Relating Tail Probabilities to the CDF

Introduction

In severity work, a tail probability measures the chance that a loss lies above a stated threshold. The wording matters: “exceeds $1,000” means $X>1{,}000$, while “is at least $1,000” means $X\geq1{,}000$. These events have the same probability only when there is no probability mass exactly at $1,000.

The cumulative distribution function and the survival function provide complementary ways to describe a severity distribution. Their relationship is straightforward for a strict exceedance, but an inclusive exceedance requires care at an atom. This distinction is useful whenever severity is discrete or can take specified values with positive probability.

Definition: For a severity random variable $X$, the cumulative distribution function is $F_X(x)=P(X\leq x)$. The survival function is $S_X(x)=P(X>x)$, the probability of a strict exceedance. The left limit $F_X(x^-)$ is the probability $P(X<x)$. Consequently, $P(X\geq x)$ is an inclusive exceedance probability.

Complement Relations

The events $\{X\leq x\}$ and $\{X>x\}$ are disjoint and together cover every possible value of $X$. Their probabilities therefore sum to 1. This gives the direct relationship between the CDF and survival function.

$$ S_X(x)=1-F_X(x) $$

This formula gives the probability of being strictly above $x$, not the probability of being at or above $x$. To obtain the inclusive tail, take the complement of $\{X<x\}$. The probability $P(X<x)$ is $F_X(x^-)$, so:

P(X\geq x)=1-F_X(x^-)

The difference between the inclusive and strict tails is the probability that $X$ equals the threshold. In terms of the CDF, this probability is the jump at $x$.

$$ P(X=x)=F_X(x)-F_X(x^-) $$

Subtracting the strict tail from the inclusive tail gives the same result:

P(X\geq x)-P(X>x) =\bigl(1-F_X(x^-)\bigr)-\bigl(1-F_X(x)\bigr) =F_X(x)-F_X(x^-) =P(X=x)

Key result: Use $1-F_X(x)$ for $P(X>x)$. Use $1-F_X(x^-)$ for $P(X\geq x)$. The two probabilities differ by $P(X=x)$, which may be positive.

Worked Tail Calculations

A reliable calculation starts by translating the wording into an event, then identifying the matching complement. If the threshold is a possible severity value, check whether probability is assigned exactly there before treating the strict and inclusive tails as equal.

Worked Example: Finding a Missing Probability and Both Tails

A hypothetical claim severity has the following probability table. The probability at $1,000 is $p$, and the probability of a claim of at least $1,000 is $0.7000$. Find $p$, $P(X>1{,}000)$, and $F_X(1{,}000)$. Verify that the table defines a probability distribution.

Method. Translate “at least $1,000” as an inclusive tail, use the stated tail condition to determine $p$, and then check the total probability before calculating the strict tail.

Severity $x$	$P(X=x)$
$0	0.1000
$400	0.2000
$1,000	$p$
$1,800	0.2500
$5,000	0.1500

The inclusive tail includes the probability at $1,000 and both probabilities above it:

P(X\geq1{,}000)=p+0.2500+0.1500=0.7000

Thus $p=0.7000-0.2500-0.1500=0.3000$. Check the table total:

$$ 0.1000+0.2000+0.3000+0.2500+0.1500=1.0000 $$

The strict tail excludes the mass at $1,000:

$$ P(X>1{,}000)=0.2500+0.1500=0.4000 $$

The CDF at $1,000$ includes the probabilities at $0$, $400$, and $1,000$:

F_X(1{,}000)=0.1000+0.2000+0.3000=0.6000

As a check, $1-F_X(1{,}000)=1-0.6000=0.4000$, which matches the strict tail. Also, $P(X\geq1{,}000)=0.4000+0.3000=0.7000$, as specified. The common mistake avoided here is omitting the threshold probability from an inclusive tail, or including it in a strict tail.

Worked Example: Using the CDF at a Severity Atom

A hypothetical severity distribution assigns probabilities $0.0800, 0.1700, 0.2500, 0.3000,$ and $0.2000$ to severities $0, $250, $700, $1,200, and $2,000, respectively. Calculate $F_X(700)$, $P(X>700)$, and $P(X\geq700)$. Explain why the two tail probabilities differ.

Method. Add all probabilities at or below the threshold for the CDF. For the strict tail, add only probabilities above it; for the inclusive tail, also include the probability at the threshold.

First, verify that the probability mass sums to 1:

$$ 0.0800+0.1700+0.2500+0.3000+0.2000=1.0000 $$

The CDF at $700$ includes the mass at $700$:

$$ F_X(700)=0.0800+0.1700+0.2500=0.5000 $$

The strict tail contains only severities of $1,200 and $2,000:

$$ P(X>700)=0.3000+0.2000=0.5000 $$

The inclusive tail also contains the $0.2500$ probability at $700:

P(X\geq700)=0.2500+0.3000+0.2000=0.7500

The complement checks give $1-F_X(700)=0.5000$ for the strict tail. Since $F_X(700^-)=0.0800+0.1700=0.2500$, the inclusive tail is $1-0.2500=0.7500$. The difference, $0.7500-0.5000=0.2500$, is exactly $P(X=700)$.

