Calculating Interval Probabilities from a CDF

Why Interval Endpoints Matter

A severity question often asks for the probability that a loss falls between two specified amounts. The calculation depends not only on the two amounts, but also on whether the endpoints are included. If the severity distribution has a point mass at an endpoint, changing a strict inequality to an inclusive one can change the answer. A reliable method is to translate each endpoint condition into the appropriate CDF value or left limit before subtracting.

This method applies to discrete, continuous, and mixed severity distributions. It is aligned with the SOA Exam ASTAM (Advanced Short-Term Actuarial Mathematics) syllabus.

Definition: The cumulative distribution function of a loss $X$ is $F_X(x)=P(X\leq x)$. Its left limit at $c$ is $F_X(c-)=\lim_{x\uparrow c}F_X(x)=P(X<c)$. The probability mass at exactly $c$ is therefore $P(X=c)=F_X(c)-F_X(c-)$.

The distinction is important: $F_X(c)$ includes a loss equal to $c$, while $F_X(c-)$ excludes it. For a continuous distribution, the two values are equal at every point, so endpoint inclusion does not affect interval probabilities. In a distribution with point masses, they may differ.

Interval Formulas From the CDF

For two distinct values $a<b$, the probability for any interval with endpoints $a$ and $b$ can be found by subtracting the cumulative probability below the interval from the cumulative probability at or below its upper edge. The endpoint convention determines which version of the CDF is used.

Theorem: For $a<b$, $$ \begin{aligned} P(a<X\leq b)&=F_X(b)-F_X(a),\\ P(a\leq X\leq b)&=F_X(b)-F_X(a-),\\ P(a<X<b)&=F_X(b-)-F_X(a),\\ P(a\leq X<b)&=F_X(b-)-F_X(a-). \end{aligned} $$

A useful way to remember the rule is to focus on the two edges separately. At the upper edge, use $F_X(b)$ if $b$ is included and $F_X(b-)$ if it is excluded. At the lower edge, subtract $F_X(a)$ if $a$ is excluded and subtract $F_X(a-)$ if it is included.

Read the lower endpoint.
If the interval includes $a$, start by excluding values strictly below $a$, which is $F_X(a-)$. If it excludes $a$, remove values at or below $a$, which is $F_X(a)$.

Read the upper endpoint.
Use $F_X(b)$ when the interval includes $b$, and $F_X(b-)$ when it excludes $b$.

Subtract the lower cumulative probability.
The resulting difference is the probability in the requested interval.

Why the Subtraction Rules Work

The formulas follow from differences between nested events. For example, when $a<b$, every outcome with $X\leq a$ is also an outcome with $X\leq b$. Removing the former event from the latter leaves exactly the outcomes satisfying $a<X\leq b$:

\{a<X\leq b\}=\{X\leq b\}\setminus\{X\leq a\}, \qquad P(a<X\leq b)=F_X(b)-F_X(a).

For an open upper edge, the event at that edge must also be removed. Since $F_X(b-)=P(X<b)$, the event $\{a<X<b\}$ is $\{X<b\}\setminus\{X\leq a\}$, giving $F_X(b-)-F_X(a)$. For a closed lower edge, the values strictly below $a$, rather than the values at or below $a$, are removed. Thus $P(a\leq X\leq b)=F_X(b)-F_X(a-)$. Applying both changes gives the formula for $[a,b)$.

When $a=b$, the four formulas above are not all used as written, because they were stated for $a<b$. The interval $[a,a]$ contains only the value $a$, so $P(a\leq X\leq a)=F_X(a)-F_X(a-)$. The intervals $(a,a]$, $[a,a)$, and $(a,a)$ are empty, so each has probability zero. In particular, $P(a\leq X<a)=0$.

Worked Example: Four Endpoint Conventions for a Discrete Loss

A hypothetical warranty loss $X$, measured in dollars, has the following severity distribution: $P(X=0)=0.10$, $P(X=100)=0.20$, $P(X=200)=0.45$, and $P(X=500)=0.25$. There is no interest or mortality assumption in this single-loss probability calculation. Find the probabilities for the four intervals with endpoints 100 and 500.

Method. First calculate the CDF values needed at each endpoint. The cumulative probability at or below 100 is $F_X(100)=P(X=0)+P(X=100)=0.10+0.20=0.30$. The left limit at 100 is $F_X(100-)=P(X=0)=0.10$. At 500, $F_X(500)=1.00$, because all possible losses are at most 500. Its left limit is $F_X(500-)=0.10+0.20+0.45=0.75$.

For $100<X\leq500$, use the CDF at both endpoints: $F_X(500)-F_X(100)=1.00-0.30=0.70$.

For $100\leq X\leq500$, include 100 and 500: $F_X(500)-F_X(100-)=1.00-0.10=0.90$.

For $100<X<500$, exclude both endpoints: $F_X(500-)-F_X(100)=0.75-0.30=0.45$.

For $100\leq X<500$, include 100 but exclude 500: $F_X(500-)-F_X(100-)=0.75-0.10=0.65$.

Check. The last probability includes the losses of 100 and 200, whose probabilities sum to $0.20+0.45=0.65$. The strictly interior interval contains only the loss of 200, with probability $0.45$. These direct checks agree with the CDF calculations. The common mistake to avoid is treating $F_X(100)$ and $F_X(100-)$ as interchangeable: they differ by the mass $0.20$ at 100.

