An Exponential Severity Model Goodness-of-Fit Case

Why Test an Exponential Severity Model?

An exponential distribution is a useful benchmark for positive loss amounts: it is mathematically simple, has a constant hazard rate, and gives a direct link between its mean and its rate parameter. Its simplicity is also a limitation. A real severity distribution may have a heavier or lighter upper tail, or may not have the exponential model’s constant hazard at all. A goodness-of-fit statistic helps quantify departures, but it does not by itself establish that a model is suitable for every actuarial purpose.

This tutorial uses the Kolmogorov-Smirnov (KS) statistic to assess an exponential model fitted to complete loss observations. It also compares an observed upper-tail proportion with the fitted model’s tail probability. That comparison can clarify the direction of a departure, while the KS statistic measures the largest difference across the full distribution. This treatment is aligned with the SOA Exam ASTAM (Advanced Short-Term Actuarial Mathematics) syllabus.

Definition: Suppose $X$ is a positive loss amount with an exponential distribution having rate parameter $\lambda>0$. Its density, cumulative distribution function, survival function, and mean are $$ f_X(x)=\lambda e^{-\lambda x},\qquad F_X(x)=1-e^{-\lambda x},\qquad S_X(x)=e^{-\lambda x},\qquad E[X]=\frac{1}{\lambda},\quad x\geq 0. $$ The scale parameter is $\theta=1/\lambda$, so $F_X(x)=1-e^{-x/\theta}$ and $E[X]=\theta$. These observations must be complete, positive, and independent with a common exponential distribution for the procedures below to apply directly.

Fitting the Rate and Constructing the KS Statistic

For complete observations $x_1,\ldots,x_n$, the likelihood under the exponential model is the product of their densities. Maximizing it gives the maximum-likelihood estimate of the rate. Substituting the fitted rate into the CDF gives the fitted distribution used to calculate the KS statistic.

L(\lambda)=\prod_{j=1}^{n}\lambda e^{-\lambda x_j} =\lambda^n e^{-\lambda\sum_{j=1}^{n}x_j},\qquad \ell(\lambda)=n\log\lambda-\lambda\sum_{j=1}^{n}x_j.

Result: The exponential maximum-likelihood estimates are $$ \widehat{\lambda}=\frac{n}{\sum_{j=1}^{n}x_j}=\frac{1}{\overline{x}}, \qquad \widehat{\theta}=\frac{1}{\widehat{\lambda}}=\overline{x}. $$ For ordered observations $x_{(1)}\leq \cdots \leq x_{(n)}$, calculate the fitted values $\widehat{F}(x_{(i)})=1-e^{-\widehat{\lambda}x_{(i)}}$. Then $$ D^+=\max_{1\leq i\leq n}\left\{\frac{i}{n}-\widehat{F}(x_{(i)})\right\},\qquad D^-=\max_{1\leq i\leq n}\left\{\widehat{F}(x_{(i)})-\frac{i-1}{n}\right\},\qquad D=\max(D^+,D^-). $$

The two one-sided terms compare the fitted CDF with the empirical distribution immediately after and immediately before each observed value. Their maximum is the largest vertical separation between the empirical CDF and the fitted CDF. A positive $D^+$ reflects an empirical CDF that gets ahead of the fitted CDF; a positive $D^-$ reflects a fitted CDF that gets ahead of the empirical CDF.

Fit the model.
Calculate the sample mean, then set $\widehat{\lambda}=1/\overline{x}$.

Evaluate the fitted CDF.
For each ordered loss, calculate $1-e^{-\widehat{\lambda}x_{(i)}}$, not the survival probability $e^{-\widehat{\lambda}x_{(i)}}$.

Find both discrepancies.
Calculate each $i/n-\widehat{F}(x_{(i)})$ and $\widehat{F}(x_{(i)})-(i-1)/n$, then take the maximum within each set.

Interpret with care.
Report $D$, identify where it occurs, and account for the fact that the rate was estimated from the same observations.

