Exponential And Weibull Models At Shape One

Why Shape One Matters

The Weibull distribution can describe severities with different hazard patterns, while the exponential distribution has a constant hazard. They are not unrelated models: the exponential distribution is exactly the Weibull distribution when its shape parameter is one. Recognizing this identity lets you translate between two common parameterizations and check that calculations use the right units.

The key distinction is between scale and rate. A Weibull scale is measured in the same units as the loss. An exponential rate is measured in reciprocal loss units. At shape one, the rate is the reciprocal of the scale, not the scale itself. This comparison is aligned with the SOA Exam ASTAM (Advanced Short-Term Actuarial Mathematics) syllabus.

Definition: A nonnegative loss $X$ has a Weibull distribution with shape $\alpha>0$ and scale $\eta>0$ if, for $x\geq 0$, $$ S_X(x)=\exp\left[-\left(\frac{x}{\eta}\right)^\alpha\right]. $$ Its density for $x>0$ is $$ f_X(x)=\frac{\alpha}{\eta}\left(\frac{x}{\eta}\right)^{\alpha-1} \exp\left[-\left(\frac{x}{\eta}\right)^\alpha\right]. $$ An exponential distribution with rate $\lambda>0$ has $S_X(x)=e^{-\lambda x}$ and $f_X(x)=\lambda e^{-\lambda x}$ for $x\geq 0$. Here $\eta$ is in loss units and $\lambda$ is in inverse loss units.

Showing the Equivalence

Set the Weibull shape to $\alpha=1$. Its survival function becomes $S_X(x)=e^{-x/\eta}$. Comparing this with the exponential survival function $e^{-\lambda x}$ shows that the two models agree for every $x\geq 0$ precisely when $\lambda=1/\eta$. Their densities agree as well: substituting $\alpha=1$ into the Weibull density gives $(1/\eta)e^{-x/\eta}$.

\boxed{\alpha=1,\qquad \lambda=\frac{1}{\eta},\qquad \eta=\frac{1}{\lambda}}

This is an equality of distributions, not just a match of one summary. With the parameter conversion, the survival probabilities, densities, quantiles, moments, and hazards all agree. A parameter called “scale” in one convention should not be mistaken for the parameter called “rate” in another.

Moments, Quantiles, and Hazards

For a Weibull random variable, the $r$th raw moment is obtained from the gamma function, defined by $\Gamma(a)=\int_0^\infty t^{a-1}e^{-t}\,dt$ for $a>0$. For $r\geq 0$,

E[X^r]=\eta^r\Gamma\left(1+\frac{r}{\alpha}\right).

In particular, $E[X]=\eta\Gamma(1+1/\alpha)$, and $\operatorname{Var}(X)=\eta^2\{\Gamma(1+2/\alpha)-[\Gamma(1+1/\alpha)]^2\}$. At shape one, $\Gamma(2)=1$ and $\Gamma(3)=2$, so $E[X]=\eta$, $E[X^2]=2\eta^2$, and $\operatorname{Var}(X)=\eta^2$. In rate form these become $E[X]=1/\lambda$ and $\operatorname{Var}(X)=1/\lambda^2$.

The quantile at cumulative probability $p$, for $0\leq p<1$, solves $F_X(x)=p$. The general Weibull quantile is $\eta[-\ln(1-p)]^{1/\alpha}$; at shape one it is $-\ln(1-p)/\lambda$. The quantile is infinite at $p=1$.

q_p=\eta[-\ln(1-p)]^{1/\alpha}, \qquad q_p\big|_{\alpha=1}=\eta[-\ln(1-p)] =\frac{-\ln(1-p)}{\lambda}.

For a continuous loss with positive survival, the hazard is $h_X(x)=f_X(x)/S_X(x)$. Thus the Weibull hazard is $(\alpha/\eta)(x/\eta)^{\alpha-1}$; at shape one it is the constant $1/\eta=\lambda$. The constant hazard means the instantaneous rate of loss occurrence per unit of loss, conditional on exceeding the current level, does not vary with that level. For $\alpha>1$ the Weibull hazard increases with $x$, while for $0<\alpha<1$ it decreases. The exponential model is the constant-hazard case.

Derivation: To obtain the Weibull moment, integrate $x^r f_X(x)$ over $x>0$, and set $t=(x/\eta)^\alpha$. Then $x=\eta t^{1/\alpha}$ and $dx=(\eta/\alpha)t^{1/\alpha-1}dt$. Substitution into the integral gives $$ E[X^r] =\int_0^\infty x^r f_X(x)\,dx =\eta^r\int_0^\infty t^{r/\alpha}e^{-t}\,dt =\eta^r\Gamma\left(1+\frac{r}{\alpha}\right). $$ When $\alpha=1$, this yields $E[X]=\eta\Gamma(2)=\eta$ and $E[X^2]=\eta^2\Gamma(3)=2\eta^2$. Subtracting the squared mean gives the variance $\eta^2$.

