Exponential Severity Estimation from Complete Data

Why Fit an Exponential Severity Model?

A severity model describes the distribution of the amount of an individual loss. The exponential distribution is a simple model for positive losses: one parameter determines its mean, its spread, and the rate at which its tail decreases. Fitting it to complete observations is a direct application of the method of moments and maximum likelihood.

The simplicity has a cost. The model assumes a constant hazard: among losses that have already exceeded a given amount, the rate of exceeding a slightly larger amount does not depend on the amount already reached. That is a strong shape assumption, not a conclusion established merely by calculating an estimate. A fitted rate summarizes the data under that assumption.

Definition: A loss $X$ has an exponential distribution with rate $\lambda>0$ if its density, distribution function, and survival function are $$ f_X(x)=\lambda e^{-\lambda x},\quad F_X(x)=1-e^{-\lambda x},\quad S_X(x)=P(X>x)=e^{-\lambda x},\qquad x\geq 0. $$ Its mean is $\theta=E[X]=1/\lambda$, and its variance is $\operatorname{Var}(X)=1/\lambda^2=\theta^2$. The rate has units reciprocal to the loss amount: if $X$ is measured in dollars, $\lambda$ is per dollar.

For estimation below, assume $X_1,\ldots,X_n$ are independent and identically distributed exact observations from this exponential model. “Complete” means each loss amount is observed exactly, rather than only known to exceed a threshold or to equal a capped payment. The estimator derived here is not the likelihood estimator for censored, truncated, rounded, or otherwise modified data.

Method of Moments

The method of moments equates a population moment to its sample counterpart. With one unknown parameter, the first raw moment is enough: the model mean is $1/\lambda$, and the sample mean is $\bar X=(1/n)\sum_{j=1}^n X_j$. Thus

\frac{1}{\hat{\lambda}_{\mathrm{MOM}}}=\bar X \quad\Longrightarrow\quad \hat{\lambda}_{\mathrm{MOM}}=\frac{1}{\bar X} =\frac{n}{\sum_{j=1}^n X_j}.

If the model is parameterized by its mean $\theta$, the moment estimate is simply $\hat{\theta}_{\mathrm{MOM}}=\bar X$. The two expressions describe the same fitted distribution because $\hat{\theta}=1/\hat{\lambda}$.

Worked Example: Fitting a Rate and a Tail Probability

A hypothetical insurer records five complete individual losses, in thousands of dollars: $1,3,4,8,9$. Assume these are independent observations from one exponential severity distribution. Estimate its rate by moments, then calculate the fitted probability that a loss exceeds 10 thousand dollars.

Method. Use the first-moment equation for the rate, then substitute the fitted rate into the exponential survival function. This avoids treating the rate itself as the sample mean.

Step 1: Calculate the sample mean. The observed total is $1+3+4+8+9=25$ thousand dollars, and $n=5$. Therefore $$ \bar X=\frac{25}{5}=5\text{ thousand dollars}. $$

Step 2: Estimate the rate. The rate is the reciprocal of the mean: $$ \hat{\lambda}_{\mathrm{MOM}}=\frac{1}{5}=0.200\text{ per thousand dollars}. $$ The corresponding fitted mean is $1/0.200=5$ thousand dollars, matching the sample mean.

Step 3: Calculate the exceedance probability. At $x=10$ thousand dollars, $$ \widehat{P(X>10)}=e^{-(0.200)(10)}=e^{-2}=0.1353. $$ The fitted model assigns probability $0.1353$ to a loss exceeding 10 thousand dollars. As a check, the exponent is dimensionless: the rate is per thousand dollars and the loss threshold is in thousands of dollars.

Maximum Likelihood from Complete Observations

For exact independent observations, the joint likelihood is the product of their densities. Taking its logarithm makes the rate estimate straightforward.

