Constructing AEP Curves from Simulated Years

From Event Losses to Annual Aggregate Loss

An event loss table records losses by event, but catastrophe pricing also requires a view of the total loss that may occur over an entire year. A simulated year can contain no events, one event, or several events. To construct an aggregate exceedance probability curve, first sum all included event losses within each simulated year, then compare the resulting annual totals with loss thresholds.

The aggregation basis must be consistent. For example, if the intended curve is for losses net of a particular insurance contract, each event loss should first reflect the applicable contract terms; annual totals are then sums of those net event losses. Do not mix gross and net losses within the same curve.

Definition: Let $L_j$ be the annual aggregate loss in simulated year $j$, for $j=1,\ldots,N$. If the year contains event losses $X_{j1},X_{j2},\ldots,X_{jn_j}$, then $L_j=\sum_{k=1}^{n_j}X_{jk}$. For a loss threshold $x$, this tutorial uses the inclusive exceedance convention $\widehat{\operatorname{AEP}}(x)=\#\{j:L_j\ge x\}/N$. The empirical AEP is the proportion of simulated years whose annual total is at least the threshold.

The inclusive convention is stated explicitly because simulations may contain tied annual totals. Some applications instead use a strict “greater than” threshold. If that convention is required, count years with $L_j>x$; at a tied threshold the result can differ. The convention should be applied consistently throughout a curve.

Constructing the Empirical Curve

A simulated year is the unit of observation for an AEP curve. The event count is not the denominator: a year with four events contributes one annual total, just as a year with one event contributes one annual total. A simulated year with no covered events contributes $L_j=0$, and must remain in the simulation set.

Identify the annual simulation set.
Confirm the number $N$ of simulated years and the loss basis being measured.

Aggregate by year.
Add all included event losses assigned to each year. Assign zero to a year with no events.

Rank annual totals.
Sort the $N$ values from largest to smallest. Preserve ties; do not turn tied years into one observation.

Count exceedances at each threshold.
For each $x$, count the annual totals satisfying $L_j\ge x$, then divide by $N$.

For a finite set of simulated years, the empirical curve is a step function. If annual totals are ranked from largest to smallest, the curve can be tabulated at each distinct annual total by counting how many simulated years have a total at least that large. With ties, all years at the tied amount are counted at that threshold.

Worked Example: Aggregating Six Simulated Years

A hypothetical portfolio has the following event losses, in thousands of dollars. Construct the annual totals, rank them, and estimate the inclusive AEP at a threshold of $80,000.

Simulated year	Event losses ($000)	Annual total ($000)
1	65, 45	110
2	40, 30	70
3	45	45
4	20, 15	35
5	20	20
6	No events	0

Method. Sum the event losses within each year: $65+45=110$, $40+30=70$, $45=45$, $20+15=35$, $20=20$, and $0$ for the no-event year. The six annual totals are therefore $110,70,45,35,20,0$, all in thousands of dollars.

In descending order, the totals are $110,70,45,35,20,0$. At the $80,000 threshold, only the year with $110,000 meets $L_j\ge80$ (in thousands). Thus the exceedance count is $1$, not the number of events in that year or the number of large event losses in the full table.

\widehat{\operatorname{AEP}}(80)=\frac{1}{6}=0.1667

The estimated inclusive AEP is $0.1667$, or $16.67\%$. A common mistake is to include the $70,000 year because it has multiple events. The threshold applies to the annual sum, and $70,000 is below $80,000. The count of annual totals at least $80,000 is exactly one.

Why the Count Divided by the Number of Years

The empirical calculation is the direct estimate of the annual exceedance probability under the simulated annual process. Suppose simulated annual totals $L_1,\ldots,L_N$ are independent and identically distributed, and let $p=P(L_j\ge x)$ for a fixed threshold $x$. Define an indicator $I_j$ to equal 1 when $L_j\ge x$ and 0 otherwise. Then the estimated exceedance probability is the average of those indicators.

\widehat{\operatorname{AEP}}(x)=\frac{1}{N}\sum_{j=1}^{N} I_j

Each indicator has expectation $E[I_j]=1\cdot p+0\cdot(1-p)=p$. By linearity of expectation, the expected value of their average is $N p/N=p$. Thus the count divided by $N$ estimates the probability, and under the stated independent, identically distributed simulation assumption it is unbiased for that probability. The estimate is still subject to simulation variation; it does not make a finite simulation identical to the full annual loss distribution.

The assumptions matter. The simulated years should represent the same annual process and be sampled in a manner appropriate to the model. If years are deliberately weighted or are not representative draws, a simple unweighted count divided by $N$ may not estimate the desired probability. In that case the simulation design’s weights and definitions must be used.

Worked Example: Tied Annual Totals and a Step Curve

A hypothetical model produces eight annual totals, in thousands of dollars, after event losses have been summed within each year: $50,0,80,25,85,50,10,90$. Construct the empirical inclusive AEP at the distinct annual totals and find the estimate at a threshold of $50,000.

