Reading and Interpolating Exceedance Curves

Reading an Exceedance Curve

A catastrophe exceedance curve summarizes how the probability of a loss relates to the size of that loss. It can be used in either direction: read an exceedance probability at a specified loss level, or estimate the loss level associated with a specified probability. Both tasks require checking what the curve represents and how its axes are scaled. A value read from a graph is an estimate, not necessarily an exact model output.

Definition: For a specified loss measure $L$, the exceedance probability at loss threshold $x$ is $p(x)=P(L>x)$. An exceedance curve plots $p(x)$ against $x$. In a tabulation, each pair $(x,p(x))$ gives a loss level and its associated exceedance probability. The loss measure and probability convention must be identified before reading or interpolating.

For an annual aggregate exceedance curve, $L$ is annual aggregate loss; for an occurrence exceedance curve, it refers to loss from an event under the curve’s stated convention. These are not interchangeable. Earlier material on occurrence and aggregate exceedance probability curves distinguishes the two measures. This tutorial focuses on reading and interpolating a curve once its measure is known.

An exceedance curve should generally decline as the loss threshold increases: a larger loss is less likely to be exceeded. Before interpolating, confirm that the two selected points belong to the same curve, use the same currency and loss basis, and bracket the requested value. A plot with logarithmic axes changes how positions between points are interpreted.

Linear-Axis Interpolation

If both axes are linear, equal horizontal distances represent equal changes in loss, and equal vertical distances represent equal changes in probability. Between two tabulated points $(x_1,p_1)$ and $(x_2,p_2)$, the straight-line estimate at a loss $x$ is:

p(x)=p_1+\frac{x-x_1}{x_2-x_1}(p_2-p_1).

This formula assumes a straight-line segment between the two points. It is appropriate for reading from a curve drawn with linear axes when the intended approximation is linear in loss and probability. It is not a claim that the underlying loss distribution is exactly linear between the points.

Worked Example: Probability at an Intermediate Loss on Linear Axes

A hypothetical annual aggregate exceedance curve uses linear axes. It gives $p_1=0.12$ at $x_1=\$20$ million and $p_2=0.03$ at $x_2=\$80$ million. Estimate the exceedance probability at $x=\$50$ million.

Method. Use linear interpolation in both loss and probability. The requested loss lies between the two given losses, so interpolation rather than extrapolation is appropriate.

First calculate the fraction of the horizontal interval:

t=\frac{50-20}{80-20}=\frac{30}{60}=0.5.

The probability changes from 0.12 to 0.03, a difference of $0.03-0.12=-0.09$. Apply half of that change:

$$ p(50)=0.12+0.5(0.03-0.12) =0.12-0.045 =0.075. $$

The estimated probability is 0.075, equivalent to 7.5%. As a check, 0.075 lies between 0.12 and 0.03, and the loss of \$50 million is halfway between \$20 million and \$80 million, so the interpolated probability is halfway between the endpoint probabilities.

Keep probability as a decimal during calculation unless the entire calculation is consistently in percent. In the example, 0.075 is the probability; 7.5% is the same quantity expressed as a percentage. Writing 7.5 as the probability would be incorrect because probabilities cannot exceed 1.

Interpolation on Logarithmic Axes

On a logarithmic axis, equal visual distances correspond to equal ratios, not equal differences. For a log-log plot, both loss $x$ and probability $p$ are on logarithmic scales. A straight segment on that plot means that the logarithms of the values are linearly interpolated. The data values must be positive for this method.

Interpolation rule: On a log-log plot, let $t$ be the fraction of the horizontal distance in log-loss coordinates. Then interpolate the logarithm of probability by the same fraction:

t=\frac{\ln x-\ln x_1}{\ln x_2-\ln x_1}, \qquad \ln p(x)=\ln p_1+t(\ln p_2-\ln p_1), \qquad p(x)=p_1\left(\frac{p_2}{p_1}\right)^t.

The logarithm base does not matter if the same base is used throughout. The formula is not the same as linear interpolation of $p$ against $x$. On a log-log plot, the horizontal fraction depends on ratios: for example, the midpoint between losses of 10 and 100 is $\sqrt{10\cdot100}$, not their arithmetic midpoint.

The same rule can be inverted to find a loss at a specified probability. First find the fraction $t$ between the endpoint probabilities in log-probability coordinates; then use that fraction between the endpoint losses in log-loss coordinates:

t=\frac{\ln p-\ln p_1}{\ln p_2-\ln p_1}, \qquad x=x_1\left(\frac{x_2}{x_1}\right)^t.

These expressions assume the plotted segment is straight in log-log coordinates and the requested value lies between the endpoints. They describe a graphical interpolation convention; they do not establish that the underlying catastrophe loss distribution follows a power law outside the displayed points.

Worked Example: Loss Level at a Target Probability on Log-Log Axes

A hypothetical annual aggregate curve is plotted with both axes logarithmic. Its points include a loss of \$10 million at exceedance probability 0.10 and a loss of \$100 million at probability 0.01. Estimate the loss at probability 0.04.

Method. Interpolate in log-probability to get the fraction $t$, then apply that fraction in log-loss. The probability 0.04 lies between 0.10 and 0.01.

Calculate the fraction between the endpoint probabilities:

t=\frac{\ln(0.04)-\ln(0.10)}{\ln(0.01)-\ln(0.10)} =\frac{\ln(0.4)}{\ln(0.1)} \approx 0.39794.

The loss ratio across the full interval is $100/10=10$. Therefore:

x=10\left(\frac{100}{10}\right)^{0.39794} =10(10^{0.39794}) =10(2.5) =\$25\text{ million}.