Worked Comparisons

The following examples use hypothetical loss models. Each states the units because a scale or rate has meaning only together with its units.

Worked Example: Translate a Weibull Model at Shape One

A hypothetical equipment-loss model is Weibull with shape $\alpha=1$ and scale $\eta=3$ thousand dollars. Find the corresponding exponential rate, the survival probability above 6 thousand dollars, the mean and variance, the 90th percentile, and the hazard.

Method. Use the shape-one parameter identity and then apply the exponential formulas. The loss is measured in thousands of dollars, so the rate is per thousand dollars.

Rate and survival. The rate is $\lambda=1/\eta=1/3=0.3333$ per thousand dollars. At $x=6$, the exponent is $-\lambda x=-(1/3)(6)=-2$. Therefore $S_X(6)=e^{-2}=0.1353$, and $F_X(6)=1-0.1353=0.8647$.

Moments. The mean is $\eta=3$ thousand dollars. The variance is $\eta^2=3^2=9$ thousand-dollars squared, and the standard deviation is $\sqrt{9}=3$ thousand dollars.

90th percentile. Set $p=0.90$. Then $q_{0.90}=3[-\ln(1-0.90)]=3[-\ln(0.10)]=3\ln(10)=6.9078$ thousand dollars.

Hazard and check. The hazard is constant at $1/\eta=1/3=0.3333$ per thousand dollars. The survival calculation also follows from the Weibull form: $\exp[-(6/3)^1]=e^{-2}$, agreeing with the exponential result. The calculations are reported to four decimal places where applicable. Common mistake avoided: using $3$ as the rate would reverse the scale-rate relationship and produce the wrong survival probability.

Worked Example: Start With an Exponential Rate

A hypothetical warranty severity is exponential with rate $\lambda=0.4$ per thousand dollars. Find its equivalent shape-one Weibull scale, mean, variance, survival probability above 4 thousand dollars, and 75th percentile.

Method. Convert rate to scale by taking the reciprocal, then calculate using either parameterization.

Scale and moments. The equivalent Weibull has $\alpha=1$ and $\eta=1/0.4=2.5$ thousand dollars. Its mean is $1/\lambda=1/0.4=2.5$ thousand dollars. Its variance is $1/\lambda^2=1/(0.4^2)=1/0.16=6.25$ thousand-dollars squared.

Survival probability. At $x=4$, $\lambda x=(0.4)(4)=1.6$, so $S_X(4)=e^{-1.6}=0.2019$.

75th percentile. Here $1-p=0.25$, so $-\ln(1-p)=-\ln(0.25)=\ln(4)$. Therefore $q_{0.75}=\ln(4)/0.4=3.4657$ thousand dollars.

Check and trap. The Weibull quantile gives $2.5\ln(4)=3.4657$ thousand dollars, the same answer. The constant hazard is $0.4$ per thousand dollars. Common mistake avoided: the scale is $2.5$ thousand dollars, not $0.4$ thousand dollars; the latter is the rate and has reciprocal units.

Worked Example: Same Scale Does Not Mean Same Distribution

Compare two hypothetical severities, both with scale $\eta=4$ thousand dollars: a Weibull with shape $\alpha=2$ and an exponential model (equivalently, a Weibull with shape one). Find each mean and variance, compare their median quantiles, and calculate the shape-two Weibull hazard at 1 and 4 thousand dollars.

Method. Use the Weibull moment and quantile formulas at each shape. The exponential has rate $\lambda=1/4=0.25$ per thousand dollars. Keeping the scale fixed does not make the shape-two Weibull exponential.

Shape-two mean and variance. Since $\Gamma(3/2)=\sqrt{\pi}/2$ and $\Gamma(2)=1$, the Weibull mean is $4\Gamma(3/2)=2\sqrt{\pi}=3.5449$ thousand dollars. Its second moment is $4^2\Gamma(2)=16$ thousand-dollars squared. Its variance is $16-(2\sqrt{\pi})^2=16-4\pi=3.4336$ thousand-dollars squared.

Exponential mean and variance. With rate $0.25$ per thousand dollars, the mean is $1/0.25=4$ thousand dollars and the variance is $1/(0.25)^2=16$ thousand-dollars squared.

