Shifted Exponential Severity With Fixed Threshold

Why the Threshold Changes the Likelihood

A severity sample may consist only of losses above a specified amount. When the threshold is fixed and known, the model describes the excess above that threshold, not the total loss measured from zero. This distinction determines the likelihood and the maximum likelihood estimate (MLE).

Let $c$ be a known fixed threshold, and let $X$ denote a loss with $X\ge c$. The shifted exponential model says that $X-c$ has an exponential distribution with rate $\lambda>0$. Equivalently, $X=c+Z$, where $Z$ is exponential with rate $\lambda$. The threshold is not an unknown parameter to be estimated.

Definition: A shifted exponential severity with known threshold $c$ and rate $\lambda>0$ has density $$ f_X(x)= \begin{cases} \lambda e^{-\lambda(x-c)}, & x\ge c,\\ 0, & x<c. \end{cases} $$ For $x\ge c$, its distribution and survival functions are $F_X(x)=1-e^{-\lambda(x-c)}$ and $S_X(x)=e^{-\lambda(x-c)}$. The mean loss is $E[X]=c+1/\lambda$, so the mean excess over the threshold is $1/\lambda$.

There are two common ways to arrive at this model. The losses might directly follow a shifted exponential distribution. Alternatively, they might be observations from an exponential severity distribution that are included only when they exceed $c$. In the second case, the density conditional on exceeding the threshold is $f_X(x\mid X>c)=f_X(x)/S_X(c)=\lambda e^{-\lambda(x-c)}$ for $x>c$. Thus, the conditional excess also has an exponential distribution. The likelihood below applies when the observed severity values are complete and the threshold is known.

Constructing the Likelihood From Excesses

Suppose $x_1,\ldots,x_n$ are independent complete observations from the shifted exponential model. Define the observed excesses by $z_i=x_i-c$, and their total by $T=\sum_{i=1}^n(x_i-c)$. Since $c$ is fixed, each density contribution is based on an excess, not on the full loss.

L(\lambda)=\prod_{i=1}^n \lambda e^{-\lambda(x_i-c)} =\lambda^n e^{-\lambda T},\qquad \ell(\lambda)=n\log\lambda-\lambda T.

The likelihood is a function of $\lambda$, with the observed losses held fixed. It contains no factor for the probability of being above $c$: these observations are modeled as shifted severities, or as a sample conditional on exceeding $c$. The number of observations $n$ is treated as fixed; this is a severity estimation problem, not an estimate of how often losses occur.

Theorem: If $n\ge1$, each observed value satisfies $x_i\ge c$, and $T=\sum_{i=1}^n(x_i-c)>0$, the MLE of the rate is $$ \widehat{\lambda}=\frac{n}{T} =\frac{1}{\frac{1}{n}\sum_{i=1}^n(x_i-c)}. $$ The fitted mean excess is $1/\widehat{\lambda}=T/n$, and the fitted mean total loss is $c+T/n$.

Why the MLE Has This Form

Maximizing the likelihood is equivalent to maximizing its logarithm, which is simpler to differentiate. The score equation is obtained by differentiating with respect to $\lambda$ and setting the result to zero.

Take the logarithm.
For the observed data, $\ell(\lambda)=n\log\lambda-\lambda T$, where $T$ is the sum of the excesses.

Set the score to zero.
The derivative is $\ell'(\lambda)=n/\lambda-T$. Thus $n/\lambda-T=0$, which gives $\lambda=n/T$.

Check that the solution is a maximum.
The second derivative is $\ell''(\lambda)=-n/\lambda^2<0$ for every $\lambda>0$. The log-likelihood is strictly concave, so the score equation gives the unique maximum when $T>0$.

This derivation also explains the estimate: a larger average excess indicates a more spread-out exponential tail and therefore a smaller rate. If $T=0$, all observations equal the threshold and $L(\lambda)=\lambda^n$ has no finite maximum over $\lambda>0$. That boundary case is different from the usual continuous-data situation, in which positive excesses occur with probability one.