The estimated loss is \$25 million. Check that it falls between \$10 million and \$100 million. Since 0.04 is closer to 0.10 than to 0.01 in log-probability coordinates, the result is closer to \$10 million than to \$100 million on the log-loss scale.

Worked Example: Probability at a Loss on Log-Log Axes

A hypothetical occurrence exceedance curve has the following points on log-log axes: probability 0.20 at a loss of \$5 million, and probability 0.0125 at a loss of \$80 million. Estimate the exceedance probability at a loss of \$20 million.

Method. Find the horizontal fraction in log-loss coordinates, then apply the same fraction to the log-probability interval. The loss ratio from \$5 million to \$20 million is $20/5=4$; the full loss ratio is $80/5=16$.

t=\frac{\ln(20/5)}{\ln(80/5)} =\frac{\ln 4}{\ln 16} =0.5.

The probability ratio across the full interval is $0.0125/0.20=0.0625=1/16$. Thus:

p(20)=0.20\left(\frac{0.0125}{0.20}\right)^{0.5} =0.20\sqrt{\frac{1}{16}} =0.20(0.25) =0.0500.

The estimated exceedance probability is 0.0500, or 5.00%. The loss of \$20 million is halfway between the endpoint losses in log coordinates because $20$ is the geometric mean of $5$ and $80$. Accordingly, the estimated probability is the geometric mean of 0.20 and 0.0125, which is 0.0500.

Why the Log-Log Formula Works

The log-log rule follows directly from straight-line interpolation after transforming both coordinates. Let $u=\ln x$ and $w=\ln p$. If the plotted segment is straight, then $w$ changes linearly with $u$. At a fraction $t$ of the horizontal distance, the vertical coordinate is the same fraction between its endpoints:

w=w_1+t(w_2-w_1), \qquad t=\frac{u-u_1}{u_2-u_1}.

Substitute $u=\ln x$ and $w=\ln p$, then exponentiate both sides:

\ln p=\ln p_1+t(\ln p_2-\ln p_1) \quad\Longrightarrow\quad p=p_1\exp\!\left(t\ln\frac{p_2}{p_1}\right) =p_1\left(\frac{p_2}{p_1}\right)^t.

Thus the log-log expression is ordinary straight-line interpolation in transformed coordinates. The exponentiation makes the interpolated probability a weighted geometric interpolation of the endpoint probabilities rather than a weighted arithmetic average.

Reading the Axes and Avoiding Traps

Some plots use a logarithmic scale on only one axis. Apply interpolation in the coordinates actually shown: transform the logarithmic axis and leave the linear axis unchanged. For example, if loss is linear and probability is logarithmic, interpolate probability geometrically at the required linear fraction in loss. If loss is logarithmic and probability is linear, interpolate probability arithmetically at the fraction measured in log-loss coordinates. Do not assume both axes are logarithmic merely because one is.

A frequent error is to use equal differences where the graph represents equal ratios, or to confuse a decimal probability with its percentage display. In the first worked example, the correct arithmetic gives $0.075$, not a probability of 7.5. The correction is to report the decimal probability as $0.075$, or explicitly convert it to 7.5%. A second common error is to use an arithmetic midpoint of losses on a log-loss axis; the correct midpoint is the geometric mean.

A disciplined reading can be organized as follows:

Identify the curve.
Confirm whether it represents annual aggregate loss, occurrence loss, or another stated measure, and note the probability convention.

Read the axis scales and units.
Determine whether each axis is linear or logarithmic, and convert losses to consistent units before calculating.

Choose bracketing points.
Use adjacent points that enclose the requested loss or probability. If the request lies outside the displayed range, interpolation is not available.

Interpolate in plotted coordinates.
Use equal differences on a linear axis and equal ratios on a logarithmic axis; transform back to the original units.

Check the result.
Confirm it lies between the endpoint values, preserves the decreasing direction of an exceedance curve, and is reported in the requested format.

Interpolation is least reliable when points are far apart, the plotted curve bends sharply between them, or the values are affected by discrete loss outcomes. In a discrete distribution, the probabilities $P(L>x)$ and $P(L\ge x)$ can differ at a loss value with positive probability. Follow the curve’s stated convention; a smooth line between plotted points does not remove that distinction. Avoid extrapolating beyond the available points unless a separate model explicitly supports it.

Reading and interpolating curves is useful when a decision requires a threshold not printed in a model table, or a probability not marked on a chart. It is aligned with the CAS Exam 9 (Risk Management for Actuaries) syllabus because it tests interpretation of catastrophe loss results, careful use of probability scales, and clear communication of estimates and assumptions.

Check Your Understanding

For each problem, identify the relevant axis convention before calculating. Report probabilities as decimals unless the question requests a percentage.

A hypothetical annual aggregate curve has linear axes and gives probabilities 0.18 at \$30 million and 0.06 at \$90 million. Estimate the probability at \$50 million. (Answer: 0.1400)
On a log-log curve, probability is 0.16 at \$4 million and 0.01 at \$64 million. Estimate the probability at \$16 million. (Answer: 0.0400)
On a log-log curve, the probability is 0.10 at \$8 million and 0.00625 at \$128 million. Estimate the loss at probability 0.025. (Answer: \$32 million)
A curve has a linear loss axis and a logarithmic probability axis. It gives probability 0.10 at \$10 million and 0.01 at \$30 million. Estimate the probability at \$20 million, rounding to four decimals. (Answer: 0.0316)

What You'll Learn