Rectangle Probabilities from CDF Increments

Why Four CDF Values Determine a Rectangle

A joint cumulative distribution function records probabilities for regions extending from the lower left of the plane. A rectangle is bounded on all four sides, so its probability is not usually a single CDF value. Instead, four corner values combine to retain the probability inside the rectangle while removing the regions outside it. This is useful whenever a problem supplies a joint CDF, particularly when a discrete variable can put positive probability directly on a boundary.

Definition: For random variables $X$ and $Y$, their joint cumulative distribution function is $$ F(x,y)=P(X\le x,\ Y\le y). $$ It gives the probability that both variables are at most their specified thresholds. Define the left-limit notation by $F(x-,y)=\lim_{t\uparrow x}F(t,y)$, $F(x,y-)=\lim_{t\uparrow y}F(x,t)$, and $F(x-,y-)=\lim_{s\uparrow x,\ t\uparrow y}F(s,t)$. These limits include values strictly below the relevant threshold rather than values equal to it.

Start with the half-open rectangle $(a,b]\times(c,d]$. The CDF value $F(b,d)$ includes all outcomes with $X\le b$ and $Y\le d$. Subtracting $F(a,d)$ removes outcomes with $X\le a$, while subtracting $F(b,c)$ removes outcomes with $Y\le c$. The lower-left region $X\le a,\ Y\le c$ was subtracted twice, so it must be added back once.

P(a<X\le b,\ c<Y\le d) =F(b,d)-F(a,d)-F(b,c)+F(a,c).

The four terms are an inclusion-exclusion calculation. Their signs are always $+,-,-,+$: add the top-right corner, subtract the two adjacent strips, and add back their overlapping lower-left region. This sign pattern is the same whether the distribution is discrete or continuous. What changes with discreteness is which threshold values or left limits belong in the four terms.

Endpoint Conventions: Values Versus Left Limits

The CDF is defined using “less than or equal to.” Consequently, $F(a,d)$ includes probability with $X=a$, whereas $F(a-,d)$ does not. That distinction determines whether a boundary is included in the rectangle. To keep the endpoint choices systematic, use the following rule separately for $X$ and $Y$.

Endpoint rule: At an upper endpoint, use the endpoint itself when the interval includes it, and its left limit when the interval excludes it. At a lower endpoint, use the endpoint itself when the interval excludes it, and its left limit when the interval includes it. Substitute the resulting thresholds into the four-corner increment, with signs $+,-,-,+$.

For example, an interval $[a,b]$ includes its lower endpoint, so the lower subtraction uses $a-$; it also includes its upper endpoint, so the upper evaluation uses $b$. An interval $(a,b)$ excludes both endpoints, so the lower subtraction uses $a$ and the upper evaluation uses $b-$. The same decisions are made independently for the $Y$-interval.

Interval for $X$	Lower threshold used in subtraction	Upper threshold used in evaluation
$(a,b]$	$a$	$b$
$[a,b]$	$a-$	$b$
$(a,b)$	$a$	$b-$
$[a,b)$	$a-$	$b-$

The table applies to $Y$ as well. In particular, for $[a,b]\times[c,d]$, the increment is $F(b,d)-F(a-,d)-F(b,c-)+F(a-,c-)$. Left limits are not a special correction applied only to an upper or lower side; they are simply the CDF thresholds required to include or exclude boundary values correctly.

Worked Example: A Rectangle from Discrete CDF Corners

A hypothetical joint model for two integer-valued measurements has the following joint probability table. Find the probability that $0<X\le2$ and $0<Y\le1$. Use the CDF increment and check it against the included table cells.

