Common Severity Models: Exponential, Lognormal, and Pareto

Why Severity Models Matter

A severity model describes the size of an individual loss, conditional on a loss occurring. Its shape matters: two models can have similar expected losses but assign very different probabilities to large claims. Exponential, lognormal, and Pareto distributions are useful standard models because they illustrate distinct patterns of tail behavior and have formulas that support direct calculation.

This tutorial focuses on probabilities and moments for these three models. The random variable \(X\) denotes a nonnegative loss, measured in a stated unit such as thousands of dollars. A parameter’s units must match the units of \(X\). Probabilities are dimensionless; means have the units of \(X\), and variances have squared units.

Definition: A severity distribution models the amount of a single loss. Its support is the set of values that can occur. Its survival function is \(S_X(x)=P(X>x)\), and its cumulative distribution function is \(F_X(x)=P(X\leq x)=1-S_X(x)\). For a continuous severity, the probability of an exact value is zero, so \(P(X>x)\) and \(P(X\geq x)\) are equal.

Exponential Severity

Use the rate parameterization \(X\sim\operatorname{Exponential}(\lambda)\), where \(\lambda>0\) is a rate per unit of loss. The model is continuous and has support \(x\geq 0\). Its density and survival function are

f_X(x)=\lambda e^{-\lambda x},\qquad S_X(x)=e^{-\lambda x},\qquad x\geq 0.

For \(x<0\), the density is zero and \(F_X(x)=0\); for \(x\geq0\), \(F_X(x)=1-e^{-\lambda x}\). The rate parameter is the reciprocal of the mean, not the mean itself. The moments are

E[X]=\frac{1}{\lambda},\qquad \operatorname{Var}(X)=\frac{1}{\lambda^2}.

The exponential distribution has a relatively light tail compared with the lognormal and Pareto models described below. It also has the memoryless property: for \(s,t\geq0\), \(P(X>s+t\mid X>s)=P(X>t)\). This follows by dividing survival probabilities: \(e^{-\lambda(s+t)}/e^{-\lambda s}=e^{-\lambda t}\). The property is specific to the exponential model and should not be assumed for other severities.

Worked Example: Exponential Tail Probability and Moments

A hypothetical portfolio models an individual claim amount \(X\) in thousands of dollars by an exponential distribution with rate \(\lambda=0.5\) per thousand dollars. Find the mean and variance, the probability that a claim exceeds 3 thousand dollars, and the probability that it is between 2 and 5 thousand dollars. Also find the expected excess above 3 thousand dollars, given that the claim exceeds that amount.

Method and units. Use the rate parameterization, keeping \(X\) in thousands of dollars throughout. The rate is \(0.5\) per thousand dollars.

Moments. The mean is \(1/0.5=2\) thousand dollars. The variance is \(1/(0.5)^2=4\) thousand-dollar-squared.

Exceeding 3. By the survival function, \(P(X>3)=e^{-0.5(3)}=e^{-1.5}\approx0.2231\).

Between 2 and 5. Subtract the survival probability at 5 from the survival probability at 2: \(P(2<X\leq5)=S_X(2)-S_X(5)=e^{-1}-e^{-2.5}\approx0.367879-0.082085=0.285794\), or 0.2858 to four decimal places.

Conditional excess. Write the excess as \(X-3\) conditional on \(X>3\). By memorylessness, \(X-3\mid X>3\) has the original exponential distribution, so its expected value is \(1/0.5=2\) thousand dollars.

Check and common mistake. The interval probability is positive and less than \(P(X>2)=e^{-1}\approx0.3679\), as it must be. A frequent error is to treat \(\lambda=0.5\) as the mean, giving a mean of 0.5. The correction is that the rate is the reciprocal of the mean, so \(E[X]=2\).

Lognormal Severity

A positive loss is lognormal when its natural logarithm is normally distributed. Specifically, \(X\) is lognormal with parameters \(\mu\) and \(\sigma^2\) if \(Y=\ln X\sim N(\mu,\sigma^2)\), with \(\sigma>0\). Here \(\sigma^2\) is the variance of the logarithm, not of the loss. The support is \(x>0\), and the density is

f_X(x)=\frac{1}{x\sigma\sqrt{2\pi}} \exp\left(-\frac{(\ln x-\mu)^2}{2\sigma^2}\right),\qquad x>0.

The cumulative probability is obtained by standardizing \(\ln X\). If \(\Phi\) denotes the standard normal cumulative distribution function, then

P(X\leq x)=\Phi\left(\frac{\ln x-\mu}{\sigma}\right),\qquad P(X>x)=1-\Phi\left(\frac{\ln x-\mu}{\sigma}\right),\qquad x>0.

