Joint Distribution Functions And Rectangle Probabilities

Why Paired-Loss Probabilities Need a Joint CDF

When an insurer tracks two losses together, a probability statement may concern both at once: for example, the chance that one loss is within a specified range while another also falls within its own range. The relevant object is the joint cumulative distribution function. Its values describe probabilities for pairs of threshold events, and differences among four such values give probabilities over rectangular regions.

Definition: For random losses $X$ and $Y$, their joint cumulative distribution function is $$ F_{X,Y}(x,y)=P(X\le x,\;Y\le y). $$ It gives the probability that both losses are at or below their respective thresholds. A joint CDF must be nondecreasing in each argument, and its values lie between 0 and 1.

The notation matters: $F_{X,Y}(x,y)$ is a probability involving both losses, not the probability that either one is below its threshold. For instance, $F_{X,Y}(x,y)$ describes the lower-left region extending up to $(x,y)$ in the plane of paired loss outcomes.

Suppose $X$ and $Y$ are paired losses under a hypothetical insurance portfolio. A rectangle probability might ask for $a<X\le b$ and $c<Y\le d$. The standard four-corner calculation uses the joint CDF at the rectangle’s upper-right, lower-right, upper-left, and lower-left corners. The specified endpoints are important: the formula below corresponds to intervals open on the lower end and closed on the upper end.

The Four-Corner Increment Formula

Theorem: For real numbers $a<b$ and $c<d$, $$ P(a<X\le b,\;c<Y\le d) =F_{X,Y}(b,d)-F_{X,Y}(a,d)-F_{X,Y}(b,c)+F_{X,Y}(a,c). $$

The signs alternate. Start with the probability below both upper thresholds, remove the two portions below a lower threshold, then add back the overlap that was removed twice. The formula applies to any joint distribution, including one with point masses; no assumption of continuity or independence is required.

Why the Formula Holds

The event $X\le b,\;Y\le d$ contains the desired rectangle, but also contains outcomes where $X\le a$ or $Y\le c$. Subtracting the event $X\le a,\;Y\le d$ removes the first unwanted part. Subtracting $X\le b,\;Y\le c$ removes the second. Their overlap, $X\le a,\;Y\le c$, has then been subtracted twice, so it must be added back once. In indicator notation, this is the identity

\mathbf{1}_{\{a<X\le b,\;c<Y\le d\}} = \bigl(\mathbf{1}_{\{X\le b\}}-\mathbf{1}_{\{X\le a\}}\bigr) \bigl(\mathbf{1}_{\{Y\le d\}}-\mathbf{1}_{\{Y\le c\}}\bigr).

Taking expectations gives the probability formula: the expectation of an event’s indicator is its probability, and each resulting joint-event probability is a joint-CDF value. This also explains why the lower-left corner has a plus sign.

Worked Example: A Rectangle From Four Joint-CDF Values

A hypothetical insurer models two losses, $X$ and $Y$, in dollars. Its valid joint CDF gives $F_{X,Y}(300,200)=0.62$, $F_{X,Y}(100,200)=0.38$, $F_{X,Y}(300,50)=0.44$, and $F_{X,Y}(100,50)=0.25$. Find the probability that $100<X\le300$ and $50<Y\le200$.

Method. Match the requested region to the four-corner formula, with $a=100$, $b=300$, $c=50$, and $d=200$.

Substitute the upper-right value and subtract the two lower-threshold values:

$$ 0.62-0.38-0.44=-0.20. $$

Add back the lower-left overlap:

P(100<X\le300,\;50<Y\le200) =0.62-0.38-0.44+0.25 =0.05.

The requested probability is $0.0500$, or 5.00%. As a check, the four-corner increment is nonnegative and is no greater than 1. The common mistake is to stop after the two subtractions; the overlap must be added back, and the intermediate value $-0.20$ is not the rectangle probability.

Reading CDF Values and Checking the Result

In an applied calculation, the joint CDF values may come from a specified probability model, a table, or a validated model output. First identify each corner from the inequalities, then substitute the values with their signs. Keep the variables and threshold units aligned: if losses are reported in thousands of dollars, a threshold of 250 means $250,000, not $250.

A rectangle probability must be between 0 and 1. In particular, the four-corner increment cannot be negative when the input values genuinely come from a valid joint CDF. A negative result usually signals a sign error, a mismatched corner, or inconsistent input values. The individual CDF values themselves should also be nondecreasing as either threshold increases.

Worked Example: Thresholds Reported in Thousands

For a hypothetical insurer, $X$ is a property loss and $Y$ is a business-interruption loss. Both are measured in thousands of dollars. A valid joint-CDF lookup gives $F_{X,Y}(600,400)=0.73$, $F_{X,Y}(250,400)=0.49$, $F_{X,Y}(600,100)=0.38$, and $F_{X,Y}(250,100)=0.22$. Find the probability that $X$ is more than $250,000 and at most $600,000, while $Y$ is more than $100,000 and at most $400,000.

Step 1: Align the thresholds and units. Since the model is expressed in thousands, the dollar thresholds correspond to $a=250$, $b=600$, $c=100$, and $d=400$ in the model’s units.