$X$	$Y=0$	$Y=1$
0	0.10	0.15
1	0.20	0.10
2	0.15	0.30

Method. Both intervals are open at the lower end and closed at the upper end, so use the ordinary half-open formula. The four needed CDF values are:

$F(2,1)=1.00$, because all table outcomes satisfy $X\le2,\ Y\le1$. Also, $F(0,1)=0.10+0.15=0.25$, $F(2,0)=0.10+0.20+0.15=0.45$, and $F(0,0)=0.10$.

\begin{aligned} P(0<X\le2,\ 0<Y\le1) &=F(2,1)-F(0,1)-F(2,0)+F(0,0)\\ &=1.00-0.25-0.45+0.10\\ &=0.40. \end{aligned}

Check. The rectangle includes the cells $(X,Y)=(1,1)$ and $(2,1)$, with probabilities $0.10$ and $0.30$. Their sum is $0.40$, matching the increment. The lower boundaries $X=0$ and $Y=0$ are excluded, so their probability must not remain in the answer. A common mistake is to use only $F(2,1)-F(0,1)$; that removes the $X\le0$ region but leaves outcomes with $Y=0$. The other subtraction and the added overlap are essential.

Deriving the Four-Corner Increment

The formula follows by partitioning the lower-left region at the rectangle’s two lower boundaries. Let $A=\{X\le b,\ Y\le d\}$, $B=\{X\le a,\ Y\le d\}$, and $C=\{X\le b,\ Y\le c\}$. Within $A$, the outcomes outside $(a,b]\times(c,d]$ are those in $B$ or $C$. The part common to both is $\{X\le a,\ Y\le c\}$. Inclusion-exclusion therefore gives:

\begin{aligned} P((a,b]\times(c,d]) &=P(A)-P(B\cup C)\\ &=P(A)-P(B)-P(C)+P(B\cap C)\\ &=F(b,d)-F(a,d)-F(b,c)+F(a,c). \end{aligned}

This argument also explains why the corner signs cannot be changed arbitrarily: $F(a,d)$ and $F(b,c)$ each remove one unwanted strip, and $F(a,c)$ restores the overlap removed twice. Replacing any threshold by a left limit changes which boundary outcomes belong to that region, but the inclusion-exclusion reasoning and signs remain the same.

Worked Example: Continuous CDF and Different Endpoint Choices

Suppose a hypothetical pair of measurements has the joint CDF $F(x,y)=\min(x,1)\min(y,1)$ for $x\ge0,\ y\ge0$, and $F(x,y)=0$ if either coordinate is negative. Find $P(0.20<X\le0.70,\ 0.10<Y\le0.80)$. Then explain what would change if the four endpoints were included or excluded differently.

Method. The requested intervals are half-open, so evaluate the CDF at the four endpoints. All four thresholds lie between 0 and 1, where $F(x,y)=xy$:

$F(0.70,0.80)=0.70(0.80)=0.56$; $F(0.20,0.80)=0.20(0.80)=0.16$; $F(0.70,0.10)=0.70(0.10)=0.07$; and $F(0.20,0.10)=0.20(0.10)=0.02$.

\begin{aligned} P(0.20<X\le0.70,\ 0.10<Y\le0.80) &=0.56-0.16-0.07+0.02\\ &=0.35. \end{aligned}

Endpoint check. This model has probability zero on each boundary line, so including or excluding any of these endpoints does not change the result. For a discrete model, the same change of endpoint convention can change the probability: replacing $F(a,d)$ by $F(a-,d)$, for instance, retains probability with $X=a$. The mistake to avoid is assuming that endpoint conventions never matter just because they do not affect this continuous example.

Worked Example: Including a Lower Boundary and Excluding an Upper Boundary

A hypothetical discrete model has the joint probability table below. Find $P(1\le X<3,\ 0<Y\le2)$. Use the endpoint rule and verify the result by adding the relevant cells.

$X$	$Y=0$	$Y=1$	$Y=2$
1	0.05	0.10	0.05
2	0.10	0.20	0.10
3	0.10	0.15	0.15

Method. The $X$-interval includes its lower endpoint $1$, so its lower subtraction uses $1-$; it excludes its upper endpoint $3$, so its upper evaluation uses $3-$. The $Y$-interval is $(0,2]$: use $0$ for the lower subtraction and $2$ for the upper evaluation.

