# Show that $F(x,y)=1$ for $x+y\ge 0$ and $F(x,y)=0$ otherwise does not define a joint CDF

Let $F(x,y)=1$ for $x+y\geq 0$ and be zero otherwise. Show that $F$ cannot possibly be the joint distribution function of a pair of random variables.

Ok so basically I need to show that there can't exist a function $f(x,y)$ such that $\int_{-\infty}^{\infty}\int_{x=-y}^{\infty}f(x,y)dxdy=1$ and that $\int_{-\infty}^{\infty}\int_{-\infty}^{x=-y}f(x,y)dxdy=0$, correct? I haven't explicitly found a counter-example to this (and I suppose if the claim is to be true then there shouldn't be one), but it seems plausible that such an $f(x,y)$ might exist, and playing around with these double integrals hasn't led to any contradictions.

So is there some other requirement on my probability density function $f(x,y)$, or is there some other requirement that the existence of random variables entails? To be honest, I find random variables a bit confusing and their role seems often superfluous and designed to confuse rather than clarify.

Thanks! =)

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Your reformulation is not correct. – GEdgar Jan 2 '13 at 20:47
Hint: What is the total probability that a joint distribution assigns to the entire space? What is the 'total probability' of this function? – Chris Janjigian Jan 2 '13 at 20:55
well 1 right??? – Thoth Jan 2 '13 at 20:56
Right. The set where $x + y \geq 0$ is a half plane, so the integral of this function is just the area of half of the plane. What is the area of half of the plane? – Chris Janjigian Jan 2 '13 at 20:57
Well the area of the half plane is infinite.. but I thought we were integrating a probability density function over the half plane, and this function may go to zero quickly in all directions right? – Thoth Jan 2 '13 at 21:01

So... Assume by way of contradiction that there exists some random variables $X$ and $Y$ such that $F(x,y)=\mathbb P(X\leqslant x,Y\leqslant y)$ for every $x$ and $y$, with $F(x,y)=\mathbf 1_{x+y\geqslant\color{red}{5}}$, say.
Since $\color{green}{6}+\color{blue}{2}\geqslant\color{red}{5}$, $F(\color{green}{6},\color{blue}{2})=F(\color{blue}{2},\color{green}{6})=1$, hence the events $[X\leqslant \color{green}{6},Y\leqslant\color{blue}{2}]$ and $[X\leqslant\color{blue}{2},Y\leqslant \color{green}{6}]$ both have full probability. Their intersection $[X\leqslant\color{blue}{2},Y\leqslant\color{blue}{2}]$ must have full probability as well, that is, $F(\color{blue}{2},\color{blue}{2})=1$. But $\color{blue}{2}+\color{blue}{2}\lt\color{red}{5}$ hence $F(\color{blue}{2},\color{blue}{2})=0$. This is absurd.
Note that this does not assume that the distribution of $(X,Y)$ has a density, since $F$ could a priori be the PDF of some discrete (or even worse...) distribution on the plane. – Did Jan 2 '13 at 21:13
@did I follow you up till $(-5) + (-5) < 2$. Do you mean $< 0$? – Calvin Lin Jan 2 '13 at 21:35
@CalvinLin No, I meant $\lt2$ because I had used $x+y\geqslant2$ to define $F$, to show there was nothing special with $x+y\geqslant0$. New version uses some "random" parameters, namely $(\color{red}{5},\color{green}{6},\color{blue}{2})$, to make the same idea still more apparent. – Did Jan 2 '13 at 22:29
@did Thanks! The colors helped a lot. I missed the $1_{x+y\geq 5}$ as I thought you were doing the original problem. – Calvin Lin Jan 2 '13 at 23:01