Why the Weibull Model Is Useful for Severity
Severity data are positive and often vary substantially in scale. The Weibull distribution is a flexible model for positive losses: its shape parameter determines whether the hazard rises, falls, or stays level as loss size increases. Estimating that shape is therefore useful both for describing a fitted model and for understanding its implications for the relative frequency of larger losses.
Here, the hazard is the instantaneous rate of loss occurrence over increasing loss size, conditional on the loss having exceeded the size already reached. It is a property of the severity distribution, not a claim frequency rate. The estimation methods below assume complete, exact observations; they do not treat claims censored by a policy limit or samples observed only above a deductible.
How Shape Determines the Hazard
The expression for the hazard separates into a positive constant, \(k/\theta^k\), multiplied by \(x^{k-1}\). Consequently, its direction as \(x\) increases is determined entirely by the exponent \(k-1\).
| Shape | Hazard as loss size increases | Interpretation |
|---|---|---|
| \(0<k<1\) | Decreases | The conditional rate of reaching still larger loss sizes declines. |
| \(k=1\) | Constant | The Weibull reduces to the exponential distribution, with hazard \(1/\theta\). |
| \(k>1\) | Increases | The conditional rate of reaching still larger loss sizes rises. |
These descriptions concern hazard, not the density itself. A rising hazard does not mean that the density rises at every loss size; the density can rise initially and then fall. Nor does a decreasing hazard mean that losses are necessarily small. The scale and shape together determine the distribution.
Worked Example: Interpreting a Decreasing Severity Hazard
A hypothetical portfolio is modeled with a Weibull severity distribution having shape \(k=0.5\) and scale \(\theta=4\), where \(X\) and \(\theta\) are measured in hundreds of dollars. Classify the hazard and compare it at loss sizes of 1, 4, and 16 hundred dollars.
Method. Substitute the parameters into the Weibull hazard and compare its values. This avoids the common mistake of interpreting the shape from the density rather than from the hazard formula.
The hazard is $$ h_X(x)=\frac{0.5}{4}\left(\frac{x}{4}\right)^{-0.5} =0.125\left(\frac{x}{4}\right)^{-0.5}. $$ Since \(k=0.5<1\), it decreases with \(x\).
At \(x=1\), $$ h_X(1)=0.125(0.25)^{-0.5} =0.125(2)=0.2500. $$ At \(x=4\), $$ h_X(4)=0.125(1)^{-0.5}=0.1250. $$ At \(x=16\), $$ h_X(16)=0.125(4)^{-0.5} =0.125(0.5)=0.0625. $$ Thus the hazard halves each time \(x\) is multiplied by 4, consistent with the power \(x^{-0.5}\). The units of each hazard value are per hundred dollars of loss size.
Check. The calculated hazards decline from \(0.2500\) to \(0.1250\) to \(0.0625\), as required when \(0<k<1\).
Maximum Likelihood from Complete Severity Data
Suppose \(x_1,\ldots,x_n\) are positive, exact observations from independent losses with the same Weibull distribution. “Complete” means each sampled loss is observed at its actual value, without censoring, truncation, or rounding. Under these assumptions, the likelihood is the product of the Weibull densities. Taking its logarithm gives $$ \ell(k,\theta)=n\ln k-nk\ln\theta+(k-1)\sum_{i=1}^n\ln x_i -\sum_{i=1}^n\left(\frac{x_i}{\theta}\right)^k. $$
For a fixed positive \(k\), differentiate with respect to \(\theta\) and set the derivative equal to zero: $$ \frac{\partial\ell}{\partial\theta} =-\frac{nk}{\theta} +\frac{k}{\theta^{k+1}}\sum_{i=1}^n x_i^k=0. $$ Rearranging yields the maximum-likelihood scale for that fixed shape: $$ \widehat{\theta}(k)= \left(\frac{1}{n}\sum_{i=1}^n x_i^k\right)^{1/k}. $$ This is a profile estimate: for each candidate shape, it selects the scale that maximizes the likelihood at that shape.
