Shifted and Threshold Severity Models

Why Thresholds Need Care

A severity model may assign probability only to losses at or above a specified minimum. That minimum is the threshold; it is not automatically a scale parameter to be estimated from the sample. When a threshold is known from the model specification, estimation proceeds with it fixed. Before calculating any parameter estimate, check that every observed loss is at least that threshold. If even one observation falls below it, the likelihood is zero for every value of the model’s remaining parameters.

This tutorial fits two continuous severity models with known thresholds: the Pareto Type I model and the shifted exponential model. Both use a threshold, but their other parameters have different meanings: the Pareto parameter controls tail shape, while the shifted exponential parameter controls the scale of the excess above threshold. The calculations assume independent, identically distributed, complete observations of individual losses, with no censoring or truncation.

Definition: A known-threshold model has a specified threshold $\theta$, held fixed during parameter estimation. Its likelihood is zero if any observed value $x_i$ is below $\theta$. A sample that satisfies $x_i\geq\theta$ for every observation is supported by the model, but this support check alone does not establish that the model describes the data well.

Known-Threshold Pareto Type I

For the Pareto Type I model, $\theta>0$ is the known minimum loss and $\alpha>0$ is the shape parameter. Its density is

f_X(x)=\frac{\alpha\theta^\alpha}{x^{\alpha+1}}, \qquad x\geq\theta.

The distribution assigns zero density below $\theta$. For $n$ independent observations $x_1,\ldots,x_n$, the likelihood is therefore zero if any $x_i<\theta$. Otherwise,

L(\alpha)=\prod_{i=1}^{n}\frac{\alpha\theta^\alpha}{x_i^{\alpha+1}}.

Taking logarithms and collecting terms gives

\ell(\alpha)=n\ln\alpha+n\alpha\ln\theta-(\alpha+1)\sum_{i=1}^{n}\ln x_i =n\ln\alpha-\alpha D-\sum_{i=1}^{n}\ln x_i, \qquad D=\sum_{i=1}^{n}\ln\left(\frac{x_i}{\theta}\right).

Each term in $D$ is nonnegative when the sample is supported. The quantity $D$ summarizes the observed excesses on a logarithmic scale. It is not the sum of the dollar excesses $x_i-\theta$.

Maximum-likelihood estimate: If all observations satisfy $x_i\geq\theta$ and at least one is strictly greater than $\theta$, then $D>0$ and the Pareto Type I maximum-likelihood estimate is $\widehat{\alpha}=n/D$. If all observations equal $\theta$, then $D=0$, the likelihood is unbounded as $\alpha\to\infty$, and there is no finite maximum-likelihood estimate.

The estimate’s behavior follows directly from the log-likelihood. When $D>0$, the score equation has a unique solution, and the second derivative is negative throughout the parameter space:

\frac{d\ell}{d\alpha}=\frac{n}{\alpha}-D=0 \quad\Longrightarrow\quad \widehat{\alpha}=\frac{n}{D}, \qquad \frac{d^2\ell}{d\alpha^2}=-\frac{n}{\alpha^2}<0.

Thus the stationary point is the maximum. If $D=0$, the log-likelihood instead reduces to $n\ln\alpha-\sum_i\ln x_i$, which increases without bound as $\alpha$ increases. It would be incorrect to apply $n/D$ in that case.

Worked Example: Fitting a Pareto Model with a Known Minimum

A hypothetical portfolio is modeled using a Pareto Type I severity distribution with known threshold $\theta=\$80$. Four complete, independent observed losses are $\$80,\$160,\$320,\$640$. Estimate $\alpha$ by maximum likelihood, and check the score at the estimate.

Step 1: Check support. The smallest observed loss is $\$80$, so every observation satisfies $x_i\geq\theta$. The likelihood is not zero on account of the threshold.

Step 2: Calculate the log-ratio sum. The ratios $x_i/\theta$ are $1,2,4,8$. Therefore,

D=\ln(1)+\ln(2)+\ln(4)+\ln(8)=\ln(64)=4.1589.

Step 3: Estimate the shape. There are $n=4$ observations, so

\widehat{\alpha}=\frac{4}{\ln(64)} =\frac{4}{4.1589} \approx 0.9618.

