Inverse Gaussian Mean-Variance Derivation

Why the Mean-Variance Relationship Matters

For a positive insurance severity, the mean describes the expected payment, while the variance describes how widely individual payments vary around that mean. The inverse Gaussian distribution is useful when a positive response has variability that grows rapidly as its mean increases. Its defining mean-variance relationship is not simply “variance is proportional to the mean”: with shape held fixed, variance is proportional to the cube of the mean.

This relationship is important when comparing hypothetical severity groups or interpreting an inverse Gaussian model. A larger fitted mean does not, by itself, tell us the variance unless we also know what is assumed about the shape parameter. The derivation below makes that condition explicit.

Definition: In this tutorial, an inverse Gaussian random variable $X$ has mean parameter $\mu>0$ and shape parameter $\lambda>0$, with density $$ f_X(x)=\sqrt{\frac{\lambda}{2\pi x^3}}\exp\left(-\frac{\lambda(x-\mu)^2}{2\mu^2x}\right),\qquad x>0. $$ The support is strictly positive. Under this parameterization, $E[X]=\mu$, and the variance is to be derived. Some applications instead use the dispersion parameter $\phi=1/\lambda$.

The parameters have different roles. The mean $\mu$ sets the expected value; the shape $\lambda$ controls dispersion at a given mean. A larger $\lambda$, with $\mu$ fixed, produces a smaller variance. If $\lambda$ is fixed and $\mu$ changes, the variance changes with $\mu^3$.

Deriving the Mean and Variance

The moment-generating function (MGF) for the stated inverse Gaussian parameterization is $$ M_X(t)=\exp\left\{\frac{\lambda}{\mu}\left[1-\sqrt{1-\frac{2\mu^2t}{\lambda}}\right]\right\},\qquad t<\frac{\lambda}{2\mu^2}. $$ The upper bound on $t$ is the region in which the square root is real and the MGF is finite. Define the cumulant-generating function $K(t)=\log M_X(t)$. Differentiating $K$ gives the mean and variance: $K'(0)=E[X]$ and $K''(0)=\operatorname{Var}(X)$.

K(t)=\frac{\lambda}{\mu}\left[1-\left(1-\frac{2\mu^2t}{\lambda}\right)^{1/2}\right].

For the first derivative, apply the chain rule to the square-root term:

\begin{aligned} K'(t) &=\frac{\lambda}{\mu}\left[-\frac{1}{2}\left(1-\frac{2\mu^2t}{\lambda}\right)^{-1/2}\left(-\frac{2\mu^2}{\lambda}\right)\right]\\ &=\mu\left(1-\frac{2\mu^2t}{\lambda}\right)^{-1/2}. \end{aligned}

At $t=0$, this becomes $K'(0)=\mu$. Thus the parameter $\mu$ is indeed the mean. Differentiate once more:

\begin{aligned} K''(t) &=\mu\left[-\frac{1}{2}\left(1-\frac{2\mu^2t}{\lambda}\right)^{-3/2}\left(-\frac{2\mu^2}{\lambda}\right)\right]\\ &=\frac{\mu^3}{\lambda}\left(1-\frac{2\mu^2t}{\lambda}\right)^{-3/2}. \end{aligned}

Evaluating at zero gives the required result:

Theorem: If $X$ has an inverse Gaussian distribution with mean $\mu>0$ and shape $\lambda>0$, then $$ E[X]=\mu,\qquad \operatorname{Var}(X)=\frac{\mu^3}{\lambda}. $$ Equivalently, using dispersion $\phi=1/\lambda$, the variance is $\phi\mu^3$. The relationship assumes the stated inverse Gaussian parameterization; it should not be transferred unchanged to a different distribution or parameter convention.

The variance function is therefore $V(\mu)=\mu^3$, multiplied by dispersion $\phi$. In a generalized linear model context, this is the inverse Gaussian mean-variance pattern: the variance is $\phi V(\mu)=\phi\mu^3$. The formula describes conditional response variability under the model; it does not say that every observed value must lie close to its mean.

What Changes When the Mean Changes?

