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For an expected value parameter $\mu >0$ and a shape parameter $\lambda >0$, the inverse Gaussian distribution has probability density functionIts mean is $\mu$ and its variance is ${\mu }^3/\lambda$.

This is the generalized Pareto distribution for the excess $X-u$, with scale $\sigma >0$ and shape $\xi$. Its support requires $1+\xi (x-u)/\sigma >0$. The Pickands-Balkema-de Haan theorem says that, for distributions in an appropriate extreme-value domain of attraction, the conditional distribution of excesses above a sufficiently high threshold approaches a generalized Pareto form. This makes it the natural peaks-over-threshold method model; it is an asymptotic justification, not an assertion that every threshold is sufficiently high.

For $\xi >0$, the tail decays as a power and has no finite upper endpoint, giving a heavy tail. For $\xi =0$, the limiting distribution is exponential, with survival probability $\exp \{-(x-u)/\sigma \}$. For $\xi <0$, there is a finite upper endpoint $u-\sigma /\xi$, giving a bounded tail. Thus the sign of the shape parameter distinguishes heavy, exponential-type and bounded tails.

Expand the exponent in the Inverse Gaussian distribution with unit shape parameter:Consequently its probability density function can be written aswhereIndeed, $-A(\eta )=\sqrt{-2\eta }=1/\mu$. This is the canonical form of a one-parameter exponential family, and its natural parameter is therefore

Here the natural statistic is $T(X)=X$. The exponential-family derivative identities giveandThus