Exponential Severity Estimation By Rate

Introduction

When exact loss amounts are observed, a parametric severity model can be fitted by maximizing the likelihood of those observations. For the exponential model, the likelihood has a simple form, but the parameter is a rate: its units are the inverse of the units used to measure loss. Keeping the rate, the observations, and their units straight is essential to obtaining an interpretable estimate.

This tutorial derives the maximum likelihood estimator (MLE) for the exponential rate using complete, independent loss observations. “Complete” means that each loss amount is observed exactly, rather than being censored or included only after passing a reporting threshold. Those data mechanisms change the likelihood and are not handled by the complete-data formula below.

Model and assumptions: Let $X_1,\ldots,X_n$ be independent and identically distributed loss amounts, each with an exponential distribution of rate $\lambda$, where $\lambda>0$. Thus, for $x\geq 0$, $$ f_X(x;\lambda)=\lambda e^{-\lambda x}, \qquad F_X(x;\lambda)=1-e^{-\lambda x}. $$ Each $X_i$ is an exact observed loss, the observations come from this same model, and no truncation, censoring, or additional selection affects which losses are recorded.

Constructing the Likelihood

For observed values $x_1,\ldots,x_n$, the likelihood is their joint density viewed as a function of the unknown parameter $\lambda$. Independence makes the joint density the product of the individual densities. The values $x_i$ are held fixed when maximizing; only $\lambda$ varies.

L(\lambda) =\prod_{i=1}^{n}\lambda e^{-\lambda x_i} =\lambda^n e^{-\lambda\sum_{i=1}^{n}x_i}, \qquad \lambda>0.

Writing $S=\sum_{i=1}^{n}x_i$, the log likelihood is $$ \ell(\lambda)=\log L(\lambda)=n\log\lambda-\lambda S. $$ The logarithm is used because it changes a product into a sum and does not change which parameter maximizes the likelihood: the logarithm is strictly increasing.

Key result: If $S=\sum_{i=1}^{n}x_i>0$, the MLE of the exponential rate is $$ \widehat{\lambda}_{\mathrm{MLE}}=\frac{n}{\sum_{i=1}^{n}x_i}. $$ Its units are the inverse units of loss. For example, if losses are measured in thousands of dollars, the estimate is a rate per thousand dollars.

Why the Rate Estimate Has This Form

Differentiate the log likelihood with respect to $\lambda$, set the score equal to zero, and check the curvature. This gives both the candidate estimate and the reason it is a maximum.

Differentiate.
For $S>0$, the log likelihood is $\ell(\lambda)=n\log\lambda-\lambda S$. Its derivative is $\ell'(\lambda)=n/\lambda-S$.

Solve the score equation.
Setting $n/\lambda-S=0$ gives $n/\lambda=S$, hence $\lambda=n/S$. Since $n>0$ and $S>0$, this solution is in the permitted parameter space.

Verify the maximum.
The second derivative is $\ell''(\lambda)=-n/\lambda^2<0$ for every $\lambda>0$, so the log likelihood is strictly concave. Also, $L(\lambda)$ tends to zero as $\lambda$ approaches zero or infinity when $S>0$. The unique stationary point is therefore the global maximum.

The numerator is $n$, not 1: each observed density contributes one factor of $\lambda$, so the product contributes $\lambda^n$. If all recorded values are zero, then $S=0$ and $L(\lambda)=\lambda^n$, which increases without bound as $\lambda$ increases. In that exceptional data configuration there is no finite MLE. Under a continuous exponential model, an exact zero has probability zero, but the mathematical boundary case is useful to recognize.

Worked Examples

Worked Example: Estimating the Rate From Five Complete Losses

A hypothetical portfolio records five exact losses, in thousands of dollars: $1,2,3,4,$ and $5$. Assume these are independent observations from one exponential severity distribution. Find the MLE of the rate and the maximized log likelihood.

Method and sufficient calculations. The sample size is $n=5$, and the total observed loss, in the stated units, is $$ S=1+2+3+4+5=15. $$ Therefore, $$ \widehat{\lambda}=\frac{n}{S}=\frac{5}{15}=\frac{1}{3} $$ per thousand dollars.

The log likelihood for any positive rate is $\ell(\lambda)=5\log\lambda-15\lambda$. At the estimate, the second term is $15(1/3)=5$, so $$ \ell(\widehat{\lambda})=5\log(1/3)-5. $$ Using $\log(1/3)=-1.098612288\ldots$, the intermediate products are $5\log(1/3)=-5.493061443\ldots$ and $5\log(1/3)-5=-10.493061443\ldots$. Rounded only at the final step, the maximized log likelihood is $-10.4931$ to four decimal places.

Check and mistake avoided. The estimate is positive and has units of inverse thousands of dollars, as required. A common error is to use $1/S=1/15$ as the rate; that omits the factor $n$ contributed by the five density terms. The correct score is $5/\lambda-15=0$, which gives $1/3$, not $1/15$.

Worked Example: Preserving the Dollar Units

A hypothetical insurer observes four complete losses of $\$500$, $\$1{,}000$, $\$1{,}500$, and $\$3{,}000$. Under an independent, identically distributed exponential model, estimate the rate per dollar. Then use the fitted model to calculate the probability that a loss exceeds $\$3{,}000$.

