Weibull Severity From Exponential Transformations

Why the Exponential Transformation Is Useful

A severity model is useful only when its parameters can be translated reliably into probabilities and moments. The Weibull model provides a flexible positive-valued distribution, and its construction from an exponential variable gives a direct route to its survival function. That route also makes the model’s shape and scale parameters easier to interpret.

This tutorial uses one parameterization throughout: shape $\alpha>0$ and scale $\theta>0$. The scale has the same units as the loss; the shape is dimensionless. Keeping these roles distinct helps avoid mixing a scale parameter with an exponential rate.

Definition: A nonnegative random loss $X$ has a Weibull distribution with shape $\alpha>0$ and scale $\theta>0$ if its survival function, for $x\geq 0$, is $S_X(x)=\exp[-(x/\theta)^\alpha]$. Its distribution function is $F_X(x)=1-\exp[-(x/\theta)^\alpha]$ for $x\geq0$, and $F_X(x)=0$ for $x<0$. The loss is continuous and has no probability mass at zero.

The scale $\theta$ is a threshold with a particularly simple interpretation: $S_X(\theta)=e^{-1}$, so $F_X(\theta)=1-e^{-1}$. The shape $\alpha$ determines how the survival probability changes as $x$ moves relative to $\theta$. The special case $\alpha=1$ is an exponential distribution with mean $\theta$.

Deriving the Survival Function From an Exponential Variable

Let $Z$ have a unit-rate exponential distribution, so $P(Z>z)=e^{-z}$ for $z\geq0$. Define $X=\theta Z^{1/\alpha}$. Since this transformation is increasing for positive $\alpha$, an event about $X$ can be rewritten as an event about $Z$.

Start with the transformation.
Let $X=\theta Z^{1/\alpha}$, where $Z\sim\operatorname{Exp}(1)$, $\theta>0$, and $\alpha>0$.

Rewrite the survival event.
For $x\geq0$, $X>x$ is equivalent to $Z>(x/\theta)^\alpha$, because raising positive quantities to the power $\alpha$ preserves their order.

Apply the exponential survival function.
Thus $P(X>x)=P[Z>(x/\theta)^\alpha]=\exp[-(x/\theta)^\alpha]$, which is the Weibull survival function.

The unit-rate assumption matters. If an exponential variable has rate $r$, its survival function is $e^{-rz}$, not $e^{-z}$. To preserve the Weibull model with scale $\theta$, one may define $X=\theta(rZ)^{1/\alpha}$ when $Z\sim\operatorname{Exp}(r)$. Substituting the threshold into the exponential survival function then gives $e^{-(x/\theta)^\alpha}$. Using $X=\theta Z^{1/\alpha}$ with an exponential variable of a different rate would change the resulting scale.

Density, Quantiles, and Moments

For $x>0$, differentiate the distribution function to obtain the density. There is no density contribution from negative values, and the continuous model has no point mass at zero.

f_X(x)=\frac{\alpha}{\theta}\left(\frac{x}{\theta}\right)^{\alpha-1} \exp\left[-\left(\frac{x}{\theta}\right)^\alpha\right],\qquad x>0.

The factor $\alpha/\theta$ supplies the inverse-loss units required of a density. The density itself need not be a probability; probabilities come from integrating it over intervals. The density can rise or fall depending on the shape parameter and the loss level.

To find the $p$-quantile, where $0<p<1$, set $F_X(x_p)=p$ and solve for $x_p$. This inversion is valid for the stated Weibull distribution and gives a unique nonnegative quantile.

x_p=\theta[-\ln(1-p)]^{1/\alpha},\qquad 0<p<1.

The same substitution that gives the survival function also gives the raw moments. Define the gamma function, for $s>0$, by $\Gamma(s)=\int_0^\infty u^{s-1}e^{-u}\,du$. For any real $r>-\alpha$, the moment calculation is finite and yields the formula below. In particular, every nonnegative integer raw moment exists.

Theorem: For a Weibull loss with shape $\alpha>0$ and scale $\theta>0$, $E[X^r]=\theta^r\Gamma(1+r/\alpha)$ for $r>-\alpha$. Consequently, $E[X]=\theta\Gamma(1+1/\alpha)$, and $\operatorname{Var}(X)=\theta^2\{\Gamma(1+2/\alpha)-[\Gamma(1+1/\alpha)]^2\}$.

