Rectangle Increments and Nonnegative Probability

Why Four CDF Values Describe a Rectangle

A bivariate CDF records probabilities of lower-left regions: the value at \((x,y)\) is the probability that both coordinates are at most their respective thresholds. To obtain the probability of a bounded rectangle, we need to remove the parts below its lower edges. This is why the calculation uses four corner values, with alternating signs.

The endpoint convention matters. The increment corresponds to a rectangle open at its lower edges and closed at its upper edges. This convention works whether or not \(X\) or \(Y\) has positive probability at a boundary. No continuity assumption is needed for the identity.

The Rectangle-Increment Identity

The identity follows by writing the rectangle event in terms of four lower-left events. The indicator argument gives both the exact probability formula and its sign.

Proof. For any outcome, let \(I_{u,v}\) be the indicator of the event \(\{X\leq u,\;Y\leq v\}\). The alternating combination of indicators is \(I_{b,d}-I_{a,d}-I_{b,c}+I_{a,c}\). We check its value by considering whether each coordinate lies in the indicated ranges.

If \(a<X\leq b\) and \(c<Y\leq d\), the first indicator is 1 and the other three are 0, so the combination is 1. If \(X\leq a\) and \(c<Y\leq d\), the first two indicators are 1 and cancel. If \(a<X\leq b\) and \(Y\leq c\), the first and third are 1 and cancel. If either \(X>b\) or \(Y>d\), all indicators are 0. Outcomes below the lower thresholds also cancel or contribute zero. Thus, for every outcome, the combination equals the indicator of \(\{a<X\leq b,\;c<Y\leq d\}\). Taking expectations gives the claimed identity, since the expectation of an event indicator is its probability. Probabilities are nonnegative, proving the inequality. \(\square\)

The theorem is sometimes called the nonnegative-increment condition for a bivariate CDF. It says more than that the function should increase when either coordinate increases: it also constrains how its changes in the two coordinates fit together. A function can pass simple one-coordinate monotonicity checks and still have a negative rectangle increment.

Worked Calculations

Additivity Across Adjacent Rectangles

The increment also behaves as a rectangle probability should when a rectangle is split into adjacent pieces. This provides a useful calculation check: the probability of the whole rectangle must equal the sum of the probabilities of its pieces.

Proof. Write \(\Delta F((u,v]\times(c,d])=F(v,d)-F(u,d)-F(v,c)+F(u,c)\). Adding the increments for the two smaller rectangles gives

The terms involving \(e\) cancel, leaving exactly the increment for the whole rectangle. Equivalently, the two smaller half-open rectangles are disjoint and their union is the larger rectangle, so the same conclusion follows from additivity of probability. \(\square\)

Repeated splitting gives the same conclusion for any finite grid partition: the increment over the whole rectangle is the sum of all cell increments. Since each cell increment must be nonnegative, such sums cannot be smaller than zero.

Using Nonnegative Increments to Check a Candidate

When a formula is proposed as a bivariate CDF, choose thresholds \(a<b\) and \(c<d\), evaluate its four corner values, and compute the alternating increment. A negative result disproves the proposal. The failed example above illustrates why this check is essential: its product term contributes a positive increment, but the added term contributes a larger negative one on the selected rectangle.

Passing this test is necessary, but by itself it does not prove that a function is a CDF. A full verification must also check the other defining requirements of a bivariate CDF, including its appropriate continuity and boundary behavior. The increment test is a particularly direct way to detect an impossible candidate because it has a concrete interpretation: every checked value must equal a probability.

Common Pitfall

A frequent error is to use only three corners, or to give the lower-left corner a positive sign. The lower-left term is essential: it corrects for subtracting the region below both lower thresholds twice. Another error is to treat \((a,b]\) as interchangeable with \([a,b]\). When boundary values can carry probability, changing the endpoints changes the event; the stated four-corner formula specifically gives \((a,b]\times(c,d]\).

A related mistake is to check only that \(F\) increases as \(x\) or \(y\) increases. Those one-coordinate checks cannot detect every failure. Always test the alternating increment itself: for every \(a<b\) and \(c<d\), it must be nonnegative.