Weibull Severity Parameterizations

Introduction

The word “scale” is not used consistently across every reference to the Weibull distribution. One source may describe a scale parameter that has the same units as loss; another may write a coefficient multiplying a power of loss in the exponent. Those parameters are related, but they are not interchangeable. A parameterization error can therefore change a calculated survival probability, density, or expected loss by a large amount.

This tutorial defines the Weibull severity model using a shape $k$ and a loss-scale parameter $\theta$, then translates that model to a common power-rate convention. It also develops the density and raw moments needed for severity calculations. These methods are aligned with the SOA Exam ASTAM (Advanced Short-Term Actuarial Mathematics) syllabus.

Definition: A nonnegative loss $X$ has a Weibull distribution with shape $k>0$ and scale $\theta>0$, written $X\sim\operatorname{Weibull}(k,\theta)$, if for $x\geq 0$, $$ S_X(x)=P(X>x)=\exp\left[-\left(\frac{x}{\theta}\right)^k\right]. $$ Here $\theta$ has the same units as $X$, while $k$ is dimensionless. For $x<0$, $F_X(x)=0$; for $x\geq 0$, $F_X(x)=1-S_X(x)$.

Shape, Scale, and the Power-Rate Convention

The scale parameter $\theta$ is a characteristic loss amount: at $x=\theta$, the exponent is one, so $S_X(\theta)=e^{-1}$. It is not the mean in general. The shape parameter $k$ controls how the exponent changes as $x$ changes. For this tutorial, the convention is always $S_X(x)=\exp[-(x/\theta)^k]$, unless another parameterization is explicitly stated.

A common alternative writes the same survival function as $S_X(x)=\exp(-c x^k)$. Comparing exponents gives $c=\theta^{-k}$, or equivalently $\theta=c^{-1/k}$. The coefficient $c$ is sometimes called a rate coefficient, but its units depend on $k$: if loss is measured in dollars, $c$ has units of dollars to the power $-k$. It is not generally an ordinary rate with units of inverse dollars.

Convention	Survival function	Parameter relationship	Units when $X$ is in dollars
Shape and scale	$\exp[-(x/\theta)^k]$	$\theta=c^{-1/k}$	$\theta$: dollars
Shape and power-rate coefficient	$\exp(-c x^k)$	$c=\theta^{-k}$	$c$: dollars$^{-k}$
Shape and characteristic life	$\exp[-(x/\eta)^\beta]$	$\eta=\theta,\ \beta=k$	$\eta$: dollars

A reliable translation begins by equating the complete exponents, not by matching parameter names. In particular, a parameter named “scale” in one reference should not be substituted for $\theta$ until its defining formula has been checked.

Worked Example: Translating a Power-Rate Parameter

A hypothetical property severity model has survival function $S_X(x)=\exp(-0.000004x^2)$, where $x$ is measured in dollars. Find the shape-and-scale parameters, the survival probability at \$300, the density at \$300, and the mean and variance. No interest discounting or mortality component is involved; the loss distribution is Weibull.

Identify the parameters. Comparing $\exp(-c x^k)$ with the standard form gives $k=2$ and $c=0.000004$ dollars$^{-2}$. Therefore, $$ \theta=c^{-1/k}=(0.000004)^{-1/2}=500\text{ dollars}. $$ The model is $X\sim\operatorname{Weibull}(2,500)$.

Survival at \$300. The dimensionless ratio is $300/500=0.6$, so $$ S_X(300)=\exp[-(0.6)^2]=e^{-0.36}=0.6976763. $$ The probability is $0.6977$ to four decimal places.

Density at \$300. The density formula derived below gives $$ f_X(300)=\frac{2}{500}(0.6)^{1}e^{-0.36} =0.0024(0.6976763) =0.001674423\text{ per dollar}. $$

Moments. For shape $2$, $E[X]=500\Gamma(3/2)$, and $\Gamma(3/2)=\sqrt{\pi}/2$. Thus $$ E[X]=500\frac{\sqrt{\pi}}{2}=443.1135\text{ dollars}. $$ The second raw moment is $E[X^2]=500^2\Gamma(2)=250{,}000$ dollars squared, since $\Gamma(2)=1$. Hence $$ \operatorname{Var}(X)=250{,}000-(443.1134625\ldots)^2 =53{,}650.4592\text{ dollars}^2. $$ Rounded to the nearest cent, the mean is \$443.11. The variance is approximately $53{,}650.46$ dollars squared.

Method and trap avoided. The power-rate coefficient was converted using $\theta=c^{-1/k}$ before using scale formulas. Treating $0.000004$ as the scale would be incorrect: it is the coefficient of $x^2$, not a loss amount.

Density and Survival Function

For $x\geq 0$, the cumulative distribution function is $F_X(x)=1-\exp[-(x/\theta)^k]$. Differentiating it gives the density. This is the ordinary density of a continuous severity distribution; probabilities over intervals are found by integrating the density or by subtracting values of the cumulative distribution function.