Why the Rate Estimate Has This Form

The log-likelihood is easier to maximize than the likelihood because the logarithm is increasing: maximizing one also maximizes the other. Differentiating the log-likelihood with respect to $\lambda$ and setting the derivative equal to zero gives the estimate. The second derivative is negative for every $\lambda>0$, confirming that the stationary point is a maximum.

\frac{d\ell}{d\lambda}=\frac{n}{\lambda}-\sum_{j=1}^{n}x_j=0 \quad\Longrightarrow\quad \widehat{\lambda}=\frac{n}{\sum_{j=1}^{n}x_j}. \qquad \frac{d^2\ell}{d\lambda^2}=-\frac{n}{\lambda^2}<0.

The fitted mean is therefore the observed sample mean. This is a consequence of the exponential likelihood, not a general rule that every severity model’s maximum-likelihood estimate equals its sample mean.

Worked Example: Calculating the KS Statistic for Four Complete Losses

A hypothetical portfolio has four ordered, complete losses, in thousands of dollars: $0.5,\ 1.5,\ 2.0,\ 4.0$. Fit an exponential model by maximum likelihood and calculate $D^+$, $D^-$, and $D$. State what the statistic measures.

Method and fitted parameter. The number of observations is $n=4$, and their sum is $0.5+1.5+2.0+4.0=8.0$. Thus $\overline{x}=8.0/4=2.0$ thousand dollars, $\widehat{\theta}=2.0$, and $\widehat{\lambda}=1/2.0=0.5$ per thousand dollars.

Fitted CDF values. Using $\widehat{F}(x)=1-e^{-0.5x}$:

At $x_{(1)}=0.5$, $\widehat{F}=1-e^{-0.25}=0.2212$.
At $x_{(2)}=1.5$, $\widehat{F}=1-e^{-0.75}=0.5276$.
At $x_{(3)}=2.0$, $\widehat{F}=1-e^{-1}=0.6321$.
At $x_{(4)}=4.0$, $\widehat{F}=1-e^{-2}=0.8647$.

Calculate the discrepancies. The $D^+$ terms are $1/4-0.2212=0.0288$, $2/4-0.5276=-0.0276$, $3/4-0.6321=0.1179$, and $4/4-0.8647=0.1353$. Therefore $D^+=0.1353$. The $D^-$ terms are $0.2212-0=0.2212$, $0.5276-1/4=0.2776$, $0.6321-2/4=0.1321$, and $0.8647-3/4=0.1147$. Therefore $D^-=0.2776$.

Result and check. $D=\max(0.1353,0.2776)=0.2776$, attained by $D^-$ at the second observation. As a check, each fitted CDF value is between 0 and 1 and increases with the loss amount. The statistic is the largest empirical-versus-fitted CDF separation; it is not, on its own, a p-value. The common mistake of using a sample size of three despite listing four losses would change every rank fraction and produce the wrong statistic.

Reading the Result and Examining the Upper Tail

The KS statistic treats discrepancies across the distribution on the same absolute probability scale. It does not give extra weight to the tail. To focus on a specified upper-tail threshold $c$, one can separately compare the observed proportion exceeding $c$ with the fitted model’s predicted probability:

\widehat{p}_{\mathrm{obs}}(c)=\frac{\#\{i:x_i>c\}}{n}, \qquad \widehat{p}_{\mathrm{model}}(c)=\widehat{S}_X(c)=e^{-\widehat{\lambda}c}.

This comparison is a diagnostic, not a replacement for a goodness-of-fit test. Its interpretation depends on the threshold selected, and a small sample can make an observed tail proportion unstable. A larger observed proportion than the fitted probability is evidence that the sample has more exceedances at that threshold than the fitted exponential model predicts. It does not alone establish that the entire severity distribution has a heavier tail.

Worked Example: Separating the KS Discrepancy From Tail Evidence

A hypothetical reinsurer records five complete losses, in thousands of dollars: $0.4,\ 0.8,\ 1.2,\ 2.0,\ 10.6$. Fit an exponential model, calculate the KS statistic, and compare the observed and fitted proportions above $8$.

