Inverse Gaussian Distributions For Severity

Introduction

A severity model describes the size of an individual loss, often conditional on a claim having occurred. The inverse Gaussian distribution is one positive, continuous model that can represent claim severities with a pronounced right tail. Its two parameters have direct actuarial interpretations: one sets the mean severity, and the other controls its variability.

For this tutorial, $X$ denotes a positive claim severity. The focus is on identifying the inverse Gaussian parameterization, interpreting its parameters, and calculating its variance. The calculations are aligned with the CAS Exam 8 (Advanced Ratemaking) syllabus.

Definition: Under the inverse Gaussian parameterization used here, $X\sim IG(\mu,\lambda)$, where $\mu>0$ is the mean parameter and $\lambda>0$ is the shape parameter. The support is $x>0$, and the density is

f_X(x)=\sqrt{\frac{\lambda}{2\pi x^3}} \exp\!\left(-\frac{\lambda(x-\mu)^2}{2\mu^2x}\right),\qquad x>0.

The moments are $E[X]=\mu$ and $\operatorname{Var}(X)=\mu^3/\lambda$. The distribution models strictly positive continuous outcomes; it does not assign probability to a zero claim.

Interpreting the Parameters

The parameter $\mu$ is the average severity in the units used for $X$. If $X$ is measured in dollars and $\mu=600$, the model’s mean loss is $600. The parameter $\lambda$ is not another mean. It is a shape parameter that determines how tightly losses are concentrated around $\mu$.

For a fixed mean, a larger $\lambda$ gives a smaller variance, because $\operatorname{Var}(X)=\mu^3/\lambda$. A smaller $\lambda$ gives greater dispersion. The model is positively supported, so negative claim severities are impossible under it; its density also has a long right tail, allowing some outcomes to be substantially larger than the mean.

Units provide a useful check. If severity is measured in dollars, then $\mu$ and $\lambda$ are both expressed in dollars under this parameterization. Consequently, $\mu^3/\lambda$ has units of dollars squared, as a variance must. If severity is expressed in thousands of dollars, both parameters must be expressed in thousands of dollars as well.

The standard deviation is the positive square root of the variance, and the coefficient of variation (CV) is the standard deviation divided by the mean. These are useful summaries of the scale of variation, but neither should be confused with the shape parameter itself.

Why the Variance Has This Form

A short moment-generating-function check explains the mean and variance formulas. For $X\sim IG(\mu,\lambda)$, its moment-generating function is

M_X(t)=\exp\!\left\{\frac{\lambda}{\mu} \left(1-\sqrt{1-\frac{2\mu^2t}{\lambda}}\right)\right\}, \qquad t<\frac{\lambda}{2\mu^2}.

Let $K(t)=\log M_X(t)$, the cumulant-generating function. Expanding the square root about $t=0$ gives the first two terms needed for the moments:

\sqrt{1-z}=1-\frac{z}{2}-\frac{z^2}{8}+O(z^3), \qquad z=\frac{2\mu^2t}{\lambda}.

Substituting this expansion into $K(t)$ yields

K(t)=\mu t+\frac{\mu^3}{2\lambda}t^2+O(t^3).

The first derivative of $K$ at zero is the mean; the second derivative at zero is the variance. Thus $K'(0)=\mu$ and $K''(0)=\mu^3/\lambda$. This check also highlights why using $\mu^2/\lambda$ for the variance is incorrect: the power of $\mu$ must be three.

Calculating Variance and Standard Deviation

For a supplied mean and shape parameter, use the following sequence. Keep units attached to the parameters through the calculation, and take the square root only after finding the variance.

Identify the model and units.
Confirm that the stated parameters are $\mu$ and $\lambda$ for an inverse Gaussian severity, and that they use the same severity unit.

Substitute into the variance formula.
Calculate $\mu^3/\lambda$, preserving the squared units in the result.

Calculate and interpret the standard deviation if needed.
Take the positive square root of the variance and compare it with the mean or other supplied quantities.

Worked Example: Variance in Dollars

A hypothetical insurer models the positive severity of a particular equipment-repair coverage as $X\sim IG(\mu,\lambda)$, with $\mu=\$600$ and $\lambda=\$2{,}400$. Assume no discounting or mortality adjustment is involved; the parameters describe the severity in dollars. Calculate the variance, standard deviation, and CV. Name the method and identify a common mistake it avoids.

Method. Apply the inverse Gaussian moments, then calculate the square root and divide by the mean. The common mistake to avoid is using $\mu^2/\lambda$ instead of $\mu^3/\lambda$.

Variance. First cube the mean:

\mu^3=600^3=600\times600\times600=216{,}000{,}000.

Then divide by the shape parameter:

\operatorname{Var}(X)=\frac{216{,}000{,}000}{2{,}400} =90{,}000\text{ dollars}^2.

Standard deviation.

\sqrt{90{,}000}=\$300.

CV. Divide the standard deviation by the mean:

\operatorname{CV}=\frac{300}{600}=0.5=50\%.

Check. The variance has squared-dollar units and its square root is $300. The CV is dimensionless. The results are consistent with the parameters and the stated units.

Worked Example: Parameters Expressed in Thousands of Dollars

For a hypothetical specialty-property portfolio, positive claim severity is modeled as $X\sim IG(\mu,\lambda)$, where both parameters are measured in thousands of dollars. Let $\mu=2.4$ and $\lambda=9.6$. Assume there is no discounting or mortality component. Find the variance in squared thousands of dollars, the standard deviation in dollars, and the CV. Explain the unit conversion.

Method. Calculate the moments in the units used for the parameters, then convert the standard deviation to dollars. The common mistake to avoid is treating a variance in squared thousands of dollars as though it were already in dollars squared.