Why the Complement Formulas Hold

The formulas follow from partitions of the possible outcomes; they do not require a continuous severity model. For any fixed threshold $x$, every outcome belongs to exactly one of $\{X\leq x\}$ and $\{X>x\}$. Because these events are disjoint and exhaustive, their probabilities add to 1:

P(X\leq x)+P(X>x)=1

By definition, $P(X\leq x)=F_X(x)$ and $P(X>x)=S_X(x)$, proving $S_X(x)=1-F_X(x)$. For the inclusive tail, the disjoint exhaustive pair is instead $\{X<x\}$ and $\{X\geq x\}$. Since $P(X<x)=F_X(x^-)$, it follows that $P(X\geq x)=1-F_X(x^-)$. This argument also explains why replacing $F_X(x^-)$ with $F_X(x)$ in the inclusive-tail formula removes the atom at $x$.

Worked Example: Reading Strict and Inclusive Tails From CDF Values

For a hypothetical discrete severity, the only possible severities are $0, $500, $1,000, $1,500, and $2,500. Suppose $F_X(999)=0.3500$, $F_X(1{,}000)=0.5500$, $F_X(1{,}499)=0.5500$, and $F_X(1{,}500)=0.8000$. Also, $P(X>2{,}500)=0$. Find the probabilities at $1,000 and $1,500, the strict tail above $1,500, and the inclusive tail at $1,500.

Method. For a discrete severity with the stated support, a CDF increase across a possible value is the probability at that value. Use the CDF complement for the strict tail, then add the threshold mass for the inclusive tail.

The probability at $1,000 is the increase from just below $1,000 to $1,000:

P(X=1{,}000)=F_X(1{,}000)-F_X(999)=0.5500-0.3500=0.2000

There is no possible severity between $1,000 and $1,500, consistent with $F_X(1{,}499)=F_X(1{,}000)=0.5500$. The probability at $1,500 is:

P(X=1{,}500)=F_X(1{,}500)-F_X(1{,}499)=0.8000-0.5500=0.2500

The strict tail above $1,500$ is:

P(X>1{,}500)=1-F_X(1{,}500)=1-0.8000=0.2000

Therefore the inclusive tail is:

P(X\geq1{,}500)=P(X=1{,}500)+P(X>1{,}500)=0.2500+0.2000=0.4500

The results are consistent: the probability below $1,500 is $0.5500$, and the inclusive tail is $0.4500$, giving a total of $1.0000$. The common mistake avoided is reading $1-F_X(1{,}500)$ as the inclusive tail; it is the strict tail.

Common Exam Trap

The most frequent error is to apply $1-F_X(x)$ to an event written as $X\geq x$. That subtraction is correct only for $X>x$, because $F_X(x)$ already includes the probability at $x$. If there is an atom at the threshold, the error understates the inclusive tail by exactly $P(X=x)$.

Exam check: Before calculating a tail, underline the inequality in the event. For $>$, use $1-F_X(x)$. For $\geq$, use $1-F_X(x^-)$, or calculate the strict tail and add $P(X=x)$. The formulas coincide only when $P(X=x)=0$.

This distinction is aligned with the SOA Exam ASTAM (Advanced Short-Term Actuarial Mathematics) syllabus because severity calculations may require precise interpretation of a threshold event. A sound final check is to verify that the strict tail equals $1-F_X(x)$, and that the inclusive tail exceeds it by the threshold mass, never by a different amount.

Check Your Understanding

For each problem, identify whether the requested event is strict or inclusive before calculating.

A hypothetical severity distribution has probabilities $0.1500$ at $0, $0.2500$ at $600, $0.3500$ at $1,400, and $0.2500$ at $3,000. Find $F_X(1{,}400)$, $P(X>1{,}400)$, and $P(X\geq1{,}400)$. (Answer: $F_X(1{,}400)=0.7500$, $P(X>1{,}400)=0.2500$, $P(X\geq1{,}400)=0.6000$.)
For a hypothetical severity, $F_X(800^-)=0.4200$ and $F_X(800)=0.5700$. Find $P(X=800)$ and $P(X\geq800)$. (Answer: $P(X=800)=0.1500$, $P(X\geq800)=0.5800$.)
A hypothetical loss model has $P(X>2{,}000)=0.1800$ and $P(X=2{,}000)=0.0700$. Find $F_X(2{,}000)$ and $P(X\geq2{,}000)$. (Answer: $F_X(2{,}000)=0.8200$, $P(X\geq2{,}000)=0.2500$.)
A hypothetical discrete severity has possible values $0, $300, $900, and $1,700, with $F_X(300)=0.4000$, $F_X(900^-)=0.4000$, $F_X(900)=0.6500$, and $F_X(1{,}700)=1.0000$. Find $P(X=900)$, $P(X>900)$, and $P(X\geq900)$. (Answer: $P(X=900)=0.2500$, $P(X>900)=0.3500$, $P(X\geq900)=0.6000$.)
At a threshold $c$, a hypothetical severity model has $P(X<c)=0.3100$, $P(X=c)=0.2400$, and $P(X>c)=0.4500$. Calculate $F_X(c)$, $S_X(c)$, and $P(X\geq c)$, and verify that the probabilities are consistent. (Answer: $F_X(c)=0.5500$, $S_X(c)=0.4500$, $P(X\geq c)=0.6900$; the three event probabilities sum to $1.0000$.)

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