Weibull Severity With Known Shape

Introduction

A severity model describes the distribution of an individual loss. When the loss observations are positive, complete, and modeled by a Weibull distribution, maximum likelihood estimation can be especially direct if the shape parameter is known. The main task is to estimate the scale without confusing it with the sample mean or changing the model’s parameterization partway through a calculation.

This tutorial derives the likelihood and score equation for that setting. It assumes the observed losses are independent and identically distributed, and that each loss is observed in full. The method is aligned with the SOA Exam ASTAM (Advanced Short-Term Actuarial Mathematics) syllabus.

Definition: Use the Weibull parameterization with known shape $k>0$ and unknown scale $\theta>0$. A loss $X$ has density $$ f_X(x;k,\theta)=\frac{k}{\theta}\left(\frac{x}{\theta}\right)^{k-1}\exp\left[-\left(\frac{x}{\theta}\right)^k\right],\qquad x>0. $$ Its survival function is $S_X(x)=\exp[-(x/\theta)^k]$ for $x\geq 0$. In this parameterization, $\theta$ is the scale: $S_X(\theta)=e^{-1}$. It is not generally the mean.

Likelihood for Complete Weibull Observations

Let $x_1,\ldots,x_n$ be observed losses, with every $x_j>0$. Assume they are independent observations from the density above, and that $k$ is fixed at its given value. The likelihood is the product of the individual densities. Define $A=\sum_{j=1}^n x_j^k$. Collecting the factors gives

L(\theta)=\prod_{j=1}^n f_X(x_j;k,\theta) =k^n\left(\prod_{j=1}^n x_j^{k-1}\right)\theta^{-nk} \exp\left(-\frac{A}{\theta^k}\right),\qquad \theta>0.

The observations must be complete: a censored record such as “the loss exceeded 10” does not contribute the same likelihood factor as an observed loss of 10. Likewise, if the data were truncated, the likelihood would need to account for the observation mechanism. Those situations are outside the complete-data model used here.

For estimating $\theta$, the factors $k^n\prod x_j^{k-1}$ are constant because both the data and $k$ are fixed. Taking logarithms preserves the maximizing value and simplifies differentiation:

\ell(\theta)=\log L(\theta) =n\log k+(k-1)\sum_{j=1}^n\log x_j-nk\log\theta-\frac{A}{\theta^k}.

Differentiating with respect to $\theta$ gives the score:

U(\theta)=\frac{d\ell}{d\theta} =-\frac{nk}{\theta}+\frac{kA}{\theta^{k+1}}.

Set the score equal to zero. Since $k>0$ and $\theta>0$, divide by $k$, multiply by $\theta^{k+1}$, and solve:

-\frac{n}{\theta}+ \frac{A}{\theta^{k+1}}=0 \quad\Longleftrightarrow\quad A=n\theta^k \quad\Longleftrightarrow\quad \widehat{\theta}=\left(\frac{1}{n}\sum_{j=1}^n x_j^k\right)^{1/k}.

Theorem: For independent, complete, positive observations from the stated Weibull model with known $k>0$, the maximum likelihood estimate of the scale is $\widehat{\theta}=(n^{-1}\sum_{j=1}^n x_j^k)^{1/k}$. This result estimates the scale only; it does not estimate the shape.

Why the Score Solution Is the Maximum

A stationary point should not be accepted without checking that it maximizes the likelihood. The transformation $t=\theta^k$ is one-to-one from positive $\theta$ to positive $t$. Ignoring the positive factor that does not depend on $\theta$, the likelihood becomes $t^{-n}\exp(-A/t)$. Its log-likelihood, up to an additive constant, is $-n\log t-A/t$. Therefore,

\frac{d\ell}{dt}=-\frac{n}{t}+\frac{A}{t^2} =\frac{A-nt}{t^2}.

Because $A>0$, this derivative is positive when $0<t<A/n$, zero at $t=A/n$, and negative when $t>A/n$. The likelihood increases up to $A/n$ and decreases afterward, so this is its unique global maximum. Translating back gives $\widehat{\theta}^k=A/n$, as in the score calculation.

