Deriving Exceedance Probabilities From Loss Distributions

From a Discrete Loss Distribution to Exceedance

A catastrophe model may report a set of possible losses and the probability assigned to each one. To answer a question such as “What is the probability the loss exceeds $50 million?”, add the probability of every outcome above $50 million. Outcomes equal to $50 million are not included when “exceeds” means strictly greater than the threshold.

This calculation turns a discrete loss distribution into an exceedance curve: a graph or table showing the probability of exceeding each loss threshold. It uses the same exceedance concept encountered in the earlier material on occurrence and aggregate exceedance probability curves, but here the emphasis is on deriving the curve directly from the probability masses.

Definition: Let $X$ be a discrete loss with possible values $x_1,\ldots,x_n$ and probabilities $p_j=P(X=x_j)$, where $p_j\geq 0$ and $\sum_{j=1}^{n}p_j=1$. The strict exceedance probability at threshold $x$ is $$ P(X>x)=\sum_{j:x_j>x}p_j. $$ The probability of meeting or exceeding the threshold is instead $P(X\geq x)=\sum_{j:x_j\geq x}p_j$. The two differ by the probability mass at $x$, if $x$ is a possible loss.

The loss distribution must be interpreted before calculating the tail. Confirm what the random variable represents, what units its values use, and whether the probabilities describe a complete distribution. For example, the outcomes might be annual aggregate losses or event-level losses; the summation rule is the same, but the interpretation of the resulting probability is not.

Why the Tail-Sum Formula Works

Each possible loss is either above the threshold or it is not. The event $X>x$ is therefore the union of the mutually exclusive outcomes $\{X=x_j\}$ for which $x_j>x$. Because those outcomes cannot occur simultaneously, their probabilities add.

\{X>x\}=\bigcup_{j:x_j>x}\{X=x_j\} \quad\Longrightarrow\quad P(X>x)=\sum_{j:x_j>x}P(X=x_j).

This also gives a useful check. If the distribution is complete, $P(X>x)$ must be between 0 and 1 and cannot increase as $x$ increases: raising the threshold can only remove outcomes from the sum. At a possible loss value, the strict exceedance curve drops by exactly the probability mass at that value.

For a finite distribution, the curve is constant between consecutive possible loss values. At a possible loss value $x_j$, the strict exceedance value $P(X>x_j)$ excludes the mass at $x_j$. Equivalently, the curve at that threshold is the value immediately after the downward step. This convention follows directly from the strict inequality, not from how a graph happens to be drawn.

Worked Calculations

Worked Example: Building a Curve From Event-Loss Probabilities

A hypothetical insurer models an event-level loss $X$, in millions of dollars, with the following distribution. Find $P(X>x)$ at each listed threshold and give the step levels of the curve.

Loss $x_j$ (millions)	Probability $P(X=x_j)$
0	0.50
20	0.25
60	0.15
120	0.10

Method. First check that the probability mass is complete: $0.50+0.25+0.15+0.10=1.00$. Then, for each threshold, add only the probabilities attached to losses strictly greater than it.

At $x=0$, the outcomes above zero are 20, 60, and 120, so $P(X>0)=0.25+0.15+0.10=0.50$.

At $x=20$, exclude both 0 and the outcome exactly equal to 20: $P(X>20)=0.15+0.10=0.25$.

At $x=50$, the outcomes above 50 are 60 and 120: $P(X>50)=0.15+0.10=0.25$.

At $x=60$, the outcome equal to 60 is excluded: $P(X>60)=0.10$. At $x=100$, only the 120 outcome exceeds the threshold, so $P(X>100)=0.10$. At $x=120$, no outcome exceeds the threshold, giving $P(X>120)=0$.

