Cumulative Distribution Functions For Loss Severity

Why the CDF Is Useful for Loss Severity

A loss amount is uncertain, so a useful severity model must describe more than the possible values of a loss. It must also say how probability accumulates as we move from smaller losses to larger ones. The cumulative distribution function, or CDF, gives that running total. It is useful for answering questions such as whether a loss falls within a specified range, and whether a particular amount has positive probability.

For a hypothetical insurance coverage, let $X$ denote the nonnegative loss amount, measured in dollars. The CDF applies whether losses take only specified values, vary over an interval, or have a combination of these behaviors. This tutorial focuses on reading the CDF itself and using it to calculate probabilities. The methods are aligned with the SOA Exam FAM (Fundamentals of Actuarial Mathematics) syllabus.

Definition: The cumulative distribution function of a random variable $X$ is $F_X(x)=P(X\le x)$. At each real number $x$, it gives the probability that the loss is no greater than $x$. The subscript is often omitted when the random variable is clear, so $F(x)$ means $F_X(x)$.

The inequality in the definition is “less than or equal to.” That detail determines how endpoints are handled: a loss exactly equal to $x$ is included in $F(x)$. In particular, $F(100)$ is the probability of a loss of at most $100, not strictly less than $100.

Basic CDF Behavior and Its Limits

Every CDF has three basic properties. It is nondecreasing, it is right-continuous, and its limiting values at the two ends of the real line are 0 and 1. These properties are useful both for interpreting a valid CDF and for checking a proposed one.

Theorem: For a CDF $F$, if $x_1\le x_2$, then $F(x_1)\le F(x_2)$; $F$ is right-continuous; and $\lim_{x\to-\infty}F(x)=0$, while $\lim_{x\to+\infty}F(x)=1$. For a nonnegative loss $X$, $F(x)=0$ for $x<0$. The value $F(0)$ may be positive if a zero loss has positive probability.

Nondecreasing means accumulated probability can never fall as the loss threshold increases. Right-continuity means that the CDF value at $x$ agrees with the values approached from the right. It does not require the CDF to be continuous from the left. In fact, a CDF can jump at a loss amount that has positive probability.

A flat section means no probability is accumulated as the threshold moves across that section. A rising section means probability is accumulated there. A jump means a positive amount of probability is assigned to that particular loss value. The CDF is therefore a cumulative record of probability across the severity values, not simply a list of possible loss amounts.

Interval Probabilities and Endpoints

Subtracting two CDF values isolates probability between the thresholds. For $a<b$, the probability of a loss strictly greater than $a$ and no greater than $b$ is $F(b)-F(a)$. The lower endpoint is excluded because $F(a)$ already includes all losses equal to $a$.

P(a<X\le b)=F(b)-F(a)

Other interval conventions require a different value at an endpoint. Write $F(c-)$ for the left-hand limit of the CDF at $c$, that is, the probability accumulated just below $c$. Then the endpoint formulas are:

\begin{aligned} P(a\le X\le b)&=F(b)-F(a-),\\ P(a<X<b)&=F(b-)-F(a),\\ P(a\le X<b)&=F(b-)-F(a-). \end{aligned}

The probability at a single value $c$ is the size of the jump there. This follows by subtracting the probability strictly below $c$ from the probability at or below $c$.

$$ P(X=c)=F(c)-F(c-) $$

If the CDF has no jump at $c$, then $F(c)=F(c-)$ and $P(X=c)=0$. If it jumps, the jump size—not the value of the CDF itself—is the probability of an exact loss of $c$.

Worked Examples

Worked Example: Reading a Discrete Loss CDF

A hypothetical device policy has a nonnegative loss $X$ with the following severity model: $X$ equals $0 with probability 0.10, $500 with probability 0.25, $1,000 with probability 0.40, and $2,000 with probability 0.25. Find (a) $F(750)$, (b) $P(500<X\le 2{,}000)$, and (c) $P(X=500)$. State how the CDF changes at each possible loss.

Method and assumption. The loss takes only the four stated values, with probabilities that sum to $0.10+0.25+0.40+0.25=1.00$. Accumulate the probabilities of values no greater than the threshold.

Step 1: Find the CDF at $750. The values no greater than $750 are $0 and $500. Therefore $F(750)=0.10+0.25=0.35$.