The parameter \(\mu\) is the mean of \(\ln X\), and \(e^\mu\) is the median of \(X\); neither is generally the mean of \(X\). The moments are \(E[X]=e^{\mu+\sigma^2/2}\) and \(\operatorname{Var}(X)=e^{2\mu+\sigma^2}(e^{\sigma^2}-1)\). All positive integer moments exist, although the right tail can be much heavier than an exponential tail.

Justification of the lognormal moment formulas: For \(Y=\ln X\sim N(\mu,\sigma^2)\), the normal moment-generating function gives \(E[e^{kY}]=e^{k\mu+k^2\sigma^2/2}\). Since \(X^k=e^{kY}\), \(E[X]=e^{\mu+\sigma^2/2}\) and \(E[X^2]=e^{2\mu+2\sigma^2}\). Subtracting the square of the mean from the second moment gives \(\operatorname{Var}(X)=e^{2\mu+2\sigma^2}-e^{2\mu+\sigma^2}=e^{2\mu+\sigma^2}(e^{\sigma^2}-1)\).

Worked Example: Lognormal Moments and a Tail Probability

A hypothetical large-claim severity \(X\), measured in thousands of dollars, satisfies \(\ln X\sim N(\ln 4,0.25)\). Find its median, mean, variance, standard deviation, and the probability that it exceeds 8 thousand dollars.

Method and parameter check. Here \(\mu=\ln4\), while the normal variance is \(\sigma^2=0.25\), so \(\sigma=0.5\). Use \(\sigma^2\), not \(\sigma\), in the moment formulas.

Median. The median is \(e^\mu=e^{\ln4}=4\) thousand dollars.

Mean. \(E[X]=e^{\ln4+0.25/2}=4e^{0.125}\approx4.5326\) thousand dollars.

Variance and standard deviation. The variance is \(\operatorname{Var}(X)=e^{2\ln4+0.25}(e^{0.25}-1)=16(e^{0.5}-e^{0.25})\). Using \(e^{0.5}\approx1.6487213\) and \(e^{0.25}\approx1.2840254\), this is approximately \(16(0.3646959)=5.8351\) thousand-dollar-squared. The standard deviation is \(\sqrt{5.8351}\approx2.4156\) thousand dollars, or 2.416 to three decimal places.

Exceeding 8. Standardize the logarithm: \(P(X>8)=1-\Phi((\ln8-\ln4)/0.5)=1-\Phi(\ln2/0.5)\). Since \(\ln2/0.5\approx1.3863\), the probability is approximately 0.0828 to four decimal places.

Check and common mistake. The mean exceeds the median, consistent with the distribution’s right skew. The variance calculation uses \(16(e^{0.5}-e^{0.25})\approx5.8351\), and the standard deviation is about 2.4156. Using the wrong standardized tail probability would distort the answer: the normal upper tail at 1.3863 is approximately 0.0828, not 0.0823.

Type I Pareto Severity

The Type I Pareto model has a positive minimum loss level, or scale parameter, \(\theta>0\), and a tail-shape parameter \(\alpha>0\). Its support is \(x\geq\theta\), and its survival and cumulative distribution functions are

S_X(x)=\left(\frac{\theta}{x}\right)^\alpha,\qquad F_X(x)=1-\left(\frac{\theta}{x}\right)^\alpha,\qquad x\geq\theta.

For \(x<\theta\), \(F_X(x)=0\) and \(S_X(x)=1\). The density for \(x\geq\theta\) is \(f_X(x)=\alpha\theta^\alpha x^{-(\alpha+1)}\). A smaller \(\alpha\) gives a more slowly declining tail. The moments require particular care: \(E[X]=\alpha\theta/(\alpha-1)\) only when \(\alpha>1\), and \(\operatorname{Var}(X)=\alpha\theta^2/((\alpha-1)^2(\alpha-2))\) only when \(\alpha>2\). If a condition fails, the corresponding moment is infinite, not a finite value given by substituting into the formula.

Worked Example: Pareto Probabilities and Moment Existence

A hypothetical claim severity in thousands of dollars follows a Type I Pareto distribution with \(\theta=2\) and \(\alpha=3\). Calculate the probability of a claim above 6, the probability it is between 2 and 4, and its mean and variance.

Method and support. The model only permits losses at least 2 thousand dollars. For values in the support, use \(S_X(x)=(2/x)^3\).

Above 6. \(P(X>6)=S_X(6)=(2/6)^3=(1/3)^3=1/27\approx0.0370\) to four decimal places.

Between 2 and 4. Since \(F_X(2)=0\), \(P(2<X\leq4)=F_X(4)-F_X(2)=1-(2/4)^3=1-1/8=0.8750\).

