The $F(b,d)-F(b,c)-F(a,d)+F(a,c)\geqslant 0$ condition of joint CDFs

Question

I'm trying to verify that a certain function of two variables $F(x,y)$ satisfies the conditions of a joint CDF. Showing that each condition holds has been fairly straightforward except, that is, for the condition that

$a<b,c<d\implies F(b,d)-F(b,c)-F(a,d)+F(a,c)\geqslant 0$.

I honestly don't know where to start for this one. It's easy enough when the task is to show that this condition is violated but showing that it holds is another matter.

For reference, the particular bivariate function I'm trying to show satisfies this condition is the following (a modification a function Jordan M. Stoyanov examines in Section 5.6 of his book Counterexamples in Probability, 3rd Edition):

Any tips about how I might proceed would be appreciated.

Answer 1

This is also called the 2-increasing property and if your function is continuously twice differentiable, you can equivalently check that

$$\frac{\partial^2 F}{\partial x \partial y} \geq 0 \; .$$

However, your function is not on the transitions from one "block" of definition to another. For those, you can just check as @Satana suggests in the comments by considering all possible combinations of elements from each block. For example, take $0 \leq a < 1$, $1 \leq b < \infty$, $0 \leq c < 1/2$ and $1/2 \leq d < 1$, then

$$F(b,d)-F(b,c)-F(a,d)+F(a,c) = d - c - a/2 + ac \\ = d - c - a(1/2-c) \geq (1/2 - c)(1-a) \geq 0 \text{, etc...}$$

Answer 2

The condition is the same as checking $\frac{F(b,d)-F(b,c)}{F(a,d)-F(a,c)}\ge 1$. Geometrically this is the slope of any cross sectional slice parallel to the xz-plane is greater if the x-value is greater.

Here are some cases that are worth considering.

$$\frac 12<d<1\\ 0<c<\frac 12\\ b>1\\ 0<a<1$$

Gives $\frac{d-c}{\frac a2-ac}\ge1$? $\frac{d-c}{a\left(\frac 12 -c\right)}\ge1$, so this is true.

Another case that is worth considering is $0<c<d<\frac 12,b>1,0<a<1$, in which case we have $\frac{d-c}{ad-ac}\ge 1$? $\frac 1a\ge 1$, so this is true.

Last case here is $0<c<d<\frac 12, 0<a<b<1$, in which case we have $\frac{bd-bc}{ad-ac}\ge1$? $\frac ba\ge1$, so this is true.

A picture of the cdf. The slope/vertical distance between any two points that are further right on the x-axis must be at least as great as the slope/vertical distance between the same two points shifted to a lesser x-value.