP was wondering how to show that $G_2(y)$ had limiting value $1$ as $y\to\infty$ meaning that he was wanting $G_1=F_X$ and $G_2=F_Y$) that the factorization $F_{X,Y}(x,y)=G_1(x)G_2(y)~\forall~x,y$ implies that $X$ and $Y$ are independent without it being necessarily true that $G_1(x) \geq 0, G_2(y) \geq 0$ for all $x, y$; $G_1(x) \leq 0, G_2(y) \leq 0$ for all $x, y$ works just as well. $\endgroup$ – Dilip Sarwate Oct 15 '15 at 15:20
  • $\begingroup$ @Dilip The $G_i$ could even have complex values if you like :-). $\endgroup$ – whuber Oct 15 '15 at 15:52
  • 1
    $\begingroup$ @KiranK. The question asked is "If a joint CDF $F_{X,Y}(x,y)$ can be expressed as $F_{X,Y}(x,y)=G_1(x)G_2(y)~\forall x,y$, then are $X$ and $Y$ independent?" to which the answer is Yes, and it requires a little bit of work to show this. It is not claimed that $G_1(x)$ and $G_2(y)$ are valid CDFs; if you insist on including this claim, the answer is trivially Yes because one of the definitions of independent RV is that the joint CDF factors into the product of the marginal CDFs. $\endgroup$ – Dilip Sarwate Oct 21 '15 at 14:35

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.