Checking the validity of Bivariate Distribution function

Question

Which of the following is /are bivariate Distribution function:

1. $F(x,y)=\begin{cases} 0 & \text{ if } x+y<0\\ 1 & \text{ if } x+y\geq0 \end{cases}$

2. $F(x,y)=\begin{cases} 1-e^{-2x} & \text{ if } x>0,y\geq1 \\ \frac{1-e^{-x}}{2} & \text{ if } x>0,0<y<1 \end{cases}$

My approach

All the conditions of Distribution function is fulfilled in the first one as $x\rightarrow \infty,y\rightarrow \infty$, then $F\rightarrow 1$ and when both values tends to $-\infty$, the distribution goes to zero.

But in the second, if we use these conditions, then the distribution is not defined for negative values of x and y. Then, how do we go about checking the conditions of Distribution function.

Any help?

EDIT.1 Also, if anyone can give me reference material for these type of problems, it would be really helpful.

Thanks

Answer 1

Help on 1)

Let $z\in\mathbb R$.

Now if $F$ is a indeed a distribution function and $P$ denotes the corresponding probability then $P((-\infty,z]\times(-\infty,-z])=F(z,-z)=1$

It can be shown however that $P((-\infty,z]\times(-\infty,-z]-\{(z,-z)\})=0$

This because $(-\infty,z]\times(-\infty,-z]-\{(z,-z)\}$ can be written as: $$\bigcup_{n=1}^{\infty}\left[(-\infty,z-\frac1n]\times(-\infty,-z]\cup(-\infty,z]\times(-\infty,-z-\frac1n]\right]$$

where $P((-\infty,z-\frac1n]\times(-\infty,-z])=F(z-\frac1n,-z)=0$ and $P((-\infty,z]\times(-\infty,-z-\frac1n])=F(z,-z-\frac1n)=0$.

This together leads to $P(\{(z,-z)\}=1$ for every $z\in\mathbb R$ which is absurd.

So we conclude that $F$ is not a distribution function.