To justify the moment formula, integrate the density and make the substitution $u=(x/\theta)^\alpha$. Then $x=\theta u^{1/\alpha}$ and $dx=(\theta/\alpha)u^{1/\alpha-1}du$. The density times $dx$ becomes $e^{-u}du$, leaving a gamma integral:

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

The integral converges at zero when $r/\alpha>-1$, or equivalently $r>-\alpha$; it converges at infinity because of the exponential factor $e^{-u}$. The mean follows by taking $r=1$. The variance follows from $\operatorname{Var}(X)=E[X^2]-[E[X]]^2$, with $r=2$ for the second raw moment.

Worked Calculations

Worked Example: Transforming an Exponential Variable

A hypothetical claim severity is modeled as $X=200\sqrt{Z}$, where the loss is measured in dollars and $Z\sim\operatorname{Exp}(1)$. Find the Weibull parameters, $P(X>300)$, $F_X(300)$, and the density at 300 dollars.

Method and parameters. Match $X=\theta Z^{1/\alpha}$ to $X=200Z^{1/2}$. Thus $\theta=200$ dollars and $1/\alpha=1/2$, so $\alpha=2$. The model is Weibull with shape 2 and scale 200 dollars.

Survival probability. The threshold relative to scale is $300/200=1.5$. Therefore $P(X>300)=\exp[-(1.5)^2]=e^{-2.25}=0.1054$, to four decimal places.

Distribution probability. Since the model is continuous, $F_X(300)=1-P(X>300)=1-0.105399\ldots=0.8946$, to four decimal places.

Density. Substitute $x=300$, $\alpha=2$, and $\theta=200$: $f_X(300)=(2/200)(1.5)e^{-2.25}=0.015(0.105399\ldots)=0.00158099$ per dollar. Rounded to four significant figures, the density is $0.001581$ per dollar.

Check and common mistake. The survival probability and distribution probability add to $0.1054+0.8946=1.0000$, as required. A common error is to use the exponential variable’s threshold $300$ directly and calculate $e^{-300}$. The transformation requires first converting the loss threshold to $Z>(300/200)^2=2.25$.

Worked Example: Mean and Variance With Shape 2

A hypothetical portfolio uses a Weibull severity model with shape $\alpha=2$ and scale $\theta=\$1{,}000$. Calculate the mean and variance, reporting money to the nearest cent and variance in squared-dollar units.

Method and assumptions. Apply the Weibull raw-moment formula. For this shape, $\Gamma(1+1/\alpha)=\Gamma(3/2)=\sqrt{\pi}/2$, and $\Gamma(1+2/\alpha)=\Gamma(2)=1$.

Mean. $E[X]=1{,}000\Gamma(3/2)=1{,}000(\sqrt{\pi}/2)=\$886.2269\ldots$, which rounds to $\$886.23$.

Second raw moment. $E[X^2]=1{,}000^2\Gamma(2)=1{,}000{,}000(1)=1{,}000{,}000$ dollars squared.

Variance. Squaring the unrounded mean gives $[E[X]]^2=1{,}000{,}000(\pi/4)=785{,}398.1634\ldots$. Hence $\operatorname{Var}(X)=1{,}000{,}000-785{,}398.1634\ldots=214{,}601.8366\ldots$ dollars squared, or $214{,}601.84$ dollars squared to the nearest cent in squared-dollar units.

Check and common mistake. Equivalently, the variance is $1{,}000{,}000(1-\pi/4)$. Since $\pi/4\approx0.7854$, the factor is $1-0.7854=0.2146$, not $0.7854$. Using the latter as the variance factor confuses the squared mean with the variance. The corrected factor gives $214{,}601.84$ dollars squared, consistent with the subtraction above.

Worked Example: Quantile and Exceedance Probability

A hypothetical equipment-repair severity follows a Weibull distribution with shape $\alpha=3$ and scale $\theta=\$2{,}000$. Find its 90th percentile and the probability that a loss exceeds $\$3{,}000$.

Method and parameters. Use the inverse distribution function for the percentile and the survival function for the exceedance probability. The scale is in dollars, so the resulting quantile is also in dollars.

90th percentile. Set $p=0.90$: $x_{0.90}=2{,}000[-\ln(1-0.90)]^{1/3} =2{,}000[-\ln(0.10)]^{1/3}$. Since $-\ln(0.10)=2.302585\ldots$, its cube root is $1.320469\ldots$. Thus $x_{0.90}=\$2{,}640.94$, rounded to the nearest cent.