Step 4: Verify the maximum. At the estimate, $n/\widehat{\alpha}=4/0.9618\approx4.1589=D$, so the score is zero. The second derivative is $-4/\widehat{\alpha}^2<0$, confirming a maximum. The estimate is less than 1, so the fitted Pareto model has no finite mean; the shape estimate should not be misread as an estimate of the threshold or as a finite expected loss.

Known-Threshold Shifted Exponential

In the shifted exponential model, $\theta$ is again a known minimum loss, but the parameter $\beta>0$ is a scale parameter measured in the same units as the loss. The model can be described by $X=\theta+Z$, where $Z$ has an exponential distribution with mean $\beta$. Its density is

f_X(x)=\frac{1}{\beta}\exp\left(-\frac{x-\theta}{\beta}\right), \qquad x\geq\theta.

For supported observations, define the total excess $T=\sum_{i=1}^{n}(x_i-\theta)$. The log-likelihood, as a function of $\beta$, is

\ell(\beta)=-n\ln\beta-\frac{T}{\beta}.

Maximum-likelihood estimate: If every observation satisfies $x_i\geq\theta$ and $T>0$, the shifted exponential maximum-likelihood estimate is $\widehat{\beta}=T/n$, the sample mean excess above the known threshold. If all observations equal $\theta$, then $T=0$, the likelihood is unbounded as $\beta\downarrow0$, and there is no finite maximum-likelihood estimate for $\beta>0$.

To justify the estimate, differentiate the log-likelihood and solve its score equation. When $T>0$, the solution is positive; the second derivative at that solution is negative:

\frac{d\ell}{d\beta}=-\frac{n}{\beta}+\frac{T}{\beta^2}=0 \quad\Longrightarrow\quad \widehat{\beta}=\frac{T}{n}, \qquad \left.\frac{d^2\ell}{d\beta^2}\right|_{\beta=T/n} =-\frac{n}{\widehat{\beta}^{\,2}}<0.

This also explains why $\theta$ and $\beta$ must not be swapped: $\theta$ sets the lower endpoint of the distribution, while $\beta$ describes the average amount above that endpoint. The fitted mean loss is $\theta+\widehat{\beta}$, not $\widehat{\beta}$ alone.

Worked Example: Estimating a Shifted Exponential Scale

Suppose a hypothetical loss portfolio is modeled by a shifted exponential distribution with known threshold $\theta=\$15$. Four complete, independent losses are $\$17,\$19,\$23,\$25$. Find the maximum-likelihood estimate of $\beta$ and the fitted mean loss.

Step 1: Check support and calculate excesses. Each loss is at least $\$15$. The excesses are $\$17-\$15=\$2$, $\$19-\$15=\$4$, $\$23-\$15=\$8$, and $\$25-\$15=\$10$.

Step 2: Sum and average the excesses. The total excess is $T=2+4+8+10=\$24$, and there are $n=4$ observations. Thus

\widehat{\beta}=\frac{T}{n}=\frac{\$24}{4}=\$6.

Step 3: Calculate the fitted mean. Since the mean of the shifted exponential model is $\theta+\beta$, the fitted mean loss is $\$15+\$6=\$21$. As a score check, $-4/6+24/6^2=0$, and the second derivative at $\beta=6$ is $-4/6^2<0$. The estimate is therefore a maximum, not just a solution of the score equation.

Comparing the Two Fits and Checking Threshold Support

Both models require an initial support check, but they use different summaries of excess severity when estimating their remaining parameters.

Model	Known threshold	Estimated parameter	Data summary in estimate	No-finite-estimate case
Pareto Type I	$\theta>0$	Shape $\alpha$	Sum of log ratios, $\sum_i\ln(x_i/\theta)$	All observations equal $\theta$
Shifted exponential	$\theta$	Scale $\beta$	Mean excess, $\sum_i(x_i-\theta)/n$	All observations equal $\theta$

For either model, any observation below the assumed threshold makes the likelihood zero. This conclusion comes before the score equation: there is no parameter value that can repair a violated support condition. The likelihood formulas above apply to complete observations. If losses are censored at policy limits or only losses above a deductible are recorded, the likelihood must account for that observation mechanism; simply inserting the recorded amounts into these complete-data formulas is not justified.

A support check is not a goodness-of-fit test. If all observed losses exceed $\theta$, the sample is compatible with the model’s stated lower bound, but it may still show a poor fit in its body or tail. Conversely, changing $\theta$ merely to make an unsupported observation fit changes the model specification; it is not the same as estimating $\alpha$ or $\beta$ with a known threshold.

