Why the marginal distribution function F_x(x) = F_xy(x,∞)?

19 Views Asked by At

I don't know if the next reasoning is all right.

If the joint distribution function $F_{xy} = (\infty,\infty)$ = $ P(-\infty<X<\infty,-\infty<Y<\infty)$. That's the probability of the region represented by a "rectangle", with sides of infinite length.

The probabiliy of the "base" of the "rectangle" is $P(-\infty<X<\infty) =1 $ and the "height" is equal to $P(-\infty<Y<\infty)$. So the probability of the region of the "rectangle" is $base*height=1*1 = 1$

If $F_{xy}=(x,\infty)$, the height "y" equals one and the base "x" is, $0<x<1$. This fact implies that $F_{xy} (x,\infty) = F_{x}(x)$.

Is this reasoning right? What are the other ways to see that $F_{x}(x)=F_{xy} (x,\infty)? $.