The resulting threshold table is:

Threshold $x$ (millions)	$P(X>x)$
0	0.50
20	0.25
50	0.25
60	0.10
100	0.10
120	0

The curve is at 0.50 from just above 0 up to 20, at 0.25 from just above 20 up to 60, at 0.10 from just above 60 up to 120, and at 0 above 120. At the listed loss values themselves, use the strict exceedance values in the table. The values are nonincreasing, and the drops are 0.25 at 20, 0.15 at 60, and 0.10 at 120, matching the probability masses there. Common mistake avoided: adding the 0.25 mass at 20 when calculating $P(X>20)$; that mass belongs in $P(X\geq20)$, not in the strict exceedance probability.

Worked Example: Summing Weighted Catalog Outcomes

A hypothetical stochastic event catalog assigns weights to modeled losses, in thousands of dollars. Treat the weights as probabilities for this calculation. Find the exceedance probability above $50,000 and above $90,000, then state the curve levels between modeled loss values.

Catalog loss (thousands)	Weight
15	0.20
35	0.30
55	0.10
90	0.25
140	0.15

Method. Verify the weights first: $0.20+0.30+0.10+0.25+0.15=1.00$. The requested thresholds are 50 and 90 thousand dollars, so identify which catalog losses are strictly above each threshold.

Losses above 50 are 55, 90, and 140. Thus $P(X>50)=0.10+0.25+0.15=0.50$. Losses above 90 include only 140, so $P(X>90)=0.15$. In contrast, $P(X\geq90)=0.25+0.15=0.40$, because it includes the weight at exactly 90.

For completeness, the strict exceedance probabilities at the catalog loss values are $P(X>15)=0.80$, $P(X>35)=0.50$, $P(X>55)=0.40$, $P(X>90)=0.15$, and $P(X>140)=0$. These are obtained by adding the weights at larger losses: for example, above 55 the weights are $0.25+0.15=0.40$. Between 35 and 55, the curve is 0.50; between 55 and 90, it is 0.40; and between 90 and 140, it is 0.15.

The curve is nonincreasing, and the drop at loss 90 is 0.25, exactly the weight assigned to that loss. Common mistake avoided: treating catalog weights as informal rankings rather than probabilities. The summation is valid here because the problem specifies that the weights form a probability distribution and they sum to one.

Worked Example: Exceedance From an Annual Aggregate Distribution

A hypothetical model has already produced the following distribution for annual aggregate net loss $X$, in millions of dollars. The losses are already net amounts after the applicable contract terms. Find the strict exceedance probabilities at $75 million, $100 million, and $150 million, and explain the relevant steps in the curve.

Annual aggregate loss (millions)	Probability
0	0.40
30	0.20
75	0.18
150	0.14
300	0.08

Method. Check completeness: $0.40+0.20+0.18+0.14+0.08=1.00$. Then sum the probabilities at annual losses strictly above each threshold.

Above 75, only 150 and 300 qualify: $P(X>75)=0.14+0.08=0.22$. Above 100, those same two outcomes qualify: $P(X>100)=0.14+0.08=0.22$. Above 150, only 300 qualifies: $P(X>150)=0.08$. The curve is flat at 0.22 between just above 75 and 150, then drops by 0.14 at 150 to 0.08. It drops by a further 0.08 at 300 to zero.

A check at the equality case makes the convention clear: $P(X\geq150)=0.14+0.08=0.22$, while $P(X>150)=0.08$. The difference, 0.14, is exactly the probability of an annual loss of 150 million. Because the stated values are already net losses, applying a deductible or limit again would change the distribution and answer the wrong question. Common mistake avoided: including the 150 outcome in the strict tail, or reapplying policy terms to losses already stated net.

A Reliable Construction Procedure

When a distribution has many outcomes, a systematic procedure reduces the risk of missing a probability mass or mishandling a boundary. Sort outcomes from smallest to largest, confirm the total probability, and calculate each tail by either summing outcomes above the threshold or subtracting the cumulative probability at or below it.

P(X>x)=1-P(X\leq x).

