3

Say we have two random variables, x and y, with $F_{xy}(x,y)$ and $f_{xy}(x,y)$ denoting their joint CDF and PDF respectively. If they can be written such that

$$F_{xy}(x,y)=G(x)H(y)$$ $$f_{xy}(x,y)=g(x)h(y)$$

is that enough to guarantee x and y are independent? My intuition suggests that it should be possible to construct a counterexample (either where none of the possible $g$/$G$ and $h$/$H$ functions are the marginal statistics or where $F$ might be factorable while $f$ is not), but so far I've had no luck. I've also had a hard time proving that it is always the case if it is in fact true.

IPoiler
  • 155

1 Answers1

4

It guarantees independence.

Let $x_0,y_0\in\mathbb R$ such that $F_{X,Y}(x_0,y_0)>0$ and observe that:

$$0<F_{X}\left(x_0\right)=\lim_{y\to\infty}F_{X,Y}\left(x_0,y\right)=G\left(x_0\right)\lim_{y\to\infty}H\left(y\right)$$

This reveals that $h:=\lim_{y\to\infty}H\left(y\right)$ exists.

Likewise it can be proved that $g:=\lim_{x\to\infty}G\left(x\right)$ exists.

This with: $$1=\lim_{z\to\infty}F_{X,Y}(z,z)=\lim_{z\to\infty}G(z)H(z)=gh$$

Now observe that for an arbitrary $x\in\mathbb R$: $$F_{X}\left(x\right)=\lim_{y\to\infty}F_{X,Y}\left(x,y\right)=G\left(x\right)\lim_{y\to\infty}H\left(y\right)=hG(x)$$

Likewise we find $F_Y(y)=gH(y)$ so we end up with:$$F_{X,Y}(x,y)=G(x)H(y)=[hG(x)][gH(y)]=F_X(x)F_Y(y)$$ showing independence.

Equation $f_{X,Y}(x,y)=g(x)h(y)$ leads directly to $F_{X,Y}(x,y)=G(x)H(y)$ where $G\left(x\right):=\int_{-\infty}^{x}g\left(u\right)du$ and $H\left(y\right):=\int_{-\infty}^{y}h\left(v\right)dv$. So we are back in the CDF case.

drhab
  • 151,093