Using CDF Summaries When the Full Distribution Is Not Given

A problem may provide only CDF values and left limits at the relevant thresholds. That is sufficient: interval probabilities depend on the cumulative probability at the interval edges, not on a complete description of the distribution between them. Before calculating, confirm that the reported values are consistent with a nondecreasing CDF and that each left limit is no larger than the CDF value at the same point.

Worked Example: Reading CDF Values and Left Limits

For a hypothetical equipment claim severity $X$, a model summary gives $F_X(300-)=0.22$, $F_X(300)=0.34$, $F_X(800-)=0.71$, and $F_X(800)=0.79$. Find the probability for each of the four interval conventions from 300 to 800. This is a single-loss severity calculation; interest and mortality are not involved.

The mass at 300 is $0.34-0.22=0.12$. The mass at 800 is $0.79-0.71=0.08$. These differences confirm why endpoint convention matters.

For $300<X\leq800$, the answer is $F_X(800)-F_X(300)=0.79-0.34=0.45$.

For $300\leq X<800$, it is $F_X(800-)-F_X(300-)=0.71-0.22=0.49$.

For $300<X<800$, it is $F_X(800-)-F_X(300)=0.71-0.34=0.37$.

For $300\leq X\leq800$, it is $F_X(800)-F_X(300-)=0.79-0.22=0.57$.

Check. The closed interval includes both endpoint masses, so its probability exceeds the strictly interior probability by $0.12+0.08=0.20$. Indeed, $0.57-0.37=0.20$. This check also confirms that the endpoint masses have been counted in the correct intervals.

Combining Disjoint Intervals

Some questions ask for a probability over more than one interval. Calculate each interval using its own endpoint conventions, then add the probabilities if the events are disjoint. Take care that adjacent ranges do not both include a shared endpoint; otherwise that endpoint’s mass would be counted twice.

Worked Example: Two Separate Severity Bands

A hypothetical cargo-loss model has CDF information $F_X(200-)=0.18$, $F_X(200)=0.26$, $F_X(900-)=0.82$, $F_X(900)=0.90$, $F_X(1200-)=0.96$, and $F_X(1200)=0.98$. Find the probability that a loss is either in the band $200\leq X<900$ or in the band $900\leq X\leq1200$. There is no interest or mortality assumption in this severity calculation.

First band. The lower endpoint is included and the upper endpoint is excluded, so $P(200\leq X<900)=F_X(900-)-F_X(200-)=0.82-0.18=0.64$.

Second band. The lower endpoint is included and the upper endpoint is included, so $P(900\leq X\leq1200)=F_X(1200)-F_X(900-)=0.98-0.82=0.16$.

Add the disjoint probabilities. The first band excludes 900 and the second includes it, so the events do not overlap. Their total probability is $0.64+0.16=0.80$.

Check. The two bands together form $200\leq X\leq1200$. The direct CDF calculation is $F_X(1200)-F_X(200-)=0.98-0.18=0.80$, which agrees. A common error would be to use $F_X(900)$ as the upper cumulative value for the first band even though that band excludes 900; doing so would count the mass $0.90-0.82=0.08$ at 900 in the first band and count it again in the second.

Common Endpoint Errors

Using $F_X(c)$ for a strict upper boundary. Because $F_X(c)$ includes $X=c$, use $F_X(c-)$ when the interval requires $X<c$.
Using $F_X(c-)$ for an excluded lower boundary. For an interval that starts with $X>c$, subtract $F_X(c)$, which removes all outcomes at or below $c$.
Assuming endpoints never matter. That is valid for continuous distributions, but not generally when there are point masses. When uncertain, calculate the mass $F_X(c)-F_X(c-)$.
Applying an interval formula when the endpoints are equal. Handle a singleton separately: $P(X=a)=F_X(a)-F_X(a-)$. An interval such as $[a,a)$ is empty and has probability zero.
Counting a shared endpoint twice. When adding adjacent ranges, assign a shared boundary to just one range, or explicitly account for its point mass.

Key takeaway: At an included upper endpoint use $F_X$; at an excluded upper endpoint use its left limit. At an included lower endpoint subtract the left limit; at an excluded lower endpoint subtract the CDF value. This rule handles discrete, continuous, and mixed severity distributions.

Check Your Understanding

For each problem, use the stated CDF values and left limits, paying attention to whether each endpoint is included.

A severity model has $F_X(40)=0.31$ and $F_X(120-)=0.76$. Find $P(40<X<120)$. (Answer: 0.4500)
A model gives $F_X(75-)=0.24$ and $F_X(75)=0.33$. What is the probability of a loss exactly equal to 75? (Answer: 0.0900)
Suppose $F_X(150)=0.21$ and $F_X(500-)=0.68$. Find $P(150<X<500)$. (Answer: 0.4700)
Suppose $F_X(100-)=0.12$, $F_X(100)=0.20$, $F_X(400-)=0.73$, and $F_X(400)=0.81$. Find $P(100\leq X\leq400)$. (Answer: 0.6900)
A hypothetical severity distribution has $F_X(250)=0.29$, $F_X(600)=0.66$, $F_X(900-)=0.78$, and $F_X(1400-)=0.95$. Find $P(250<X\leq600\text{ or }900\leq X<1400)$. (Answer: 0.5400)

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