Suppose first that $\lambda$ is held fixed. If the mean is multiplied by a positive factor $c$, then

\operatorname{Var}(X_{\mathrm{new}}) =\frac{(c\mu)^3}{\lambda} =c^3\frac{\mu^3}{\lambda} =c^3\operatorname{Var}(X_{\mathrm{old}}).

The variance scales by $c^3$, and the standard deviation scales by $c^{3/2}$. The coefficient of variation (standard deviation divided by mean) is

\operatorname{CV}(X) =\frac{\sqrt{\operatorname{Var}(X)}}{E[X]} =\frac{\sqrt{\mu^3/\lambda}}{\mu} =\sqrt{\frac{\mu}{\lambda}}.

Consequently, with fixed shape, a higher mean also means a higher coefficient of variation. That conclusion changes if the shape is adjusted as the mean changes. For example, keeping the coefficient of variation fixed requires $\lambda$ to increase in proportion to $\mu$. Keeping the variance fixed requires $\lambda$ to increase in proportion to $\mu^3$.

Worked Example: Calculate Variance and Relative Variability

A hypothetical portfolio models individual claim severities $X$, measured in thousands of dollars, using an inverse Gaussian distribution with $\mu=2$ and $\lambda=8$. Find the variance, standard deviation, and coefficient of variation. No interest or mortality assumptions apply: these are undiscounted claim severities, not life-contingent payments.

Method. Use $\operatorname{Var}(X)=\mu^3/\lambda$, then calculate standard deviation and divide by the mean for the coefficient of variation.

First cube the mean: $\mu^3=2^3=8$. Then divide by the shape: $$ \operatorname{Var}(X)=\frac{8}{8}=1\text{ thousand-dollar-squared}. $$

The standard deviation is $\sqrt{1}=1$ thousand dollars. The coefficient of variation is $1/2=0.5$, or 50%.

Check. The variance has squared monetary units, while the standard deviation has the same units as the mean. The answer is consistent with $\lambda=8$: increasing the shape would reduce the variance if the mean stayed at 2. A common mistake avoided here is reporting the variance, 1, as though it were the standard deviation; taking the square root is required.

Worked Example: Compare Two Means at Fixed Shape

Two hypothetical severity segments are modeled with inverse Gaussian distributions in thousands of dollars. Segment A has $\mu=2$; Segment B has $\mu=4$. Both use $\lambda=8$. Calculate each variance, compare their standard deviations, and compare their coefficients of variation. No interest or mortality assumptions apply.

Method. Since shape is fixed, calculate $\mu^3/\lambda$ for each segment. Do not assume that doubling the mean merely doubles the variance.

For Segment A: $$ \operatorname{Var}(X_A)=\frac{2^3}{8}=\frac{8}{8}=1. $$ Its standard deviation is $\sqrt{1}=1$ thousand dollars, and its coefficient of variation is $1/2=0.5$.

For Segment B: $$ \operatorname{Var}(X_B)=\frac{4^3}{8}=\frac{64}{8}=8. $$ Its standard deviation is $\sqrt{8}=2.8284$ thousand dollars, rounded to four decimal places. Its coefficient of variation is $\sqrt{8}/4=\sqrt{0.5}=0.7071$, also rounded to four decimal places.

The mean doubled, but the variance increased by a factor of $8/1=8$, as predicted by $2^3=8$. The standard deviation increased by $\sqrt{8}=2.8284$, not by 8. The coefficient of variation increased from 0.5 to 0.7071. A common mistake avoided is comparing variance to mean as if they scaled at the same rate; at fixed shape, the mean-cubed formula governs.

Worked Example: Choose Shape for a Target Coefficient of Variation

A hypothetical severity model in thousands of dollars has mean $\mu=5$. The modeling objective is a coefficient of variation of 0.4. Find the required inverse Gaussian shape parameter and variance. Then determine the variance for a second hypothetical segment with mean 10 if it has the same coefficient of variation. No interest or mortality assumptions apply.

Method. Start with $\operatorname{CV}=\sqrt{\mu/\lambda}$ and solve for $\lambda$; then use $\operatorname{Var}(X)=\mu^3/\lambda$.

