Show that the function it is a cumulative distribution function of some random vector Show that the function
$$F(x,y)=\left\{\begin{array}{l}
(1-e^{-x})(1-e^{-y}) & \text {if} x \geqslant 0, y \geqslant 0\\
0 & \text {in another case}
\end{array}\right.$$
it is a cumulative distribution function of some random vector
Normally it is asked to calculate the distribution function, here how to show that it actually comes from a random vector
 A: First, whitout doing calculations, observe that your $F_{XY}$ is the joint CDF of two exponential iid with parameter 1.
But to solve the question, simply apply the properties of any cumulative bivariate CDF:
$$F_{XY}(-\infty;y)=0$$
$$F_{XY}(x;-\infty)=0$$
$$F_{XY}(x;+\infty)=F_X(x)$$
$$F_{XY}(+\infty;y)=F_Y(y)$$
$$F_{XY}(+\infty;+\infty)=1$$
A: To show that a function is a cdf for a one dimensional random variable you need to show the following things:-
Method 1:-
1:- $F$ is non decreasing
2:-$\lim_{x\to\infty}F(x)=1$
3:-$\lim_{x\to -\infty}F(x)=0$
4:-$F$ is right continuous at all points. That is for a fixed $x\in\mathbb{R}$ . If you have a sequence of real numbers $x_{n}\to x$ such that $x_{n}\geq x\forall n\in\mathbb{N}$. Then $\lim_{n\to\infty}F(x_{n})=F(x)$.
You can easily verify these properties for $F(x)= 1-e^{-x}$.
Now you use that if $Y$ is another random variable which is iid as $X$. Then their joint cdf will be the product of the individual cdf's . As you have verified that $F$ is a cdf. Hence you can conclude that the joint cdf found by multiplying is also a cdf.
Method 2:-
Otherwise you just show the analogues of these 4 conditions in 2 dimension.
1.:- $F(x,y)\leq F(x,y')$ if $y\leq y'$
$F(x,y)\leq F(x',y)$ if $x\leq x'$
2.$F_{X}(x)=\lim_{y\to\infty}F(x,y)$ . Where $F_{X}(\cdot)$ denotes the marginal cdf of $X$. Similarly $F_{Y}(y)=\lim_{x\to\infty}F(x,y)$.
3.$\lim_{x\to-\infty}F(x,y)=0$ and $\lim_{y\to-\infty}F(x,y)=0$
4.$\lim_{(x,y)\to(\infty,\infty)}F(x,y)=1$
5.If $x_{n}\geq x$ and $y_{n}\geq y$ be two sequences such that $x_{n}\to x$ and $y_{n}\to y$. Then $\lim_{n\to\infty}F(x_{n},y_{n})=F(x,y)$
6.$F(x',y')-F(x',y)-F(x,y')+F(x,y)=P(x<X\leq x' , y<Y\leq y')$ where $x'\geq x$ and $y'\geq y$.
The verification of 1-5 is mostly the same. The only slightly harder one is 6. Now you can go by either way.
These holds because of a theorem that a function satisfying these 6 properties must be a joint-cdf for some random variables. And for the 1 D case , any function satisfying the four proerties must be a cdf of some random variable.