Variance. Cube the mean parameter:

2.4^3=2.4\times2.4\times2.4=13.824.

Divide by $\lambda=9.6$:

\operatorname{Var}(X)=\frac{13.824}{9.6} =1.44\text{ (thousand dollars)}^2.

Standard deviation. In thousands of dollars, it is

\sqrt{1.44}=1.2\text{ thousand dollars}=\$1{,}200.

CV. The mean is 2.4 thousand dollars, so

\operatorname{CV}=\frac{1.2}{2.4}=0.5=50\%.

Check. The standard deviation and mean use the same unit when computing the CV, and the ratio has no units. Squaring $1,200 gives $1,440,000 dollars squared, equal to $1.44\times1{,}000{,}000$ dollars squared, as required.

Worked Example: Solving for the Shape Parameter

A hypothetical liability severity model uses an inverse Gaussian distribution with mean $\mu=\$800$. Its target variance is $\$160{,}000$ squared. Assume parameters are expressed in dollars and no discounting or mortality adjustment is applied. Find the shape parameter $\lambda$, then state what would happen to the variance if $\lambda$ were doubled while the mean remained fixed.

Method. Rearrange the inverse Gaussian variance formula to isolate $\lambda$. The common mistake to avoid is multiplying by the variance rather than dividing $\mu^3$ by it.

Starting with $\operatorname{Var}(X)=\mu^3/\lambda$, multiply both sides by $\lambda$, then divide by $\operatorname{Var}(X)$:

\lambda=\frac{\mu^3}{\operatorname{Var}(X)}.

Cube the mean:

800^3=800\times800\times800=512{,}000{,}000.

Substitute the target variance:

\lambda=\frac{512{,}000{,}000}{160{,}000} =3{,}200\text{ dollars}.

If $\lambda$ doubles to $6,400 while $\mu$ stays at $800, the variance is halved because $\mu^3$ is unchanged and the denominator doubles:

\operatorname{Var}(X)=\frac{800^3}{6{,}400} =\frac{512{,}000{,}000}{6{,}400} =80{,}000\text{ dollars}^2.

Check. Substituting the original $\lambda=\$3{,}200$ gives $512{,}000{,}000/3{,}200=160{,}000$ dollars squared, the target variance. The doubled shape parameter produces half that variance.

Common Mistake: Using the Wrong Power or Units

The most important arithmetic trap is omitting one factor of $\mu$. For the first worked example, the incorrect calculation would be

\frac{\mu^2}{\lambda} =\frac{600^2}{2{,}400} =\frac{360{,}000}{2{,}400} =150.

This is not the inverse Gaussian variance. It also has units of dollars, because dollars squared divided by dollars leaves dollars. A variance must have squared units. The correct calculation is

\frac{\mu^3}{\lambda} =\frac{600^3}{2{,}400} =90{,}000\text{ dollars}^2.

A second trap is mixing scales. If the mean is entered in thousands of dollars, the shape parameter must also be in thousands of dollars. Converting only one parameter changes the calculated variance. A final check of units and of the relationship between the parameters can often expose such errors before an answer is reported.

Using the Model Carefully

The inverse Gaussian is a candidate model for positive, continuous severity values when right-skewed outcomes are plausible. Its mean specifies the center in an average-loss sense, while its shape parameter governs dispersion around that mean. This makes the model’s parameters useful for both interpretation and basic moment calculations.

The model’s support matters. Since $X>0$, it is suitable for positive severity, not a response variable that includes a point mass at zero without further modeling. Whether the model fits a particular portfolio is an empirical question; the parameter formulas alone do not establish goodness of fit. In a ratemaking analysis, distribution choice should be consistent with how the response was defined and with the observed data.

For calculations, retain the parameterization $IG(\mu,\lambda)$ and the formula $\operatorname{Var}(X)=\mu^3/\lambda$. Check the units, cube $\mu$, divide by $\lambda$, and only then calculate a standard deviation if requested.

Key takeaway: For $X\sim IG(\mu,\lambda)$, severity is positive, $E[X]=\mu$, and $\operatorname{Var}(X)=\mu^3/\lambda$. At fixed mean, increasing $\lambda$ decreases variance. Keep both parameters on the same measurement scale and report variance in squared units.

Check Your Understanding

Work each problem using the inverse Gaussian parameterization $X\sim IG(\mu,\lambda)$. Unless a question says otherwise, assume the parameters describe positive claim severity without discounting or mortality adjustment.

A hypothetical water-damage severity has $\mu=\$300$ and $\lambda=\$900$. Calculate the variance and standard deviation. (Answer: variance $=\$30{,}000$ squared; standard deviation $=\$173.21$, rounded to cents.)
A positive severity is modeled with $\mu=1.2$ and $\lambda=3.6$, both in thousands of dollars. Find the variance in squared thousands of dollars and the standard deviation in dollars. (Answer: variance $=0.48$ (thousand dollars)$^2$; standard deviation $=\$692.82$, rounded to cents.)
A severity model has mean $\$500$ and variance $\$125{,}000$ squared. Solve for $\lambda$, in dollars, and verify the result by substitution. (Answer: $\lambda=\$1{,}000$; $500^3/1{,}000=125{,}000$ dollars squared.)
Two hypothetical severity models both have mean $\$400$. Model A has $\lambda=\$800$, and Model B has $\lambda=\$3{,}200$. Calculate both variances and state their ratio, A to B. (Answer: Model A variance $=80{,}000$ dollars squared; Model B variance $=20{,}000$ dollars squared; ratio $=4:1$.)
For a positive loss with $\mu=\$900$ and $\lambda=\$3{,}600$, calculate the variance, standard deviation, and CV. (Answer: variance $=202{,}500$ dollars squared; standard deviation $=\$450$; CV $=0.5$, or $50\%$.)

What You'll Learn