Step 2: Calculate the interval probability. Since the interval is $500<X\le2{,}000$, the $500 outcome is excluded and the $1,000 and $2,000 outcomes are included. Equivalently, $F(2{,}000)-F(500)=1.00-(0.10+0.25)=1.00-0.35=0.65$.

Step 3: Find the probability of exactly $500. Immediately below $500, the CDF has accumulated only the probability at $0, so $F(500-)=0.10$. At $500, $F(500)=0.10+0.25=0.35$. The jump is $0.35-0.10=0.25$, giving $P(X=500)=0.25$.

The cumulative values are 0 below $0, 0.10 from $0 up to (but not including) $500, 0.35 from $500 up to (but not including) $1,000, 0.75 from $1,000 up to (but not including) $2,000, and 1.00 from $2,000 onward. Each increase equals the probability at that loss. Check: the interval result is also $0.40+0.25=0.65$, matching the subtraction.

Common mistake avoided: Using $F(500)=0.35$ as the probability of exactly $500 would include the $0 outcome as well. The point probability is the jump $F(500)-F(500-)$, not the cumulative value $F(500)$.

Worked Example: Calculating Probabilities From a Piecewise CDF

A hypothetical warranty loss $X$, in dollars, has the following CDF:

F(x)= \begin{cases} 0, & x<0,\\ 0.20, & 0\le x<100,\\ 0.20+0.004(x-100), & 100\le x<300,\\ 1, & x\ge300. \end{cases}

Find (a) $P(0\le X\le150)$, (b) $P(100<X\le250)$, and (c) $P(X=100)$. Assume the stated function describes the entire severity distribution.

Step 1: Check the relevant CDF values. At $x=0$, $F(0)=0.20$, while $F(0-)=0$. Thus there is a jump of 0.20 at zero. At $x=100$, the middle expression gives $F(100)=0.20+0.004(100-100)=0.20$, so there is no jump at $100. At the upper end of the rising portion, $F(300-)=0.20+0.004(300-100)=1.00$, consistent with $F(300)=1$.

Step 2: Include the zero endpoint for part (a). The probability from 0 through $150 is $F(150)-F(0-)$. Calculate $F(150)=0.20+0.004(150-100)=0.20+0.20=0.40$. Hence $P(0\le X\le150)=0.40-0=0.40$.

Step 3: Exclude the lower endpoint for part (b). Calculate $F(250)=0.20+0.004(250-100)=0.20+0.60=0.80$. Since $F(100)=0.20$, $P(100<X\le250)=0.80-0.20=0.60$.

Step 4: Check for a jump at $100. The left-hand value is $F(100-)=0.20$, and the value at $100 is $F(100)=0.20$. Therefore $P(X=100)=0.20-0.20=0$.

Check: Part (a) includes the 0.20 probability at zero and a further 0.20 accumulated between $100 and $150, for a total of 0.40. Part (b) covers a rise of 0.60. Common mistake avoided: treating the CDF value at $100 as a point probability would incorrectly assign probability 0.20 to exactly $100. The left-hand limit shows that the CDF does not jump there.

Worked Example: Separating a Jump From a Flat Section

A hypothetical repair-cost severity $X$, measured in dollars, is specified by

F(x)= \begin{cases} 0, & x<0,\\ 0.20+0.004x, & 0\le x<50,\\ 0.60, & 50\le x<100,\\ 0.60+0.004(x-100), & 100\le x<200,\\ 1, & x\ge200. \end{cases}

Find (a) the probability of a loss exactly $50, (b) $P(25<X\le125)$, and (c) $P(50<X\le100)$. Interpret the flat section.

Step 1: Find the values at and just below $50. The left-hand limit is $F(50-)=0.20+0.004(50)=0.40$. The value at $50 is $F(50)=0.60$. Therefore the jump is $0.60-0.40=0.20$, so $P(X=50)=0.20$.

Step 2: Calculate the probability over the first interval. Since $25$ is in the rising section, $F(25)=0.20+0.004(25)=0.20+0.10=0.30$. Since $125$ is in the later rising section, $F(125)=0.60+0.004(125-100)=0.60+0.10=0.70$. Thus $P(25<X\le125)=0.70-0.30=0.40$.