Result: If $X\sim\operatorname{Weibull}(k,\theta)$, then for $x\geq 0$, $$ F_X(x)=1-\exp\left[-\left(\frac{x}{\theta}\right)^k\right], \qquad f_X(x)=\frac{k}{\theta}\left(\frac{x}{\theta}\right)^{k-1} \exp\left[-\left(\frac{x}{\theta}\right)^k\right]. $$ For $x<0$, $F_X(x)=0$ and $f_X(x)=0$.

Derivation: Let $z=(x/\theta)^k$, so $F_X(x)=1-e^{-z}$. For $x>0$, the chain rule gives $$ \frac{dF_X(x)}{dx}=e^{-z}\frac{dz}{dx}, \qquad \frac{dz}{dx}=\frac{k}{\theta}\left(\frac{x}{\theta}\right)^{k-1}. $$ Multiplying these derivatives yields the stated density. The density integrates to one because it is the derivative of a distribution function that increases from zero to one.

The density has units of inverse loss, such as dollars$^{-1}$, while $S_X(x)$ and $F_X(x)$ are probabilities and have no units. This is a useful check on calculations. In the power-rate convention, differentiating $\exp(-c x^k)$ gives $f_X(x)=ckx^{k-1}\exp(-c x^k)$, which agrees with the scale form after substituting $c=\theta^{-k}$.

Worked Example: Density, Survival, and a Moment Check

A hypothetical equipment portfolio uses $X\sim\operatorname{Weibull}(2,1000)$, with loss measured in dollars. Calculate $S_X(600)$, $f_X(600)$, the mean, the second raw moment, and the standard deviation. No interest or mortality assumptions enter this severity calculation.

Survival. First, $600/1000=0.6$, and $0.6^2=0.36$. Thus $$ S_X(600)=e^{-0.36}=0.6976763. $$ The probability of a loss exceeding \$600 is $0.6977$ to four decimal places.

Density. Substituting $k=2$, $\theta=1000$, and $x=600$, $$ f_X(600)=\frac{2}{1000}(0.6)e^{-0.36} =0.0012(0.6976763) =0.0008372116\text{ per dollar}. $$

Mean and second raw moment. The moment formula established below gives $$ E[X]=1000\Gamma(3/2)=1000\frac{\sqrt{\pi}}{2} =886.2269255\ldots\text{ dollars}, $$ and $$ E[X^2]=1000^2\Gamma(2)=1{,}000{,}000\text{ dollars}^2. $$ Therefore, $$ \operatorname{Var}(X)=E[X^2]-E[X]^2 =1{,}000{,}000\left(1-\frac{\pi}{4}\right) =214{,}601.8366\ldots\text{ dollars}^2. $$ Taking the square root of the unrounded variance gives $$ \sqrt{214{,}601.83660255\ldots}=463.2513752\ldots\text{ dollars}. $$ The standard deviation is \$463.25 to the nearest cent. As a check, the variance is positive and the standard deviation is smaller than the mean, as these computed values indicate.

Method and trap avoided. Use the density formula for a density and the survival formula for a probability. A density value such as $0.0008372$ is not itself a probability; it has units of inverse dollars.

Raw Moments and Variance

The gamma function provides a compact expression for the Weibull raw moments. For $a>0$, define $\Gamma(a)=\int_0^\infty t^{a-1}e^{-t}\,dt$. For each nonnegative real $r$, a Weibull severity has a finite $r$th raw moment:

Moment formula: If $X\sim\operatorname{Weibull}(k,\theta)$, $k>0$, $\theta>0$, and $r\geq 0$, then $$ E[X^r]=\theta^r\Gamma\left(1+\frac{r}{k}\right). $$ In particular, $$ E[X]=\theta\Gamma\left(1+\frac{1}{k}\right), \qquad E[X^2]=\theta^2\Gamma\left(1+\frac{2}{k}\right), $$ and $$ \operatorname{Var}(X)=\theta^2\left[ \Gamma\left(1+\frac{2}{k}\right) -\Gamma\left(1+\frac{1}{k}\right)^2 \right]. $$

Here is the justification for the general moment expression. Starting from the density and substituting $t=(x/\theta)^k$, we have $x=\theta t^{1/k}$ and $dx=(\theta/k)t^{1/k-1}dt$. Thus

\begin{aligned} E[X^r] &=\int_0^\infty x^r f_X(x)\,dx\\ &=\int_0^\infty x^r\frac{k}{\theta} \left(\frac{x}{\theta}\right)^{k-1}e^{-(x/\theta)^k}\,dx\\ &=\theta^r\int_0^\infty t^{r/k}e^{-t}\,dt\\ &=\theta^r\Gamma\left(1+\frac{r}{k}\right). \end{aligned}

The final equality follows directly from the defining gamma integral with exponent $t^{a-1}$, where $a=1+r/k$. The variance formula then follows from $\operatorname{Var}(X)=E[X^2]-E[X]^2$. This expression shows why the scale must be raised to the same power as the moment: $E[X^r]$ has units of loss to the power $r$.