Write the likelihood.
For observed positive amounts $x_1,\ldots,x_n$, independence gives $L(\lambda)=\prod_{j=1}^n \lambda e^{-\lambda x_j}=\lambda^n e^{-\lambda\sum_j x_j}$.

Take the log and differentiate.
$\ell(\lambda)=n\log\lambda-\lambda\sum_j x_j$, so $\ell'(\lambda)=n/\lambda-\sum_j x_j$.

Solve and verify a maximum.
Setting the score to zero gives $\hat{\lambda}=n/\sum_j x_j$. Since $\ell''(\lambda)=-n/\lambda^2<0$ for $\lambda>0$, this stationary point is a maximum.

The likelihood derivation explains why the moment and maximum-likelihood estimates agree:

\hat{\lambda}_{\mathrm{MLE}} =\frac{n}{\sum_{j=1}^n x_j} =\frac{1}{\bar X} =\hat{\lambda}_{\mathrm{MOM}}.

This agreement follows from the exponential likelihood’s dependence on the data through their sum and the model’s mean $1/\lambda$. It is specific to this model and these complete independent observations; it is not a general rule that method-of-moments and likelihood estimates coincide.

Worked Example: Likelihood Estimation and Conditional Survival

A hypothetical portfolio produces four complete losses, each measured in thousands of dollars: $2,4,4,6$. Assume independent observations from an exponential distribution with a common rate. Find the maximum-likelihood rate and the fitted probability that a loss exceeds 8, given that it already exceeds 3.

Method. Maximize the complete-data likelihood for the rate, then use the ratio of survival probabilities for the conditional probability. This avoids confusing a conditional tail probability with an unconditional one.

Step 1: Find the estimate. The sum is $2+4+4+6=16$, so $n=4$ and $$ \hat{\lambda}_{\mathrm{MLE}}=\frac{4}{16}=0.250\text{ per thousand dollars}. $$ The fitted mean is $1/0.250=4$ thousand dollars.

Step 2: Form the conditional probability. By the definition of conditional probability and the exponential survival function, $$ P(X>8\mid X>3)=\frac{P(X>8)}{P(X>3)} =\frac{e^{-8\lambda}}{e^{-3\lambda}} =e^{-5\lambda}. $$

Step 3: Substitute the fitted rate. $$ \widehat{P(X>8\mid X>3)} =e^{-(5)(0.250)} =e^{-1.25}=0.2865. $$ The fitted conditional probability is $0.2865$. As a numerical check, the unconditional probabilities are $P(X>8)=e^{-2}=0.1353$ and $P(X>3)=e^{-0.75}=0.4724$; their ratio is approximately $0.2865$.

What Constant Hazard Means

For a continuous positive loss with density $f_X(x)$ and survival function $S_X(x)>0$, the hazard at amount $x$ is $h(x)=f_X(x)/S_X(x)$. It describes the instantaneous rate of crossing amount $x$, conditional on having reached that amount. For the exponential model,

h(x)=\frac{\lambda e^{-\lambda x}}{e^{-\lambda x}}=\lambda.

The exponential hazard is therefore constant at every amount. A related implication can be derived directly: for $s,t\geq 0$,

P(X>s+t\mid X>s) =\frac{S_X(s+t)}{S_X(s)} =\frac{e^{-\lambda(s+t)}}{e^{-\lambda s}} =e^{-\lambda t} =P(X>t).

Thus, conditional on a loss having exceeded $s$, its remaining excess $X-s$ has the same exponential survival function as a new loss. This memoryless property is another expression of the constant hazard assumption. It does not say that all observed losses are similar, nor does it establish that real losses have this property. It says what the fitted model implies.

Worked Example: Converting Units Before Estimating

A hypothetical set of four complete losses is recorded in dollars as $600,1000,1400,2000$. Assume the losses are independent and exponential. Estimate the rate in both per-dollar and per-thousand-dollar units, then calculate the fitted probability that a loss is between 500 and 1500 dollars.