Substituting this scale into the likelihood and differentiating with respect to \(k\) gives the shape score equation $$ \frac{1}{k}+\frac{1}{n}\sum_{i=1}^n\ln x_i -\frac{\sum_{i=1}^n x_i^k\ln x_i}{\sum_{i=1}^n x_i^k}=0. $$ The equation generally has no closed-form solution, so a numerical root-finding method is used. The following justification explains why the root is well behaved for a nonconstant sample.
Let \(g(k)\) be the left side of the shape score equation. Its weighted-log term uses weights \(w_i=x_i^k/\sum_j x_j^k\), which are positive and sum to 1.
The derivative of the weighted mean of \(\ln x_i\) is the weighted variance of the values \(\ln x_i\). Therefore, $$ g'(k)=-\frac{1}{k^2}-\operatorname{Var}_w(\ln x_i)<0. $$ The inequality holds for every \(k>0\).
As \(k\) approaches zero from above, \(g(k)\) tends to positive infinity. For a nonconstant sample, as \(k\) tends to infinity, the weights concentrate on observations equal to the sample maximum, so $$ g(k)\longrightarrow \frac{1}{n}\sum_{i=1}^n\ln x_i-\ln x_{\max}<0. $$
Continuity, the opposite signs near the two limits, and strict decrease imply exactly one positive solution to the shape score equation for a nonconstant sample.
The calculation uses logarithms of loss values, so the numerical units must be consistent throughout. Changing all observations from thousands of dollars to dollars changes the fitted scale by a factor of 1,000, but does not change the fitted shape.
Worked Example: Fitting Both Weibull Parameters by Likelihood
A hypothetical set of three complete severities, measured in thousands of dollars, is \(1,2,4\). Assume independent observations from one Weibull distribution. Estimate the shape and scale by maximum likelihood.
Method. Solve the shape score equation numerically, then substitute the resulting shape into the conditional scale estimate. This avoids treating the scale estimate as independent of the shape.
The sample mean of the logarithms is $$ \frac{\ln 1+\ln 2+\ln 4}{3} =\frac{0+\ln 2+2\ln 2}{3} =\ln 2\approx0.693147. $$ At a trial value \(k=2.012\), the powers are \(2^k\approx4.03341\) and \(4^k\approx16.26839\). The weighted mean of the logarithms is $$ \frac{2^k\ln 2+4^k\ln 4}{1+2^k+4^k} =\ln 2\left(\frac{4.03341+2(16.26839)} {1+4.03341+16.26839}\right) \approx1.18997. $$ The score function at this trial value is therefore $$ g(2.012)\approx\frac{1}{2.012}+0.693147-1.18997 \approx0.00019. $$ The score is already close to zero. Numerical root-finding gives \(\widehat{k}\approx2.0125\).
At this fitted shape, \(2^{\widehat{k}}\approx4.03481\) and \(4^{\widehat{k}}\approx16.27968\). Thus the fitted scale is $$ \widehat{\theta} =\left(\frac{1+2^{\widehat{k}}+4^{\widehat{k}}}{3}\right)^{1/\widehat{k}} \approx\left(\frac{1+4.03481+16.27968}{3}\right)^{1/2.0125} \approx2.6493. $$ To three decimal places, the estimates are \(\widehat{k}=2.013\) and \(\widehat{\theta}=2.649\) thousand dollars. The fitted hazard increases because \(\widehat{k}>1\).
Check. The positive score at \(2.012\) and the strictly decreasing score imply that the root is slightly larger; the numerical value \(2.0125\) is consistent with that direction. The scale is in thousands of dollars, the same units as the data.
Estimating Scale When Shape Is Specified
Sometimes a problem specifies the shape rather than asking for both parameters. The scale formula still follows from the likelihood, but it is then a conditional maximum-likelihood estimate: it is optimal among scales at that specified shape. It is not, by itself, a joint fit of the shape and scale.
Worked Example: Profile Scale Estimate at a Specified Shape
A hypothetical sample of three complete losses, measured in hundreds of dollars, is \(1,2,3\). Assume independent Weibull observations with the shape fixed at \(k=2\). Find the maximum-likelihood scale and verify that it improves the log-likelihood relative to \(\theta=2\).