Step 2: Apply the rectangle increment.

\begin{aligned} P(250<X\le600,\;100<Y\le400) &=F_{X,Y}(600,400)-F_{X,Y}(250,400)\\ &\quad-F_{X,Y}(600,100)+F_{X,Y}(250,100)\\ &=0.73-0.49-0.38+0.22\\ &=0.08. \end{aligned}

The probability is $0.0800$, or 8.00%. The corner values increase appropriately as either threshold increases, and the resulting increment is nonnegative. The common mistake this setup avoids is mixing dollars and thousands of dollars when looking up the CDF values.

Unbounded Rectangles and Endpoint Conventions

The same method applies when a rectangle extends without bound. Define $F_{X,Y}(\infty,d)$ as the limit of $F_{X,Y}(x,d)$ as $x$ tends to infinity; it represents $P(Y\le d)$. Likewise, $F_{X,Y}(b,\infty)=P(X\le b)$, and $F_{X,Y}(\infty,\infty)=1$. These limit values allow an infinite edge to be handled as the limit of a finite one.

Worked Example: A Rectangle With an Infinite Edge

For a hypothetical pair of insurance losses, the valid joint CDF and its limits give $F_{X,Y}(\infty,150)=0.82$, $F_{X,Y}(200,150)=0.55$, $F_{X,Y}(\infty,50)=0.46$, and $F_{X,Y}(200,50)=0.31$. Find the probability that $X>200$ and $50<Y\le150$.

Method. Use the rectangle increment with the upper bound for $X$ replaced by infinity:

\begin{aligned} P(X>200,\;50<Y\le150) &=F_{X,Y}(\infty,150)-F_{X,Y}(200,150)\\ &\quad-F_{X,Y}(\infty,50)+F_{X,Y}(200,50)\\ &=0.82-0.55-0.46+0.31\\ &=0.12. \end{aligned}

The probability is $0.1200$, or 12.00%. This calculation subtracts the part with $X\le200$ and the part with $Y\le50$, then restores their overlap. A common mistake is to treat $F_{X,Y}(\infty,150)$ as 1; it is $P(Y\le150)$, which here is 0.82.

Endpoint conventions cannot be ignored when a loss can have an atom, meaning positive probability at a particular value. For example, $F_{X,Y}(a,d)$ includes outcomes with $X=a$, so the standard formula for $(a,b]\times(c,d]$ excludes them. If the question includes a lower endpoint, use the appropriate left-hand CDF limit, such as $F_{X,Y}(a-,y)=\lim_{x\uparrow a}F_{X,Y}(x,y)$. For example:

P(a\le X\le b,\;c<Y\le d) =F_{X,Y}(b,d)-F_{X,Y}(a-,d)-F_{X,Y}(b,c)+F_{X,Y}(a-,c).

If the joint distribution is known to have no probability concentrated on the rectangle’s boundary, changing an endpoint from open to closed does not change the probability. Without that condition, use the requested endpoint convention rather than assuming boundary probabilities vanish.

Common Exam Traps

Using three terms instead of four. The lower-left corner restores the overlap removed twice. Leaving it out can even produce a negative result.
Putting a minus sign on every lower corner. The signs are plus, minus, minus, plus when read as upper-right, lower-right, upper-left, lower-left.
Confusing a joint event with an either-or event. $F_{X,Y}(x,y)$ means both thresholds are met, not at least one.
Ignoring whether the boundary is included. The formula stated here is for $(a,b]\times(c,d]$. Atoms can make other endpoint choices yield different probabilities.
Using inconsistent units or invalid CDF inputs. Check the scale of every threshold and confirm that a computed rectangle increment is nonnegative.

Key takeaway: For the rectangle $(a,b]\times(c,d]$, calculate the joint CDF at its four corners as upper-right minus lower-right minus upper-left plus lower-left. Check the endpoint convention and confirm the resulting probability is between 0 and 1.

Check Your Understanding

For each problem, use the stated joint-CDF values as values from a valid model, and apply the endpoint convention shown.

A hypothetical paired-loss model gives $F_{X,Y}(500,300)=0.68$, $F_{X,Y}(200,300)=0.41$, $F_{X,Y}(500,80)=0.39$, and $F_{X,Y}(200,80)=0.20$. Find $P(200<X\le500,\;80<Y\le300)$. (Answer: 0.0800)
For thresholds in thousands of dollars, the values are $F_{X,Y}(700,450)=0.91$, $F_{X,Y}(300,450)=0.64$, $F_{X,Y}(700,120)=0.52$, and $F_{X,Y}(300,120)=0.33$. Find the probability that the first loss is above 300 and at most 700 while the second is above 120 and at most 450. (Answer: 0.0800)
A joint CDF and its limit give $F_{X,Y}(\infty,80)=0.76$, $F_{X,Y}(300,80)=0.48$, $F_{X,Y}(\infty,20)=0.35$, and $F_{X,Y}(300,20)=0.18$. Find $P(X>300,\;20<Y\le80)$. (Answer: 0.1100)
A model provides $F_{X,Y}(400,250)=0.70$, $F_{X,Y}(100,250)=0.42$, $F_{X,Y}(400,0)=0.30$, and $F_{X,Y}(100,0)=0.12$. Find $P(100<X\le400,\;0<Y\le250)$. (Answer: 0.1000)
A valid model gives $F_{X,Y}(600,350)=0.84$, $F_{X,Y}(250,350)=0.57$, $F_{X,Y}(600,100)=0.46$, and $F_{X,Y}(250,100)=0.24$. Find $P(250<X\le600,\;100<Y\le350)$, and state whether the result passes the nonnegativity check. (Answer: 0.0500; yes)

What You'll Learn