Method. The sample size is $n=4$, and the total is $$ S=500+1{,}000+1{,}500+3{,}000=\$6{,}000. $$ Thus the rate estimate is $$ \widehat{\lambda}=\frac{4}{6{,}000} =\frac{1}{1{,}500} =0.000666666\ldots $$ per dollar. To four significant figures, this is $0.0006667$ per dollar.

For an exponential distribution, $P(X>t)=e^{-\lambda t}$ for $t\geq0$, since $1-F_X(t)=e^{-\lambda t}$. Substituting the fitted rate and the dollar threshold gives $$ P_{\widehat{\lambda}}(X>\$3{,}000) =e^{-(1/1{,}500)(3{,}000)} =e^{-2} =0.135335\ldots. $$ The fitted probability is $0.1353$ to four decimal places.

Check and mistake avoided. The product of the rate per dollar and the dollar threshold is dimensionless, and the resulting probability is between 0 and 1. Do not insert $3$ as the threshold while retaining a rate per dollar: that would silently treat the threshold as $\$3$, not $\$3{,}000$. Equivalently, the losses and threshold could all be expressed in thousands of dollars, but the estimated rate would then be per thousand dollars.

Worked Example: Comparing Candidate Rates With the Log Likelihood

A hypothetical specialty coverage program has four complete observed losses, measured in thousands of dollars: $0.5,1.0,1.5,$ and $2.0$. Assume one exponential rate applies to all four losses. Estimate the rate and verify that its log likelihood exceeds the log likelihood at rates $0.4$ and $1.2$, both expressed per thousand dollars.

Estimate. Here $n=4$, and $$ S=0.5+1.0+1.5+2.0=5.0. $$ The score equation gives $$ \widehat{\lambda}=\frac{4}{5.0}=0.8 $$ per thousand dollars. The log likelihood at any candidate rate is $\ell(\lambda)=4\log\lambda-5\lambda$.

At the estimate, $$ \ell(0.8)=4\log(0.8)-5(0.8) =4(-0.223143551\ldots)-4 =-4.892574204\ldots. $$ At the first comparison rate, $$ \ell(0.4)=4\log(0.4)-5(0.4) =4(-0.916290732\ldots)-2 =-5.665162928\ldots. $$ At the second comparison rate, $$ \ell(1.2)=4\log(1.2)-5(1.2) =4(0.182321557\ldots)-6 =-5.270713773\ldots. $$ Thus, rounded to four decimal places, the log likelihoods are $-4.8926$, $-5.6652$, and $-5.2707$, respectively. The estimate gives the largest log likelihood among these values.

Check and mistake avoided. The analytical concavity argument establishes that $0.8$ is the unique global maximizer, not merely the best of three tested values. Comparing likelihoods is valid here because they use the same observations and model; comparing candidate rates without evaluating the likelihood would not verify which one fits better.

Units, Interpretation, and Scope

A rate is measured per unit of loss. If the unit changes, the numerical rate changes inversely, while the fitted distribution for the same monetary amounts remains unchanged. For example, an estimate stated per thousand dollars must be divided by $1{,}000$ to express it per dollar. A practical check is that $\widehat{\lambda}x$ must be dimensionless whenever it appears in an exponential such as $e^{-\widehat{\lambda}x}$.

The sample total $S$ is sufficient for this MLE calculation: once $n$ and $S$ are known, the rate estimate is determined. This does not mean that the individual observations are irrelevant to every possible analysis; it means only that the likelihood for this model depends on them through their sum.

The complete-data assumption matters. If a loss is recorded only when it exceeds a deductible, or if a record says only that a loss exceeded a limit, the observed contribution to the likelihood is not simply the exponential density at an exact loss amount. Applying $n/S$ without accounting for the observation mechanism can then produce an inappropriate estimate. For the complete, exact observations considered here, the density product and the derived MLE apply directly.

Key takeaway: For $n$ independent, complete exponential losses with positive total $S$, the rate MLE is $n/S$. First check that the model and data assumptions apply, then sum losses in consistent units, retain the factor $n$, and report the rate with inverse-loss units.

Check Your Understanding

For each problem, assume the stated exact losses are independent observations from one exponential model. Keep track of the measurement units when finding the rate.

A hypothetical equipment insurer observes three complete losses of $2,4,$ and $6$ hundreds of dollars. Find the MLE of the rate in units of inverse hundreds of dollars. (Answer: $0.25$ per hundred dollars.)
Five exact losses are $\$100,\$200,\$300,\$400,$ and $\$500$. Find the rate MLE per dollar and state the total loss used. (Answer: $S=\$1{,}500$, $\widehat{\lambda}=1/300=0.003333\ldots$ per dollar.)
Four complete losses, in thousands of dollars, are $0.5,1.0,1.5,$ and $3.0$. Find the rate MLE and then the fitted probability that a loss exceeds $1.5$ thousand dollars. Give the probability to four decimal places. (Answer: $\widehat{\lambda}=2/3$ per thousand dollars; $P(X>1.5)=e^{-1}=0.3679$.)
For three complete losses with total $9$ dollars, calculate the rate MLE and the maximized log likelihood. (Answer: $\widehat{\lambda}=1/3$ per dollar; $\ell(\widehat{\lambda})=3\log(1/3)-3=-6.2958$ to four decimal places.)
Six complete losses have a total of $24$ thousands of dollars. Find the rate estimate and the maximized log likelihood. (Answer: $\widehat{\lambda}=0.25$ per thousand dollars; $\ell(\widehat{\lambda})=6\log(0.25)-6=-14.3178$ to four decimal places.)

What You'll Learn