Exceedance probability. The relative threshold is $3{,}000/2{,}000=1.5$, so $P(X>3{,}000)=\exp[-(1.5)^3]=e^{-3.375}=0.0342$, to four decimal places.

Check and common mistake. Substitution of the quantile back into the survival function gives $\exp[-(2{,}640.94/2{,}000)^3]\approx0.1000$, as required for a 90th percentile. A common error is to use $p$, rather than $1-p$, inside the negative logarithm. The quantile equation is $1-e^{-(x/\theta)^\alpha}=p$, so the survival probability at the $p$-quantile is $1-p$.

Choosing and Checking a Weibull Calculation

The formulas are compact, but most errors come from a mismatch between the stated model and the calculation. Before substituting values, identify the shape and scale, confirm that the loss threshold and scale use the same units, and decide whether the question asks for a survival probability, a distribution probability, a density, a quantile, or a moment.

Quantity	Weibull expression	Useful check
Survival probability	$S_X(x)=e^{-(x/\theta)^\alpha}$	At $x=\theta$, it equals $e^{-1}$
Distribution probability	$F_X(x)=1-S_X(x)$	$F_X(x)+S_X(x)=1$
Density	$f_X(x)=(\alpha/\theta)(x/\theta)^{\alpha-1}e^{-(x/\theta)^\alpha}$	Its units are inverse loss units
$p$-quantile	$x_p=\theta[-\ln(1-p)]^{1/\alpha}$	$S_X(x_p)=1-p$
Raw moment	$E[X^r]=\theta^r\Gamma(1+r/\alpha)$	The result has units of loss to the power $r$

A density value is not an exceedance probability, and a raw second moment is not a variance. For the variance, subtract the square of the mean from the second raw moment. Keep extra precision through that subtraction and round only the final result. These checks are especially important when the gamma function has non-integer arguments.

Key takeaway: With $Z\sim\operatorname{Exp}(1)$, the transformation $X=\theta Z^{1/\alpha}$ produces a Weibull loss with survival $e^{-(x/\theta)^\alpha}$. From that function, differentiation gives the density, inversion gives quantiles, and a gamma-integral substitution gives the raw moments.

Check Your Understanding

For each hypothetical model, use the stated shape and scale convention. Probabilities are reported to four decimal places and money to the nearest cent where relevant.

A loss has Weibull shape $\alpha=1.5$ and scale $\theta=\$400$. Find $S_X(400)$ and $F_X(400)$. (Answer: $S_X(400)=0.3679$, $F_X(400)=0.6321$.)
A severity has shape $\alpha=2$ and scale $\theta=\$300$. Calculate $f_X(300)$ per dollar and $P(X>600)$. (Answer: $f_X(300)=0.002453$ per dollar; $P(X>600)=0.0183$.)
A hypothetical exponential severity has shape $\alpha=1$ and scale $\theta=\$500$, which is the Weibull special case. Find its 75th percentile using the Weibull quantile formula. (Answer: $\$693.15$.)
A loss has shape $\alpha=2$ and scale $\theta=\$500$. Calculate its mean and variance. (Answer: mean $=\$443.11$; variance $=53{,}650.46$ dollars squared.)
Let $Z\sim\operatorname{Exp}(4)$, with rate 4, and define $X=300(4Z)^{1/2}$, where $X$ is measured in dollars. Identify the Weibull shape and scale, then calculate $P(X>600)$. (Answer: shape $2$, scale $\$300$, and $P(X>600)=0.0183$.)

Quantity	Weibull expression	Useful check
Survival probability	\(S_X(x)=e^{-(x/\theta)^\alpha}\)	At \(x=\theta\), it equals \(e^{-1}\)
Distribution probability	\(F_X(x)=1-S_X(x)\)	\(F_X(x)+S_X(x)=1\)
Density	\(f_X(x)=(\alpha/\theta)(x/\theta)^{\alpha-1}e^{-(x/\theta)^\alpha}\)	Its units are inverse loss units
\(p\)-quantile	\(x_p=\theta[-\ln(1-p)]^{1/\alpha}\)	\(S_X(x_p)=1-p\)
Raw moment	\(E[X^r]=\theta^r\Gamma(1+r/\alpha)\)	The result has units of loss to the power \(r\)

What You'll Learn