# 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?

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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 $x_2 = y_2 = \infty$.

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You mean the random vector $(X,Y)$ would only take the value $(0,0)$, right? – user191919 Jan 16 '14 at 18:35
Yes. I've fixed it. – Albert Steppi Jan 16 '14 at 18:44
Thanks. I can't see why the left hand side is equal to the right hand side of the new equation you put in... – user191919 Jan 16 '14 at 19:04