Worked Example: A Threshold That the Data Do Not Support

An analyst proposes a shifted exponential model with known threshold $\theta=\$30$ for a hypothetical set of three complete losses: $\$28,\$45,\$60$. Determine whether a maximum-likelihood estimate of $\beta$ can be obtained under this specified model.

Step 1: Compare every observation with the threshold. The first loss is $\$28$, and $\$28<\$30$. The other losses exceed the threshold, but support requires every observation to satisfy $x_i\geq\theta$.

Step 2: Evaluate the likelihood implication. The shifted exponential density is zero for $x<\$30$. Therefore the joint likelihood includes the factor $f_X(28)=0$, making the likelihood zero for every $\beta>0$.

Conclusion and check. There is no maximum-likelihood estimate of $\beta$ for this sample under the stated threshold model. Calculating the mean of the excesses for only the two supported observations would silently discard data and would not be the likelihood estimate for the stated sample.

Common Exam Traps

Using dollar excesses in a Pareto estimate. The Pareto denominator is $\sum_i\ln(x_i/\theta)$, not $\sum_i(x_i-\theta)$. The logarithm applies to dimensionless ratios.
Using log ratios for the shifted exponential scale. Its estimate is the arithmetic mean excess, $\sum_i(x_i-\theta)/n$.
Treating the threshold as the scale. In the shifted exponential model, $\beta$ is the mean excess, while the mean loss is $\theta+\beta$. In the Pareto model, $\alpha$ controls shape and $\theta$ sets the lower bound.
Ignoring support. If any observed loss is below the fixed threshold, do not proceed as though the likelihood were positive.
Dividing by zero in a boundary sample. If every observation equals $\theta$, the Pareto denominator $D$ and the shifted exponential total excess $T$ are both zero. Neither formula yields a finite estimate; in each model the likelihood is unbounded at a parameter boundary.

Key takeaway: With a known threshold, first verify that every observation lies in the model’s support. For a supported sample with at least one loss strictly above threshold, estimate the Pareto shape from log ratios and the shifted exponential scale from mean excesses. If all observations equal the threshold, neither model has a finite maximum-likelihood estimate for its remaining parameter.

Check Your Understanding

For each problem, assume the stated observations are independent, complete losses from the specified model. Check support before estimating.

A Pareto Type I model has known threshold $\theta=\$25$, and the observed losses are $\$25,\$50,\$100$. Find the maximum-likelihood estimate of $\alpha$, to four decimals. (Answer: $\widehat{\alpha}=1.4427$.)
A Pareto Type I model has known threshold $\theta=\$40$, and the observed losses are $\$80,\$160,\$320$. Find $\widehat{\alpha}$, to four decimals, and state whether the fitted model has a finite mean. (Answer: $\widehat{\alpha}=0.7213$; the fitted mean is infinite because $\widehat{\alpha}<1$.)
A shifted exponential model has known threshold $\theta=\$8$, and the observed losses are $\$10,\$14,\$20$. Find $\widehat{\beta}$ and the fitted mean loss, to the nearest cent. (Answer: $\widehat{\beta}=\$6.67$; fitted mean $=\$14.67$.)
A shifted exponential model has known threshold $\theta=\$50$, and the observed losses are $\$49,\$55,\$65$. Can a maximum-likelihood estimate of $\beta$ be obtained under the specified model? (Answer: No; the $\$49$ observation is below the threshold, so the likelihood is zero for every $\beta>0$.)
A Pareto Type I model has known threshold $\theta=\$12$, and all three observed losses equal $\$12$. Is there a finite maximum-likelihood estimate of $\alpha$? (Answer: No; $D=0$, and the likelihood is unbounded as $\alpha\to\infty$.)

Model	Known threshold	Estimated parameter	Data summary in estimate	No-finite-estimate case
Pareto Type I	\(\theta>0\)	Shape \(\alpha\)	Sum of log ratios, \(\sum_i\ln(x_i/\theta)\)	All observations equal \(\theta\)
Shifted exponential	\(\theta\)	Scale \(\beta\)	Mean excess, \(\sum_i(x_i-\theta)/n\)	All observations equal \(\theta\)

What You'll Learn