Method. Use the conditional scale estimate and substitute it into the log-likelihood. This avoids the mistake of estimating the shape when the problem has fixed it.
The scale estimate is $$ \widehat{\theta} =\left(\frac{1^2+2^2+3^2}{3}\right)^{1/2} =\sqrt{\frac{14}{3}} \approx2.1602 $$ hundreds of dollars.
For this sample and \(k=2\), the log-likelihood is $$ \ell(2,\theta)=3\ln2-6\ln\theta+\ln1+\ln2+\ln3 -\frac{1^2+2^2+3^2}{\theta^2}. $$ At \(\widehat{\theta}=\sqrt{14/3}\), the last term is \(-14/(14/3)=-3\). Using \(\ln2\approx0.693147\), \(\ln3\approx1.098612\), and \(\ln\widehat{\theta}\approx0.770222\), $$ \ell(2,\widehat{\theta})\approx2.079442-4.621332+1.791759-3 =-3.750131. $$ At the comparison value \(\theta=2\), the final term is \(-14/4=-3.5\), so $$ \ell(2,2)\approx2.079442-4.158883+1.791759-3.5 =-3.787682. $$ The fitted scale has the larger log-likelihood, as expected. Rounded to three decimals, the two log-likelihoods are \(-3.750\) and \(-3.788\), respectively.
Check. The conditional estimate is positive and has the same units as the losses. Its log-likelihood exceeds the value at the comparison scale.
Assumptions and Common Exam Traps
The likelihood equations above apply to independent, identically distributed, positive, fully observed Weibull losses. The observations must be on a common measurement scale. If some losses are only known to exceed a policy limit, or if data are observed only when losses exceed a deductible, the likelihood must account for that observation mechanism. Applying the complete-data density product directly would then be incorrect.
- Confusing shape with scale. Shape controls the direction of the hazard; scale changes the horizontal size of the loss distribution. A larger scale does not change whether the hazard increases or decreases.
- Using the wrong weighted mean. The shape score weights \(\ln x_i\) by \(x_i^k\). An unweighted average of log losses is only one term in the equation, not a replacement for the weighted term.
- Assuming the shape equation has an algebraic solution. In general, it must be solved numerically. For a nonconstant sample its score function is strictly decreasing, which supports a reliable bracketed root search.
- Mixing units within one calculation. If data are converted to dollars, every observation must be converted. The fitted shape remains unchanged, while the scale converts by the same unit factor.
- Calling an incomplete-data estimate a complete-data MLE. Censoring and truncation alter the likelihood contribution. They cannot be handled merely by inserting observed payments into the complete-data formula.
Check Your Understanding
For each problem, assume any stated complete observations are independent and positive. Give numerical answers to a sensible precision.
- A hypothetical sample of losses measured in thousands of dollars is \(1,2,4\). Its fitted Weibull shape is approximately \(2.0125\). Use the conditional scale formula to find the fitted scale, and state whether the fitted hazard increases or decreases. (Answer: \(\widehat{\theta}\approx2.649\) thousand dollars; the hazard increases.)
- A Weibull model has shape \(k=1.5\). By what factor does its hazard change when loss size doubles? State whether it rises or falls. (Answer: It is multiplied by \(2^{0.5}\approx1.4142\), so it rises by a factor of about 1.4142.)
- For complete losses \(2\) and \(4\), measured in hundreds of dollars, the shape is fixed at \(k=2\). Calculate the conditional maximum-likelihood scale and classify the hazard. (Answer: \(\widehat{\theta}=\sqrt{(2^2+4^2)/2}=\sqrt{10}\approx3.162\) hundred dollars; the hazard increases.)
- A sample has all observations equal to the same positive value. Explain why the nonconstant-sample argument for a unique finite shape root does not apply. (Answer: The limiting score as \(k\to\infty\) is no longer strictly negative; the data provide no variation in log losses, so the stated uniqueness argument for a nonconstant sample cannot be used.)