Fit the model. The sum is $0.4+0.8+1.2+2.0+10.6=15.0$, so $\overline{x}=15.0/5=3.0$, $\widehat{\theta}=3.0$, and $\widehat{\lambda}=1/3$ per thousand dollars.

Evaluate the fitted CDF. The values at the five observations are $1-e^{-0.4/3}=0.1248$, $1-e^{-0.8/3}=0.2341$, $1-e^{-1.2/3}=0.3297$, $1-e^{-2.0/3}=0.4866$, and $1-e^{-10.6/3}=0.9708$, to four decimals.

Calculate $D^+$ and $D^-$. The $D^+$ terms are $0.2000-0.1248=0.0752$, $0.4000-0.2341=0.1659$, $0.6000-0.3297=0.2703$, $0.8000-0.4866=0.3134$, and $1.0000-0.9708=0.0292$. Thus $D^+=0.3134$. The $D^-$ terms are $0.1248-0=0.1248$, $0.2341-0.2000=0.0341$, $0.3297-0.4000=-0.0703$, $0.4866-0.6000=-0.1134$, and $0.9708-0.8000=0.1708$. Thus $D^-=0.1708$, and $D=0.3134$.

Assess the upper tail. One of the five losses exceeds $8$, so $\widehat{p}_{\mathrm{obs}}(8)=1/5=0.2000$. The fitted exponential probability is $e^{-(1/3)8}=e^{-8/3}=0.0695$, to four decimals. The observed exceedance proportion is greater by $0.2000-0.0695=0.1305$. This is a tail warning in this small sample. However, the largest KS discrepancy occurs at the fourth loss, not at the maximum. The KS statistic and the threshold comparison answer related but different questions.

Method and trap avoided. The method is to fit first, then calculate the full-sample KS discrepancy and a separate threshold-based tail diagnostic. Do not interpret the $0.2000$ observed proportion as the KS statistic, or claim that the single large loss proves a heavy-tailed population.

Worked Example: A Modest Tail Difference and the Full-Sample Statistic

A hypothetical equipment insurer observes five complete losses, in thousands of dollars: $0.5,\ 1.0,\ 1.5,\ 2.0,\ 5.0$. Calculate the fitted exponential model and KS statistic, then compare observed and fitted probabilities of a loss exceeding $4$.

Fit the model. The sum is $0.5+1.0+1.5+2.0+5.0=10.0$, so $\overline{x}=10.0/5=2.0$, $\widehat{\theta}=2.0$, and $\widehat{\lambda}=0.5$ per thousand dollars.

Calculate fitted CDF values. At $0.5,\ 1.0,\ 1.5,\ 2.0,$ and $5.0$, respectively, the fitted CDF values are $0.2212,\ 0.3935,\ 0.5276,\ 0.6321,$ and $0.9179$.

Find the KS statistic. The $D^+$ terms are $0.2000-0.2212=-0.0212$, $0.4000-0.3935=0.0065$, $0.6000-0.5276=0.0724$, $0.8000-0.6321=0.1679$, and $1.0000-0.9179=0.0821$. Hence $D^+=0.1679$. The $D^-$ terms are $0.2212,\ 0.3935-0.2000=0.1935,\ 0.5276-0.4000=0.1276,\ 0.6321-0.6000=0.0321,$ and $0.9179-0.8000=0.1179$. Hence $D^-=0.2212$ and $D=0.2212$.

Compare the upper tail. One of the five losses exceeds $4$, giving an observed proportion of $1/5=0.2000$. The fitted probability is $e^{-0.5(4)}=e^{-2}=0.1353$. The observed proportion exceeds the fitted probability by $0.0647$. This modest difference is worth noting, but it is not enough to conclude that the model is inadequate without an appropriate inferential assessment and consideration of the sample size.