From the table, $F(3-,2)=0.60$, the sum of the first two $X$-rows. There are no $X$-values below 1, so $F(1-,2)=0$. Also, $F(3-,0)=0.05+0.10=0.15$, and $F(1-,0)=0$. Thus:

\begin{aligned} P(1\le X<3,\ 0<Y\le2) &=F(3-,2)-F(1-,2)-F(3-,0)+F(1-,0)\\ &=0.60-0-0.15+0\\ &=0.45. \end{aligned}

Check. The included cells have $X=1$ or $X=2$ and $Y=1$ or $Y=2$. Their probabilities sum to $0.10+0.05+0.20+0.10=0.45$. Using $F(3,2)$ instead of $F(3-,2)$ would incorrectly include the $X=3$ row, which is excluded by the strict upper bound.

When the Formula Matters

A rectangle probability can represent a pair of simultaneous threshold conditions: for example, a measurement falling within a specified range while another measurement also lies within its own range. A joint CDF provides the probability of each lower-left region; the increment isolates the range of interest. The method applies to discrete, continuous, and other joint distributions, provided the CDF values or required one-sided limits are available.

A quick check is to confirm that the four terms produce a number between 0 and 1 and that the result is not negative. If the event is a subset of a larger region whose probability is known, its probability cannot exceed that larger probability. For discrete distributions, an additional check is often possible by summing the included point probabilities. If that direct sum disagrees with the increment, first recheck the endpoints and then the signs.

For $(a,b]\times(c,d]$, use $F(b,d)-F(a,d)-F(b,c)+F(a,c)$.
Use a left limit at an upper endpoint that is excluded.
Use a left limit at a lower endpoint that is included.
When boundary probability is zero, changing endpoint inclusion does not change the probability; when there are atoms, it may.

Common exam trap: Applying the half-open formula automatically to a closed or open rectangle can omit or include boundary mass. Before calculating, write the event’s endpoint symbols explicitly, choose each threshold or left limit using the endpoint rule, and only then apply the $+,-,-,+$ corner signs.

Check Your Understanding

For each problem, choose the correct corner thresholds before calculating. Answers are given at the end of each question.

A joint CDF has $F(4,7)=0.82$, $F(1,7)=0.31$, $F(4,2)=0.26$, and $F(1,2)=0.08$. Find $P(1<X\le4,\ 2<Y\le7)$. (Answer: 0.3300)
For a discrete pair, $F(5,3)=0.91$, $F(2-,3)=0.28$, $F(5,0)=0.22$, and $F(2-,0)=0.07$. Find $P(2\le X\le5,\ 0<Y\le3)$. (Answer: 0.4800)
For a pair with $F(x,y)=xy$ on the unit square and zero probability on boundary lines, find $P(0.15<X<0.65,\ 0.25\le Y\le0.90)$. (Answer: 0.3250)
A discrete joint table has probabilities $P(X=1,Y=0)=0.12$, $P(X=1,Y=1)=0.18$, $P(X=2,Y=0)=0.20$, and $P(X=2,Y=1)=0.50$. Find $P(1\le X<2,\ 0\le Y\le1)$. (Answer: 0.3000)
A discrete model has $F(3-,4-)=0.54$, $F(1,4-)=0.18$, $F(3-,2)=0.21$, and $F(1,2)=0.06$. Find $P(1<X<3,\ 2<Y<4)$. (Answer: 0.2100)

Interval for \(X\)	Lower threshold used in subtraction	Upper threshold used in evaluation
\((a,b]\)	\(a\)	\(b\)
\([a,b]\)	\(a-\)	\(b\)
\((a,b)\)	\(a\)	\(b-\)
\([a,b)\)	\(a-\)	\(b-\)

What You'll Learn