For the first segment, square the coefficient of variation: $$ 0.4^2=0.16=\frac{\mu}{\lambda}=\frac{5}{\lambda}. $$ Therefore $\lambda=5/0.16=31.25$. Its variance is $$ \operatorname{Var}(X)=\frac{5^3}{31.25}=\frac{125}{31.25}=4 $$ thousand-dollar-squared, and its standard deviation is $\sqrt{4}=2$ thousand dollars. Checking the target, $2/5=0.4$.

For the second segment, the mean is 10 and the coefficient of variation remains 0.4. Thus $$ 0.16=\frac{10}{\lambda},\qquad \lambda=\frac{10}{0.16}=62.5. $$ Its variance is $10^3/62.5=1000/62.5=16$ thousand-dollar-squared. The variance is four times the first segment’s variance, while the mean is twice as large. This is consistent: equal coefficients of variation make variance proportional to the square of the mean. The shape had to change; treating it as fixed would not preserve the target coefficient of variation.

Common Exam Trap and Practical Interpretation

A frequent error is to use $\operatorname{Var}(X)=\mu^2/\lambda$, perhaps because the square on $\mu$ in the density exponent is remembered but the moment relationship is not. The correct power is three:

\text{Wrong: }\operatorname{Var}(X)=\frac{\mu^2}{\lambda} \qquad\qquad \text{Correct: }\operatorname{Var}(X)=\frac{\mu^3}{\lambda}.

For example, with $\mu=4$ and $\lambda=8$, the wrong calculation gives $16/8=2$, while the correct calculation gives $64/8=8$. The discrepancy is material. Another trap is silently changing the comparison: “same shape” and “same coefficient of variation” are not equivalent conditions. Under fixed shape, $\operatorname{CV}=\sqrt{\mu/\lambda}$ increases with the mean. Under fixed coefficient of variation, shape must rise in proportion to the mean.

When interpreting a fitted model, keep three questions separate: what is the mean, what is the dispersion or shape, and what quantity is being compared? With $\phi=1/\lambda$, increasing dispersion $\phi$ increases variance; with $\lambda$, increasing shape decreases variance. The reciprocal convention is an easy source of sign-of-effect mistakes.

Identify the parameterization.
Confirm that $\mu$ is the mean and $\lambda$ is the shape in the density or model statement.

Write the correct variance formula.
Use $\operatorname{Var}(X)=\mu^3/\lambda$, or $\phi\mu^3$ when $\phi=1/\lambda$.

State what is fixed.
For comparisons, specify whether shape, dispersion, variance, or coefficient of variation is held constant.

Check units and scale.
Variance has squared units, standard deviation has the units of the mean, and ratios such as the coefficient of variation are unit-free.

Key takeaway: For an inverse Gaussian response with mean $\mu$ and shape $\lambda$, the variance is $\mu^3/\lambda$. Holding shape fixed makes variance grow cubically with the mean; holding the coefficient of variation fixed requires shape to grow linearly with the mean.

Check Your Understanding

For each hypothetical severity model, use the inverse Gaussian mean-and-shape parameterization. Severity amounts are in thousands of dollars, and no interest or mortality assumptions apply.

A model has $\mu=3$ and $\lambda=9$. Calculate its variance and standard deviation. (Answer: variance $=3$ thousand-dollar-squared; standard deviation $=\sqrt{3}=1.7321$ thousand dollars.)
At fixed shape, a severity mean increases from 2 to 6. By what factor does its variance change? (Answer: $3^3=27$, so variance increases by a factor of 27.)
A model has $\mu=4$ and $\lambda=16$. Calculate its variance, standard deviation, and coefficient of variation. (Answer: variance $=4$, standard deviation $=2$ thousand dollars, coefficient of variation $=0.5$.)
A hypothetical severity has $\mu=2$ and a target variance of 0.5. Find the required shape parameter and check the resulting variance. (Answer: $\lambda=2^3/0.5=16$; variance $=8/16=0.5$ thousand-dollar-squared.)
Two models have $(\mu,\lambda)=(3,12)$ and $(6,24)$. Calculate their variances and the ratio of the second variance to the first. (Answer: variances $=27/12=2.25$ and $=216/24=9$ thousand-dollar-squared; the ratio is $9/2.25=4$.)

What You'll Learn