Worked Calculations

Worked Example: Estimating Scale with Shape Two

A hypothetical insurer observes three complete losses of 1, 2, and 2 thousand dollars. Assume the losses are independent Weibull observations with known shape $k=2$. Estimate the scale, show the likelihood and score, and check the estimate.

Method. Apply the known-shape scale estimator to the complete observations. The loss unit throughout the calculation is one thousand dollars.

First compute the powered observations and their sum: $1^2=1$, $2^2=4$, and $2^2=4$, so $A=1+4+4=9$. With $n=3$,

\widehat{\theta}=\left(\frac{9}{3}\right)^{1/2} =\sqrt{3}=1.7321\text{ thousand dollars}

to four decimal places. This is approximately $1.7321$ thousand dollars, or $\$1,732.05$, to the nearest cent.

For a direct likelihood check, the data-dependent product is $1^{1}\cdot2^{1}\cdot2^{1}=4$, and $k^n=2^3=8$. Thus $L(\theta)=32\theta^{-6}\exp(-9/\theta^2)$. The score is $U(\theta)=-6/\theta+18/\theta^3$. At $\theta=\sqrt{3}$, the first term is $-6/\sqrt{3}$ and the second is $18/(3\sqrt{3})=6/\sqrt{3}$; their sum is zero. The global-maximum argument above confirms that this stationary point is the MLE.

Check and trap avoided. The estimated scale is positive and has the same units as the observations. Taking the arithmetic mean, $5/3$ thousand dollars, would not give the MLE for shape 2; the required calculation averages the squared losses before taking a square root.

Worked Example: Fitted Survival Probability in Dollar Units

A hypothetical portfolio has three complete losses of $\$1,000$, $\$2,000$, and $\$3,000$. Assume independent Weibull losses with known shape $k=3$. Estimate the scale and use it to find the fitted probability that a loss exceeds $\$3,000$.

Method. Work in thousands of dollars, so the observations are $1,2,3$ and the estimated scale will also be in thousands of dollars. Compute the cubes: $1^3=1$, $2^3=8$, and $3^3=27$. Thus $A=36$, and

\widehat{\theta}=\left(\frac{36}{3}\right)^{1/3} =\sqrt[3]{12}=2.2894\text{ thousand dollars}

to four decimal places. This corresponds to approximately $\$2,289.43$ to the nearest cent.

The fitted survival probability at $x=3$ thousand dollars is

\widehat{S}_X(3) =\exp\left[-\left(\frac{3}{\sqrt[3]{12}}\right)^3\right] =\exp\left(-\frac{27}{12}\right) =e^{-2.25} \approx 0.1054.

The probability is reported to four decimal places. As a check, the exponent uses the cube of the ratio because the known shape is 3; using a square or omitting the ratio would not be the Weibull survival formula.

Check and trap avoided. The estimate is in thousands of dollars because the data were entered in thousands. The survival calculation uses $x/\widehat{\theta}$, so the loss threshold and scale must use the same unit.

Worked Example: A Decreasing-Hazard Shape and the Scale

A hypothetical coverage has three complete losses of $\$400$, $\$900$, and $\$1,600$. Assume independent Weibull losses with known shape $k=0.5$. Estimate the scale and calculate the fitted survival probability at $\$900$.

Method. Use hundreds of dollars as the loss unit, giving observations $4,9,16$. Raising each observation to the known power $0.5$ gives $\sqrt{4}=2$, $\sqrt{9}=3$, and $\sqrt{16}=4$. Therefore $A=2+3+4=9$, and

\widehat{\theta}=\left(\frac{9}{3}\right)^{1/0.5} =3^2=9\text{ hundreds of dollars}

or $\$900$. At $x=9$ hundreds of dollars, the fitted survival probability is

\widehat{S}_X(9) =\exp\left[-\left(\frac{9}{9}\right)^{0.5}\right] =e^{-1} \approx 0.3679.

Check and trap avoided. Substitution of $x=\widehat{\theta}$ must give $e^{-1}$ for every positive shape in this parameterization. A common error is to treat the scale estimate $9$ hundreds of dollars as 9 dollars, despite having entered the data in hundreds.