Method. Rank the totals from largest to smallest: $90,85,80,50,50,25,10,0$. There are eight simulated years, including the year with zero loss.

Threshold $x$ ($000)	Number of annual totals at least $x$	Estimated inclusive AEP
90	1	1/8 = 0.1250
85	2	2/8 = 0.2500
80	3	3/8 = 0.3750
50	5	5/8 = 0.6250
25	6	6/8 = 0.7500
10	7	7/8 = 0.8750
0	8	8/8 = 1.0000

At $x=50$, the annual totals satisfying $L_j\ge50$ are $90,85,80,50,50$, which is five years. Therefore the estimate is $5/8=0.6250$, or $62.50\%$.

The two years tied at $50,000 remain two separate simulated years. Under the inclusive convention they both count at the $50,000 threshold. Under the strict convention $L_j>50$, only $90,85,80$ count, giving $3/8=0.3750$. Do not silently switch conventions to obtain a preferred answer.

Worked Example: Building Totals From a Year-by-Year Event Listing

Worked Example: Three Events in One Simulated Year

A hypothetical catastrophe simulation reports event losses, in thousands of dollars, for six years: Year 1 has losses of 35, 20, and 15; Year 2 has 80; Year 3 has 45 and 30; Year 4 has no events; Year 5 has 60 and 55; Year 6 has 25, 25, and 20. Find all annual totals, rank them, and estimate the inclusive AEP at $75,000.

Method. Add losses separately within each year. Year 1 is $35+20+15=70$. Year 2 is $80$. Year 3 is $45+30=75$. Year 4 is $0$. Year 5 is $60+55=115$. Year 6 is $25+25+20=70$. The annual totals are $70,80,75,0,115,70$, in thousands of dollars.

Rank the six totals: $115,80,75,70,70,0$. At a threshold of $75,000, the qualifying annual totals are $115,80,75$. The two years totaling $70,000 do not qualify, and the zero-loss year does not qualify.

\widehat{\operatorname{AEP}}(75)=\frac{3}{6}=0.5000

The estimated inclusive AEP is $0.5000$, or $50.00\%$. The common error is to count the three Year 1 events as three exceedance observations. AEP counts simulated years: Year 1 contributes one total of $70,000 and therefore contributes no exceedance at the $75,000 threshold.

Interpretation and Common Errors

An AEP curve describes the simulated distribution of annual aggregate loss, not the distribution of individual event losses. Its threshold count must therefore be performed only after annual aggregation. The same event catalog can yield different AEP curves if event losses are assigned to different simulated years, because the annual sums change.

Counting events instead of years. The denominator is $N$, the number of simulated years. Multiple events in one year still produce one annual observation.
Dropping zero-loss years. Omitting no-event years changes the denominator and generally overstates the frequency of positive annual losses.
Ranking events rather than annual totals. Sort the completed $L_j$ values, not the individual event amounts.
mishandling ties. Retain every year in the sample and apply the stated strict or inclusive convention consistently.
Changing the loss basis mid-calculation. Aggregate losses measured on a consistent gross or net basis; do not combine incompatible event-level amounts.
Reading a finite curve as exact. A simulated curve is an estimate based on a finite number of years. Its displayed steps reflect the sample, so additional decimal places do not remove simulation uncertainty.

Key takeaway: An empirical AEP curve begins with one annual aggregate loss per simulated year. Sum within years, keep every year in the denominator, rank the annual totals, and estimate each exceedance probability by counting years that meet the explicitly stated threshold rule.

Check Your Understanding

For each problem, use the inclusive convention $L_j\ge x$ and count simulated years, not events.

Five simulated years have event losses, in thousands of dollars: Year 1: 18 and 12; Year 2: none; Year 3: 45 and 20; Year 4: 25; Year 5: 30 and 15. Find the annual totals and the AEP at $x=45$. (Answer: Totals $30,0,65,25,45$; $2/5=0.4000$.)
Four simulated years have annual totals of $40,60,60,10$, in thousands. Find the inclusive AEP at $x=60$, and state the strict exceedance estimate at the same threshold. (Answer: Inclusive $2/4=0.5000$; strict $0/4=0.0000$.)
Six years have event losses, in thousands: $Y_1: 30+25$, $Y_2: 70$, $Y_3: 20+20$, $Y_4:$ none, $Y_5: 50+30$, and $Y_6: 45$. Find the ranked annual totals and inclusive AEP at $x=70$. (Answer: Ranked totals $80,70,55,45,40,0$; $2/6=0.3333$.)
An eight-year simulation has ranked annual totals, in thousands, of $120,100,90,90,65,40,15,0$. Calculate the inclusive AEP at $x=90$, and explain how the tied totals affect the count. (Answer: $4/8=0.5000$; both years at 90 count, along with 120 and 100.)
A model lists 12 event records across six simulated years. After grouping and summing, the annual totals are $95,75,75,50,20,0$, in thousands. Find the AEP at $x=75$, and identify the correct denominator. (Answer: $3/6=0.5000$; the denominator is six simulated years, not 12 events.)

What You'll Learn