Worked Example: Moments With a Noninteger Shape

A hypothetical repair-cost severity is $X\sim\operatorname{Weibull}(3,2000)$, with loss in dollars. Find the survival probability and density at \$1,000, and calculate the mean, second raw moment, and variance. No interest or mortality assumption is part of the severity model.

Survival at \$1,000. The ratio is $1000/2000=0.5$, and $0.5^3=0.125$. Therefore, $$ S_X(1000)=e^{-0.125}=0.8824969. $$

Density at \$1,000. Using $k/\theta=3/2000=0.0015$ and $(0.5)^{3-1}=0.25$, $$ f_X(1000)=0.0015(0.25)e^{-0.125} =0.000375(0.8824969) =0.0003309363\text{ per dollar}. $$

Mean. Since $1+1/k=4/3$ and $\Gamma(4/3)\approx0.8929795116$, $$ E[X]=2000\Gamma(4/3) =2000(0.8929795116) =1785.9590\text{ dollars}. $$

Second moment and variance. Since $1+2/k=5/3$ and $\Gamma(5/3)\approx0.9027452930$, $$ E[X^2]=2000^2\Gamma(5/3) =4{,}000{,}000(0.9027452930) =3{,}610{,}981.172\text{ dollars}^2. $$ Using the unrounded mean, $$ \operatorname{Var}(X)=3{,}610{,}981.172-(1785.9590232\ldots)^2 \approx421{,}331.54\text{ dollars}^2. $$ The mean is \$1,785.96 to the nearest cent; the variance is approximately $421{,}331.54$ dollars squared. The second raw moment exceeds the squared mean, as required for a nonnegative variance.

Method and trap avoided. The gamma-function argument is $1+r/k$, not $r/k$. Omitting the leading one would give the wrong moments. Keep the scale in dollars and the corresponding moments in dollars or dollars squared.

Common Parameterization Traps

Weibull formulas are short, but several checks prevent common errors:

Confusing scale with rate coefficient. If the survival function is $\exp(-c x^k)$, convert with $\theta=c^{-1/k}$. Do not use $c$ as though it were a loss-scale parameter.
Dropping the power on the scale. The exponent is $(x/\theta)^k$, not $x/\theta^k$. The ratio $x/\theta$ is dimensionless before it is raised to $k$.
Using the wrong moment argument. The $r$th raw moment uses $\Gamma(1+r/k)$, and the variance requires subtracting the square of the mean from the second raw moment.
Calling a density a probability. The density has units of inverse loss. To obtain a probability over a range, use $F_X(b)-F_X(a)$ for $a<X\leq b$.

The scale-and-shape form is particularly useful because its parameters have clear dimensional roles, and moment calculations follow from the same pair $k,\theta$. When a problem supplies a different convention, rewrite its survival function in the standard form before calculating.

Check Your Understanding

For each problem, state the relevant parameterization before calculating.

A hypothetical severity has $X\sim\operatorname{Weibull}(2,400)$, with losses in dollars. Find $S_X(200)$ and $f_X(200)$. (Answer: $S_X(200)=0.7788$; $f_X(200)=0.001947$ per dollar.)
A model is given by $S_X(x)=\exp(-c x^{3/2})$, and its scale is 60 dollars. Find $c$ and $E[X]$. (Answer: $c=60^{-3/2}\approx0.0021517$ dollars$^{-3/2}$; $E[X]=60\Gamma(5/3)\approx\$54.16$.)
For $X\sim\operatorname{Weibull}(2,100)$, calculate the mean, second raw moment, and variance. (Answer: $E[X]\approx\$88.62$; $E[X^2]=10{,}000$ dollars squared; $\operatorname{Var}(X)\approx2{,}146.02$ dollars squared.)
A severity satisfies $S_X(100)=e^{-0.25}$ and $S_X(200)=e^{-1}$. Assuming a Weibull model, identify $k$ and $\theta$, then calculate $f_X(100)$. (Answer: $k=2,\ \theta=200$ dollars; $f_X(100)\approx0.003894$ per dollar.)
A hypothetical loss has $X\sim\operatorname{Weibull}(3,500)$, with loss in dollars. Calculate $E[X]$ and $E[X^3]$. (Answer: $E[X]=500\Gamma(4/3)\approx\$446.49$; $E[X^3]=500^3\Gamma(2)=125{,}000{,}000$ dollars cubed.)

Convention	Survival function	Parameter relationship	Units when \(X\) is in dollars
Shape and scale	\(\exp[-(x/\theta)^k]\)	\(\theta=c^{-1/k}\)	\(\theta\): dollars
Shape and power-rate coefficient	\(\exp(-c x^k)\)	\(c=\theta^{-k}\)	\(c\): dollars\(^{-k}\)
Shape and characteristic life	\(\exp[-(x/\eta)^\beta]\)	\(\eta=\theta,\ \beta=k\)	\(\eta\): dollars

What You'll Learn