Interpreting the Scale and Avoiding Parameterization Errors

The scale controls the horizontal spread of the distribution, while the known shape controls its form. For example, the survival formula at the scale is $S_X(\theta)=\exp(-1)=e^{-1}$, which provides a useful substitution check. The scale need not equal the expected loss. For this parameterization, the mean is

E[X]=\theta\,\Gamma\left(1+\frac{1}{k}\right), \qquad \Gamma(a)=\int_0^\infty t^{a-1}e^{-t}\,dt \quad (a>0).

This mean formula follows by substituting $t=(x/\theta)^k$ in the integral of $x f_X(x;k,\theta)$. In particular, when $k=1$, $\Gamma(2)=1$, so the mean equals $\theta$. For other shapes, the factor $\Gamma(1+1/k)$ generally differs from 1. The arithmetic mean of the data is therefore not a general substitute for the scale MLE.

Common exam trap: The Weibull formula depends on the parameterization. Here the exponential term is $\exp[-(x/\theta)^k]$, and the density contains $\theta^{-nk}$ in the sample likelihood. Do not use a formula from a parameterization that defines a rate, or that places a different constant in the exponent, without first converting it to the stated model.

A second trap is to use the known shape as though it were an estimate. In this problem, $k$ is supplied and held fixed throughout the likelihood calculation. Only $\theta$ is varied when differentiating, and only $\theta$ is estimated. If the shape were also unknown, the likelihood would require a different estimation procedure.

The calculation is reliable when its assumptions match the data: observations are positive, complete, independent, and governed by the same Weibull distribution. It does not automatically handle censoring, truncation, or different distributions across risk groups. Before using the formula, check the observation type, the specified shape, and the units of every loss.

Confirm the model.
Write the density convention and verify that the shape $k>0$ is known and the scale is the only unknown parameter.

Check the observations.
Confirm that all $n$ observations are positive, complete losses from the same independent model, and express them in consistent units.

Compute the powered sum.
Calculate $A=\sum x_j^k$, keeping the specified shape in every power.

Estimate and check.
Use $\widehat{\theta}=(A/n)^{1/k}$, then verify $n\widehat{\theta}^k=A$ and retain the observation units.

Key takeaway: With complete independent Weibull observations and known shape $k$, the scale MLE is the $k$-th root of the sample mean of the $k$-th powers. The power, the root, the parameterization, and the loss units must all agree.

Check Your Understanding

For each hypothetical setting, use the stated Weibull parameterization and assume independent complete observations. Answers are given so you can check the scale estimate and any requested fitted probability.

A portfolio has losses $3$ and $4$ thousand dollars, with known shape $k=2$. Find the scale MLE in thousands of dollars and convert it to dollars to the nearest cent. (Answer: $\widehat{\theta}=\sqrt{12.5}=3.5355$ thousand dollars to four decimal places, or $\$3,535.53$ to the nearest cent.)
Complete losses are $1,4,9$ hundreds of dollars, and the known shape is $k=1.5$. Find the scale MLE. (Answer: $\widehat{\theta}=12^{2/3}\approx5.2415$ hundreds of dollars.)
For losses $1,4,9,16$ in a consistent unit and known shape $k=0.5$, find the scale MLE and fitted probability that $X$ exceeds that scale. (Answer: $\widehat{\theta}=6.25$ units and $\widehat{S}_X(\widehat{\theta})=e^{-1}\approx0.3679$.)
Losses are $0.5,1.5,2$ thousand dollars, with known shape $k=2$. Find the scale MLE and the fitted probability that a loss exceeds 2 thousand dollars. (Answer: $\widehat{\theta}=\sqrt{13/6}\approx1.4720$ thousand dollars and $\widehat{S}_X(2)=e^{-24/13}\approx0.1579$.)
Complete losses are $2,4,6$ in a common monetary unit, with known shape $k=1$. Find the scale MLE and the fitted probability that a loss exceeds 4 units. (Answer: $\widehat{\theta}=4$ units and $\widehat{S}_X(4)=e^{-1}\approx0.3679$.)

What You'll Learn