Legitimacy of a Bivariate Distribution Function I am asked to explain why the following function is not a legitimate (multivariate) distribution function.
$$
F(x,y) = 1 - e^{-x-y}, x,y \geq 0
$$
I am tempted to reason as follows:
The function has right continuity on each of its variables, so its not that. Monotonicity doesnt pose a problem either. The function also varies between $0$ and $1$, so basically what remains out of the basic properties is the limit one, which says:
$$
\lim_{x_{1},...,x_{n}\rightarrow+\infty}F(x_{1},...,x_{n})=1
$$
Maybe I am missing something on this multivariate limit, but I suppose it works as well for the given cumulative function. What I thought I could argue is that we know the distribution function by FTC would satisfy:
$$
\frac{\partial^2F}{\partial x\, \partial y} = f_{X,Y}(x,y)
$$
However:
$$
\frac{\partial^2F}{\partial x\, \partial y} = - e^{-x-y} \leq 0
$$
Since the joint density cannot assume negative values. Is that it? Would there be another reason for it not being legitimate?
 A: Suppose $F$ is the joint cumulative distribution function of the random variables $X$ and $Y$, so that $P(X \leq x, Y \leq y) = F(x,y)$.
Fix $x = 0$ and consider
$$\lim_{y\rightarrow \infty} F(0,y) = \lim_{y\rightarrow \infty}1 - e^{-y} = 1$$
For $x_{0} < 0$
$$\lim_{y\rightarrow \infty}F(x_{0}, y) = 0$$
follows from $F(x,y)$ being equal to zero outside the first quadrant.
It follows that $P(X = 0) = 1$.
A similar argument shows that $P(Y = 0) = 1$.
But then the random vector $(X,Y)$ can only take the value $(0,0)$, which is inconsistent with our alleged cumulative distribution function. Hence $F$ can't really be a cumulative distribution function after all.
Another way to see it is that all bivariate joint cumulative distribution functions must satisfy the property
For all $x_1 < x_2 \in \mathbb{R}$ and $y_1 < y_2 \in \mathbb{R}$
$$F(x_2,y_2) - F(x_2,y_1) - F(x_1, y_2) + F(x_1,y_1) = P(x_1 <  X \leq x_2, y_2 < Y \leq y_2) \geq 0$$
To see why the left hand side gives the probability in the middle first consider the region $\left\{(x,y) : x \leq x_2, y \leq y_2\right\}$. $F(x_2, y_2)$ gives the probability $(X,Y)$ lies in this region. $F(x_1, y_2) = P(X \leq x_1, y \leq y_2)$, while $F(x_2, y_1) = P(X \leq x_2, y \leq y_1)$. If we subtract by $F(x_1, y_2)$ and $F(x_2, y_1)$, we almost have $P(x_1 < X \leq x_2, y_1 < Y \leq Y_2)$, except we've subtracted out the probability for the region $\left\{(x,y) : x \leq x_1, y \leq y_1\right\}$ twice. The probability $(X,Y)$ lies in this last region is $F(x_1, y_1)$, adding $F(x_1, y_1)$ back in to make up for the double subtraction we did gives us 
$$P(x_1 < X \leq x_2, y_1 < Y \leq y_2) = F(x_2, y_2) - F(x_1, y_2) - F(x_2, y_1) + F(x_1, y_1)$$
It can easily be seen that $F(x,y) = 1 - e^{-x-y}$ doesn't satisfy this property by choosing $x_1 = y_1 = 0$ and $x_2 = y_2 = \infty$.
A: It suffices (unter the assumption of existence) to check that the "would-be" bivariate pdf  
$$\frac{\partial^2}{\partial x \partial y} F_{X,Y}(x,y)=-e^{-x-y}$$
(for any $x,y>0$) violates the positivity condition for a pdf.
See as well a similar question : Joint exponential distribution seems ok until I take derivitive