Step 3: Apply the lower-endpoint convention for part (c). The requested interval excludes $50 and includes $100. The CDF has $F(100)=0.60$, so $P(50<X\le100)=F(100)-F(50)=0.60-0.60=0$.

The CDF is flat from $50 through values below $100. Its value does not increase over that range, so no probability is accumulated for losses strictly greater than $50 and less than or equal to $100. The jump at $50 is separate: it assigns probability to exactly $50. Check: the jump at $50 is 0.20, while the interval probability in part (c) is 0; these answers refer to distinct sets of outcomes.

Common mistake avoided: Subtracting $F(50-)$ instead of $F(50)$ in part (c) would include the loss exactly equal to $50, producing 0.20 for the wrong interval.

A Short Justification for the Interval Formula

For $a<b$, the event $X\le b$ consists of two disjoint events: $X\le a$, and $a<X\le b$. Their probabilities therefore add:

P(X\le b)=P(X\le a)+P(a<X\le b).

Rearranging and using $F(x)=P(X\le x)$ gives

P(a<X\le b)=P(X\le b)-P(X\le a)=F(b)-F(a).

This argument explains why the lower endpoint is excluded in this formula. To include $a$, subtract the probability strictly below $a$, which is $F(a-)$, instead. The formulas follow from the events being counted; they are not interchangeable shortcuts.

Common CDF Checks and Exam Traps

Subtracting the wrong lower endpoint. $F(b)-F(a)$ gives $P(a<X\le b)$, not $P(a\le X\le b)$. For the latter, use $F(b)-F(a-)$.
Confusing a CDF value with a point probability. $F(c)$ includes every loss up to and including $c$. The probability of exactly $c$ is $F(c)-F(c-)$.
Assuming every CDF is smooth. A valid CDF can jump. Right-continuity does not rule out jumps; it specifies which value the CDF takes at the jump.
Overlooking flat sections. A flat CDF section accumulates no probability as the threshold moves through that section. A jump at its boundary can still assign probability to that boundary value.
Accepting a decreasing or incorrectly bounded function. A CDF cannot decrease, fall below 0, exceed 1, or have limits other than 0 and 1 at negative and positive infinity, respectively.

When reading a severity CDF, first identify which piece contains each threshold. Then evaluate the CDF at the required endpoints, using a left-hand limit wherever an endpoint must be excluded or a point probability must be isolated. Finally, check that the resulting probability is between 0 and 1 and that the interval convention matches the event in the question.

Key takeaway: The CDF $F(x)=P(X\le x)$ records accumulated probability. Differences of CDF values give interval probabilities, while the size of a jump gives the probability at one exact loss amount. Endpoint notation determines which formula to use.

Check Your Understanding

For each hypothetical loss model, identify the relevant CDF values or limits before calculating. The answers are provided at the end of each question.

A severity $X$ equals $0, $200, $500, or $900 with probabilities 0.15, 0.25, 0.35, and 0.25, respectively. Find $P(200<X\le900)$ and $P(X=200)$. (Answer: $0.60$ and $0.25$)
A nonnegative loss has $F(x)=0$ for $x<0$, $F(x)=0.10+0.01x$ for $0\le x<20$, $F(x)=0.50$ for $20\le x<60$, and $F(x)=1$ for $x\ge60$. Find $P(0\le X\le20)$ and $P(X=20)$. (Answer: $0.50$ and $0.20$)
A loss is $0, $100, $250, or $400 with probabilities 0.10, 0.20, 0.30, and 0.40. Find $P(100\le X<400)$. (Answer: $0.50$)
A CDF has $F(30)=0.42$, $F(30-)=0.30$, and $F(80-)=0.75$. Find $P(30<X<80)$ and $P(X=30)$. (Answer: $0.33$ and $0.12$)
A proposed CDF is $G(x)=0$ for $x<0$, $G(x)=0.40$ for $0\le x\le10$, $G(x)=0.70$ for $10<x\le20$, and $G(x)=1$ for $x>20$. Explain why it is invalid, then give the correct CDF for a loss that equals $0, $10, and $20 with probabilities 0.40, 0.30, and 0.30. (Answer: $G$ is not right-continuous at 10; the correct CDF is 0 for $x<0$, 0.40 for $0\le x<10$, 0.70 for $10\le x<20$, and 1 for $x\ge20$)

What You'll Learn