Worked Example: Estimating a Rate From Dollar Excesses

A hypothetical severity sample follows a shifted exponential distribution with known threshold $c=\$1{,}000$ and unknown rate $\lambda$ per dollar. The four complete losses are $\$1{,}250$, $\$1{,}600$, $\$2{,}050$, and $\$1{,}100$. Find the MLE of $\lambda$, the fitted mean excess, and the fitted mean loss.

Method and assumptions. Use the shifted exponential likelihood for independent complete loss observations with fixed threshold $\$1{,}000$. No interest or mortality assumption is involved.

Calculate the excesses. Subtract the threshold from each loss: $\$250$, $\$600$, $\$1{,}050$, and $\$100$. Their total is $T=250+600+1{,}050+100=\$2{,}000$, and $n=4$.

Estimate the rate. The MLE is $\widehat{\lambda}=n/T=4/2{,}000=0.002$ per dollar.

Interpret the estimate. The fitted mean excess is $1/\widehat{\lambda}=1/0.002=\$500$. The fitted mean total loss is $c+1/\widehat{\lambda}=\$1{,}000+\$500=\$1{,}500$.

Check. The observed average excess is $\$2{,}000/4=\$500$, matching the fitted mean excess. The fitted mean total loss is therefore the threshold plus that average excess, not the excess alone. No rounding is needed.

Common mistake avoided: Using $\sum x_i=\$6{,}000$ in place of $T$ would give $4/6{,}000$, which is not the likelihood estimate for this shifted model. The likelihood uses the excess total $\$2{,}000$.

Units, Interpretation, and a Conditional Probability

The rate has units reciprocal to those used for the loss amounts. If losses are recorded in dollars, $\lambda$ is per dollar; if recorded in thousands of dollars, it is per thousand dollars. Changing units changes the numerical rate but not the underlying fitted severity distribution. Keeping the threshold, observations, and rate in consistent units is essential.

Once $\lambda$ is estimated, probabilities for future losses follow from the fitted survival function. For a level $y\ge c$, the fitted probability that a future loss exceeds $y$ is $e^{-\widehat{\lambda}(y-c)}$. The amount in the exponent is again the distance from the threshold. This fitted probability is conditional on the shifted severity model; it does not estimate the frequency of losses.

Worked Example: Recording Losses in Thousands

A hypothetical portfolio is modeled by a shifted exponential severity with known threshold $c=5$, where all monetary amounts are in thousands of dollars. Five complete losses are $6.2$, $7.1$, $5.4$, $8.3$, and $8.0$. Estimate the rate in the stated units and the probability that a future loss exceeds $7.5$ thousand dollars.

Method and assumptions. Treat the five values as independent complete observations from the shifted exponential distribution, with unknown rate per thousand dollars and fixed threshold $5$. No interest or mortality assumption is involved.

Calculate the excess total. The excesses, in thousands of dollars, are $1.2$, $2.1$, $0.4$, $3.3$, and $3.0$. Their sum is $T=1.2+2.1+0.4+3.3+3.0=10.0$, with $n=5$.

Estimate the rate. $\widehat{\lambda}=5/10.0=0.5$ per thousand dollars. The estimated mean excess is $1/0.5=2.0$ thousand dollars, and the estimated mean total loss is $5+2.0=7.0$ thousand dollars.

Calculate the probability. The level $7.5$ is $7.5-5=2.5$ thousand dollars above the threshold. Therefore, $P(X>7.5)$ under the fitted model is $e^{-0.5(2.5)}=e^{-1.25}\approx0.2865$, to four decimal places.

Check. The fitted mean excess $2.0$ equals the observed average excess $10.0/5$. The probability calculation uses $2.5$, the distance above the threshold, and not $7.5$, the total loss level.

Common mistake avoided: Entering the dollar values as though they were dollars would create a rate with inconsistent units. Here all amounts are in thousands, so the estimated rate is $0.5$ per thousand dollars.

Worked Example: Likelihood and Tail Probability

A hypothetical severity sample has known threshold $c=\$1{,}000$ and follows a shifted exponential model with unknown rate per dollar. The four complete losses are $\$1{,}200$, $\$1{,}400$, $\$1{,}600$, and $\$1{,}800$. Find the rate MLE, the likelihood evaluated at that estimate, and the fitted probability of a loss above $\$1{,}500$.