Method. Estimate using the observations in dollars, then express the resulting rate and thresholds consistently. This avoids inserting a dollar threshold into an exponent whose rate is per thousand dollars.

Step 1: Calculate the mean and rate in dollars. The sum is $600+1000+1400+2000=5000$ dollars. Hence $$ \bar X=\frac{5000}{4}=1250\text{ dollars}, \qquad \hat{\lambda}=\frac{1}{1250}=0.000800\text{ per dollar}. $$

Step 2: Convert the rate. One thousand dollars contains 1000 dollars, so the rate per thousand dollars is $1000(0.000800)=0.800$. Equivalently, the fitted mean is $1/0.800=1.25$ thousand dollars.

Step 3: Calculate the interval probability. In dollars, $$ \widehat{P(500<X\leq1500)} =F_X(1500)-F_X(500) =e^{-(0.000800)(500)}-e^{-(0.000800)(1500)}. $$ The two survival terms are $e^{-0.4}=0.6703$ and $e^{-1.2}=0.3010$, so $$ \widehat{P(500<X\leq1500)}=0.6703-0.3010=0.3693. $$ The interval probability is $0.3693$. Using thousands instead gives $e^{-(0.800)(0.5)}-e^{-(0.800)(1.5)}$, the same result.

Common Exam Traps and Interpretation

Using the mean as the rate. The sample mean estimates $\theta=1/\lambda$, not $\lambda$. If the mean is 5 thousand dollars, the rate is $0.200$ per thousand dollars, not 5 per thousand dollars.
Mixing units. The product $\lambda x$ in an exponential must be dimensionless. Convert either the rate or the loss amount before evaluating the exponential.
Applying the complete-data formula to altered observations. The likelihood above uses exact observed amounts from the stated distribution. An amount recorded only as “above a threshold” does not contribute the same density term as an exact amount.
Reading a fitted assumption as a data fact. The constant hazard is imposed by the exponential model. Estimating $\lambda$ does not prove that the true hazard is constant.

The estimate is useful because it turns a sample into a compact severity model that can produce fitted means and tail probabilities. Its interpretation remains conditional on the model and data assumptions: independent, identically distributed, complete observations and a common exponential distribution.

Check Your Understanding

For each problem, assume complete independent observations from a common exponential distribution unless stated otherwise. Rates and probabilities are rounded to four significant digits where needed.

A hypothetical sample of losses, in thousands of dollars, is $3,5,7$. Find the moment estimate of the rate and the fitted probability of a loss exceeding 10 thousand dollars. (Answer: $\hat{\lambda}=0.200$ per thousand dollars; $P(X>10)=0.1353$.)
Complete losses are $500,900,1100,1500$ dollars. Estimate the exponential rate per dollar and find the probability that a loss is between 500 and 1500 dollars. (Answer: $\hat{\lambda}=0.001000$ per dollar; $P(500<X\leq1500)=0.3834$.)
A complete sample, in thousands of dollars, is $1,2,3,6$. Estimate the rate by maximum likelihood and find the probability that a loss exceeds 6 given that it exceeds 2. (Answer: $\hat{\lambda}=0.3333$ per thousand dollars; $P(X>6\mid X>2)=e^{-4/3}=0.2636$.)
A hypothetical complete sample, in thousands of dollars, is $2,3,7,8$. Find the common moment and maximum-likelihood rate estimates, then calculate the fitted median loss. (Answer: $\hat{\lambda}=0.200$ per thousand dollars by both methods; median $=(\ln 2)/0.200=3.466$ thousand dollars.)
Complete losses are $500,1500,2500$ dollars. Estimate the rate and calculate the probability that a loss exceeds 3000 dollars given that it exceeds 1500 dollars. State why the conditional result does not depend on the threshold already exceeded. (Answer: $\hat{\lambda}=1/1500=0.0006667$ per dollar; conditional probability $=e^{-1}=0.3679$, independent of the initial threshold under the exponential model’s memoryless property.)

What You'll Learn