The complement works because the events $\{X>x\}$ and $\{X\leq x\}$ are disjoint and together cover every possible outcome. Notice that the cumulative probability in this formula includes an outcome equal to $x$. Using $1-P(X<x)$ instead would produce $P(X\geq x)$, which includes the mass at the threshold.

Define the loss variable and units.
State whether the outcomes are event-level, annual aggregate, gross, or net losses as specified by the model. Keep thresholds in the same units.

Validate the distribution.
Check that each probability is nonnegative and that all probabilities sum to one, allowing only for stated rounding.

Apply the threshold condition literally.
For “exceeds $x$,” include $x_j$ only when $x_j>x$. For “at least $x$,” include outcomes with $x_j\geq x$.

Check and describe the curve.
Confirm that tail probabilities are between zero and one and nonincreasing. At each possible loss, the drop equals its probability mass.

Common Errors and Why the Curve Matters

An exceedance curve summarizes how probability is distributed across loss thresholds. It lets a reader compare the chance of losses of different sizes without repeatedly scanning every row of a discrete distribution. In catastrophe risk work, getting the tail calculation right is essential: a small boundary or unit error can materially change the reported probability for a selected loss threshold.

Confusing “exceeds” with “equals or exceeds.” For a threshold that is itself a possible loss, the mass at that loss is excluded from $P(X>x)$ and included in $P(X\geq x)$.
Summing the wrong side of the threshold. Include probabilities for losses greater than the threshold, not losses less than it. Alternatively, subtract $P(X\leq x)$ from one.
Ignoring units. A threshold of 50 is not interpretable until it is clear whether losses are in dollars, thousands, or millions. Convert the threshold or the loss values before comparing them.
Failing to validate weights. If probabilities do not sum to one, first determine whether the distribution is incomplete, rounded, or uses weights requiring normalization. Do not silently treat arbitrary weights as probabilities.
Expecting a sloped curve for a finite discrete distribution. A finite set of loss values produces flat segments and downward jumps, not a continuous decline. The jump at each loss value corresponds to its probability mass.

Key takeaway: For a discrete loss $X$, calculate $P(X>x)$ by adding the probability of every outcome strictly greater than $x$. The resulting step curve is nonincreasing, and its drop at a possible loss equals the probability mass at that loss.

Check Your Understanding

For each problem, use the strict definition of exceedance unless the wording explicitly asks for “at least.”

A hypothetical loss $X$, in millions of dollars, equals 0 with probability 0.60, 25 with probability 0.25, and 80 with probability 0.15. Find $P(X>25)$ and $P(X\geq25)$. (Answer: $P(X>25)=0.15$; $P(X\geq25)=0.40$.)
A discrete loss distribution assigns probabilities 0.30, 0.20, 0.35, and 0.15 to losses of 10, 40, 70, and 100 thousand dollars, respectively. Find $P(X>40)$ and $P(X>70)$. (Answer: $P(X>40)=0.50$; $P(X>70)=0.15$.)
A hypothetical annual loss distribution, in millions, is $X=0, 50, 200$ with probabilities 0.50, 0.30, and 0.20. Find the strict exceedance probability at 0 and at 50, and state the drop at 50. (Answer: $P(X>0)=0.50$; $P(X>50)=0.20$; the drop at 50 is 0.30.)
A weighted event catalog lists losses of 20, 60, 110, and 170 thousand dollars with respective weights 0.10, 0.35, 0.30, and 0.25. Find $P(X>60)$, then compare it with $P(X\geq60)$. (Answer: $P(X>60)=0.55$; $P(X\geq60)=0.90$.)
A complete distribution has probabilities 0.25 at loss 0, 0.30 at loss 40, 0.25 at loss 100, and 0.20 at loss 160. Find the strict exceedance values at thresholds 40 and 100, and the probability mass represented by the drop at 100. (Answer: $P(X>40)=0.45$; $P(X>100)=0.20$; the drop at 100 is 0.25.)

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