Check and trap avoided. The fitted CDF at $5$ is $0.9179$, so the fitted probability of exceeding $5$ is its complement, $1-0.9179=0.0821$. Using $0.9179$ as the upper-tail probability would confuse the CDF with the survival function. The correct threshold calculation at $4$ uses $S_X(4)=e^{-2}$.

Significance, Assumptions, and Common Mistakes

The calculations above describe how far the empirical CDF lies from the fitted exponential CDF. A significance conclusion requires a reference distribution for the statistic under the null model. Because $\lambda$ was estimated from the same observations, ordinary KS critical values for a fully specified, fixed distribution do not automatically give a valid test. Use a calibration that accounts for parameter estimation, as in the earlier material on calibrating tests after parameter estimation. Do not report a standard fixed-parameter KS p-value as if the rate had been known in advance.

The direct procedure also relies on the stated data conditions. If losses are censored, truncated, dependent, or rounded in a way that creates ties, the complete continuous-observation calculation may not be appropriate without adjustment. A deductible or policy limit can also change the distribution of observed payments relative to the underlying loss distribution. A model fitted to such observations must reflect how they were selected or modified; treating them as unrestricted complete losses can distort both the estimate and the goodness-of-fit assessment.

Several checks help prevent avoidable errors:

Keep rate and scale distinct. If $\overline{x}=2$, then $\widehat{\theta}=2$ and $\widehat{\lambda}=0.5$, not the reverse.
Use the CDF for KS calculations. The CDF is $1-e^{-\widehat{\lambda}x}$; $e^{-\widehat{\lambda}x}$ is the survival function.
Use all observations and correct ranks. The rank fractions use the actual sample size $n$. A mismatch between the stated number of losses and the listed observations changes the statistic.
Separate description from inference. A numerical $D$ is a discrepancy measure. It is not automatically a p-value or a pass/fail decision.
Do not equate a tail check with a whole-distribution test. A threshold exceedance comparison focuses on one point; KS considers the maximum discrepancy across the distribution.

Key takeaway: For complete independent exponential losses, estimate $\lambda$ as the reciprocal of the sample mean, calculate the fitted CDF at every ordered loss, and take the larger of $D^+$ and $D^-$. Report where the largest discrepancy occurs, examine upper-tail evidence separately when relevant, and use a significance calibration that accounts for estimation.

Check Your Understanding

For each problem, use complete observations unless the problem states otherwise. Round probabilities and KS statistics to four decimals.

A hypothetical service plan has three complete losses of $1,\ 2,$ and $3$ thousand dollars. Find the exponential maximum-likelihood rate and the fitted CDF at $x=2$. (Answer: $\widehat{\lambda}=0.5000$ per thousand dollars; $\widehat{F}(2)=0.6321$.)
Four ordered complete losses are $1,\ 2,\ 3,$ and $6$ thousand dollars. Find the maximum-likelihood rate and the fitted KS statistic. (Answer: $\widehat{\lambda}=0.3333$ per thousand dollars; $D=0.2835$.)
Four complete losses are $1,\ 1,\ 1,$ and $5$ thousand dollars. Fit the exponential model and calculate $D^+$, $D^-$, and $D$, using the ordered observations as given. (Answer: $\widehat{\lambda}=0.5000$; $D^+=0.3565$, $D^-=0.3935$, $D=0.3935$.)
A fitted exponential model has $\widehat{\theta}=4$ thousand dollars. In a sample of 10 complete losses, two exceed $8$ thousand dollars. Compare the observed exceedance proportion with the model probability. (Answer: observed $=0.2000$; fitted $=e^{-2}=0.1353$; observed proportion is higher by $0.0647$.)
Losses are recorded only up to a policy cap of $20$ thousand dollars, so amounts above the cap are recorded as $20$. Can the recorded values be treated directly as complete observations in the fitted exponential KS procedure? (Answer: No. Values at the cap are censored or capped observations, not complete loss amounts; the observation mechanism must be accounted for.)

What You'll Learn