Medians. For the shape-two Weibull, $q_{0.50}=4[-\ln(0.50)]^{1/2}=4\sqrt{\ln(2)}=3.3302$ thousand dollars. For the exponential, $q_{0.50}=-\ln(0.50)/0.25=4\ln(2)=2.7726$ thousand dollars.

Hazards. At $x=1$, the shape-two hazard is $(2/4)(1/4)=0.1250$ per thousand dollars. At $x=4$, it is $(2/4)(4/4)=0.5000$ per thousand dollars. The exponential hazard is $0.2500$ per thousand dollars at both levels.

Check and trap. At $x=\eta=4$, both models have survival $e^{-1}=0.3679$, but their different shapes give different means, medians, and hazards. Common mistake avoided: equality of scale and survival at one loss level does not establish that two distributions are the same.

Common Mistakes and Practical Checks

A parameter comparison is reliable only if it keeps track of the model, units, and quantity being compared. The table summarizes the shape-one identities.

Quantity	Weibull at $\alpha=1$	Exponential with rate $\lambda$
Parameter conversion	$\eta=1/\lambda$	$\lambda=1/\eta$
Survival	$e^{-x/\eta}$	$e^{-\lambda x}$
Mean	$\eta$	$1/\lambda$
Variance	$\eta^2$	$1/\lambda^2$
Hazard	$1/\eta$	$\lambda$

Confusing rate and scale: a rate of $0.4$ per thousand dollars corresponds to a scale of $2.5$ thousand dollars, not $0.4$ thousand dollars.
Dropping the units: if $x$ is in dollars, $\lambda$ must be per dollar; if $x$ is in thousands of dollars, $\lambda$ must be per thousand dollars. The product $\lambda x$ must be dimensionless.
Assuming every Weibull is exponential: the exact equivalence requires $\alpha=1$. With any other shape, the hazard varies with $x$, and the distribution is not exponential.
Comparing scale as if it were the mean: the Weibull mean is $\eta\Gamma(1+1/\alpha)$. It equals $\eta$ at shape one, but generally not at other shapes.
Over-interpreting one matching value: models can share survival at $x=\eta$ while differing elsewhere. Compare the defining functions or parameters, not just one calculation.

Key takeaway: A Weibull severity with shape one is exponential, with rate equal to the reciprocal of the Weibull scale. Its mean is the scale, its variance is the scale squared, its hazard is constant, and its quantiles reduce to the exponential quantile formula. For other shapes, use the Weibull formulas rather than assuming those exponential results still hold.

Check Your Understanding

For each hypothetical model, identify the parameter conversion and calculate the requested quantities. Answers use the stated loss units.

A Weibull severity has shape $1$ and scale $8$ thousand dollars. Find the equivalent exponential rate, mean, variance, and constant hazard. (Answer: rate $0.125$ per thousand dollars; mean $8$ thousand dollars; variance $64$ thousand-dollars squared; hazard $0.125$ per thousand dollars.)
An exponential severity has rate $0.5$ per thousand dollars. Find the equivalent Weibull scale, the survival probability above $3$ thousand dollars, and the 90th percentile. (Answer: scale $2$ thousand dollars; $S_X(3)=e^{-1.5}=0.2231$; $q_{0.90}=2\ln(10)=4.6052$ thousand dollars.)
A Weibull has shape $1$ and scale $\$2{,}500$. Find its median and hazard, with the hazard in inverse dollars. (Answer: median $2{,}500\ln(2)=\$1{,}732.87$; hazard $1/2{,}500=0.0004$ per dollar.)
A Weibull severity has shape $2$ and scale $3$ thousand dollars. Find its mean, variance, and hazard at $1.5$ thousand dollars. (Answer: mean $3\Gamma(3/2)=2.6587$ thousand dollars; variance $9-9\pi/4=1.9314$ thousand-dollars squared; hazard $1/3=0.3333$ per thousand dollars.)
A shape-one Weibull severity has scale $5$ thousand dollars. Find the probability that the loss is greater than $2$ but no greater than $7$ thousand dollars, and find its median. (Answer: $e^{-2/5}-e^{-7/5}=0.4237$; median $5\ln(2)=3.4657$ thousand dollars.)

Quantity	Weibull at \(\alpha=1\)	Exponential with rate \(\lambda\)
Parameter conversion	\(\eta=1/\lambda\)	\(\lambda=1/\eta\)
Survival	\(e^{-x/\eta}\)	\(e^{-\lambda x}\)
Mean	\(\eta\)	\(1/\lambda\)
Variance	\(\eta^2\)	\(1/\lambda^2\)
Hazard	\(1/\eta\)	\(\lambda\)

What You'll Learn