Method and assumptions. Use the product of the four shifted exponential densities. The threshold is fixed, and the losses are complete and independent. No interest or mortality assumption is involved.

Calculate the excesses and estimate. The excesses are $\$200$, $\$400$, $\$600$, and $\$800$, so $T=\$2{,}000$ and $n=4$. Thus $\widehat{\lambda}=4/2{,}000=0.002$ per dollar.

Evaluate the likelihood. At the MLE, the likelihood is $L(\widehat{\lambda})=(0.002)^4e^{-0.002(2{,}000)}=(0.002)^4e^{-4}$. Since $(0.002)^4=0.000000000016$ and $e^{-4}\approx0.0183156$, this is approximately $2.9305\times10^{-13}$. This density likelihood is not a probability.

Calculate the tail probability. The level $\$1{,}500$ is $\$500$ above the threshold, so the fitted probability is $e^{-0.002(500)}=e^{-1}\approx0.3679$, to four decimal places.

Check. The sample mean excess is $\$2{,}000/4=\$500$, and the fitted mean total loss is $\$1{,}500$. The probability above that mean is $e^{-1}$, consistent with the shifted exponential survival formula.

Common mistake avoided: The likelihood is the product of density values and can be much smaller than one; it should not be interpreted as the probability of observing exactly these four continuous losses.

Common Exam Traps

Using total losses in the exponential term. The exponent is $-\lambda(x_i-c)$, not $-\lambda x_i$. The model starts at $c$.
Estimating the threshold when it is fixed. The threshold is supplied as known. It is subtracted from each observation, not estimated along with $\lambda$.
Mixing monetary units. If the losses and threshold are in thousands, the excess total is in thousands and the rate is per thousand. Convert all amounts together if changing units.
Adding a frequency component. The likelihood uses $n$ observed severity values as a fixed sample size. Modeling how many losses occur is a separate frequency problem.
Using the wrong tail distance. For $y\ge c$, the survival probability is $e^{-\widehat{\lambda}(y-c)}$, not $e^{-\widehat{\lambda}y}$.

Key takeaway: For complete shifted exponential losses above a known fixed threshold, subtract the threshold from every loss, sum those excesses, and estimate the rate as the number of observations divided by that sum. Check that the threshold and all losses use the same units.

Check Your Understanding

For each problem, use the shifted exponential likelihood with the stated fixed threshold. Treat the observations as complete and independent, and report probabilities to four decimal places.

A hypothetical loss sample has threshold $\$400$ and losses $\$700$, $\$900$, and $\$1{,}000$. Find the rate MLE per dollar. (Answer: $3/1{,}400\approx0.002143$ per dollar.)
With threshold $\$2{,}500$, four losses are $\$3{,}000$, $\$3{,}900$, $\$4{,}100$, and $\$5{,}000$. Find the rate MLE and the fitted mean total loss. (Answer: $\widehat{\lambda}=1/1{,}500\approx0.0006667$ per dollar; fitted mean total loss $=\$4{,}000$.)
With threshold $\$100$, a sample consists of $\$220$, $\$370$, $\$610$, and $\$800$. Estimate the rate and the fitted probability that a future loss exceeds $\$500$. (Answer: $\widehat{\lambda}=0.0025$ per dollar; probability $e^{-1}\approx0.3679$.)
Six complete losses have known threshold $\$1{,}000$, and their total is $\$9{,}600$. Find the rate MLE. (Answer: $6/(9{,}600-6{,}000)=1/600\approx0.001667$ per dollar.)
A sample has threshold $\$800$ and losses $\$1{,}000$, $\$1{,}200$, $\$1{,}500$, and $\$1{,}900$. Estimate the rate and the fitted probability that a future loss exceeds $\$2{,}000$. (Answer: $\widehat{\lambda}=1/600\approx0.001667$ per dollar; probability $e^{-2}\approx0.1353$.)

What You'll Learn