Mean and variance. Since \(\alpha=3>1\), the mean exists and is \(3(2)/(3-1)=3\) thousand dollars. Since \(\alpha=3>2\), the variance exists and is \(3(2^2)/((3-1)^2(3-2))=12/4=3\) thousand-dollar-squared.

Check and common mistake. The probability above 6 is smaller than the probability above 4, which is \((2/4)^3=0.125\). A common error is to use the Pareto moment formulas without checking their conditions. Here both moments exist; if the shape parameter were at most 2, the variance would not be finite.

Comparing the Models

The models differ in their supports, parameter meanings, and tail behavior. These distinctions guide model selection and prevent formulas from being transferred incorrectly from one distribution to another.

Model	Support	Parameter convention	Moment and tail features
Exponential	\(x\geq0\)	Rate \(\lambda\); mean \(1/\lambda\)	Finite mean and variance; memoryless; relatively light tail
Lognormal	\(x>0\)	\(\ln X\sim N(\mu,\sigma^2)\)	All positive integer moments finite; right-skewed with a heavier tail than exponential
Type I Pareto	\(x\geq\theta\)	Scale \(\theta\), shape \(\alpha\)	Mean requires \(\alpha>1\); variance requires \(\alpha>2\); polynomially declining tail

A tail comparison can be made by looking at how \(S_X(x)\) declines as \(x\) grows. The exponential survival function declines exponentially in \(x\), whereas the Pareto survival function declines as a power of \(x\). A lognormal tail is also heavier than an exponential tail, but it is not a power-law tail. These broad descriptions do not establish that one model always assigns a larger probability at every finite threshold: parameter values matter, and the models can cross over particular ranges.

For actuarial work, the expected severity alone may be insufficient when large losses materially affect the risk profile. A model’s tail probability and moment conditions can influence estimates of extreme claims and the stability of calculations that use variance. This comparison is aligned with the CAS Exam 9 (Risk Management for Actuaries) syllabus.

Common Parameter and Calculation Traps

Confusing an exponential rate with a scale. With rate \(\lambda\), the mean is \(1/\lambda\). State the parameter convention before calculating.
Treating lognormal parameters as loss moments. \(\mu\) and \(\sigma^2\) describe \(\ln X\), not \(X\). The loss mean is \(e^{\mu+\sigma^2/2}\).
Ignoring the Pareto support. Type I Pareto losses cannot be below \(\theta\). For thresholds below \(\theta\), the survival probability is 1.
Using a moment formula outside its range. Check \(\alpha>1\) for a finite Pareto mean and \(\alpha>2\) for a finite variance.
Mixing loss units. If \(X\) is measured in thousands of dollars, its mean is in thousands and its variance is in squared thousands. Convert units only after completing the calculation.

Key takeaway: Identify the support and parameter convention first, then select the matching probability or moment formula. For Pareto severities, verify moment-existence conditions before reporting a mean or variance; for lognormal severities, remember that the normal model applies to \(\ln X\).

Check Your Understanding

For each hypothetical severity, calculate the requested quantities using the stated parameterization and units.

A loss \(X\), in thousands of dollars, is exponential with rate \(0.4\) per thousand dollars. Find its mean, variance, and \(P(X>5)\). (Answer: mean \(2.5\) thousand dollars; variance \(6.25\) thousand-dollar-squared; probability \(e^{-2}\approx0.1353\).)
A severity satisfies \(\ln X\sim N(\ln3,0.16)\), with \(X\) in thousands of dollars. Find its median, mean, and \(P(X>6)\). (Answer: median \(3\); mean \(3e^{0.08}\approx3.2499\); probability \(1-\Phi(1.7329)\approx0.0416\).)
A Type I Pareto severity has \(\theta=5\) and \(\alpha=4\). Find \(P(X>10)\), the mean, and the variance. (Answer: probability \(0.0625\); mean \(20/3\approx6.6667\); variance \(50/9\approx5.5556\).)
A severity in thousands of dollars satisfies \(\ln X\sim N(\ln2,0.09)\). Find its mean, variance, and \(P(X>4)\). (Answer: mean \(2e^{0.045}\approx2.0921\); variance \(4e^{0.09}(e^{0.09}-1)\approx0.4122\); probability \(1-\Phi(2.3105)\approx0.0104\).)
A Type I Pareto severity has \(\theta=3\) and \(\alpha=1.8\). Find \(P(X>6)\), its mean, and whether its variance is finite. (Answer: probability \(2^{-1.8}\approx0.2872\); mean \(6.75\); variance is infinite because \(\alpha\leq2\).)

What You'll Learn