Inverse Gaussian distribution

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 ${\mu }^3/\lambda$.

For an independent statistical sample from an Inverse Gaussian distribution with mean $\mu >0$ and known shape $\lambda >0$, the maximum-likelihood estimator of $\mu$ is the sample mean. To prove this, expand the log-likelihood function to get $\ell (\mu )=C-\lambda {\sum }_i{y}_i/(2{\mu }^2)+n\lambda /\mu$. Its score function is $\lambda ({\sum }_i{y}_i-n\mu )/{\mu }^3$, positive below $\bar{y}$ and negative above it. Taking the negative expected second derivative gives one-observation Fisher information $I_1(\mu )=\lambda /{\mu }^3$. Thus the regular asymptotic normality of a maximum likelihood estimator yields $\sqrt{n}(\bar{Y}-\mu )\stackrel{d}{\to }N(0,{\mu }^3/\lambda )$. The same conclusion follows from the central limit theorem and the distribution's variance ${\mu }^3/\lambda$.

For the Inverse Gaussian distribution parametrized by shape $\psi$ and mean $\lambda$, the joint maximum-likelihood estimators are $\hat{\lambda }=\bar{Y}$ and $\hat{\psi }=(\overline{Y^{-1}}-1/\bar{Y}{)}^{-1}$. With known $\lambda$, the shape estimator is instead $(\overline{Y^{-1}}-2/\lambda +\bar{Y}/{\lambda }^2{)}^{-1}$. Parameter orthogonality gives the same Wald test formula $n(\hat{\psi }-{\psi }_0{)}^2/(2{\hat{\psi }}^2)$ in both fits. The two estimators differ by $O_p(n^{-1})$ under regular model sampling, so the test statistics agree to first order under the null; they need not be numerically equal in finite samples.

In the parametrization with density proportional to $y^{-3/2}\exp [-(\lambda /y+\varphi y)/2]$, the Inverse Gaussian distribution has cumulant-generating function $K(t)=\sqrt{\lambda \varphi }-\sqrt{\lambda (\varphi -2t)}$. For $n$ independent copies, the average has this same family with parameters $(n\varphi ,n\lambda )$, while the sum has parameters $(\varphi ,n^2\lambda )$. These assertions follow by multiplying moment-generating functions and scaling the density. The more usual mean parameter is $\mu =\sqrt{\lambda /\varphi }$, so the sum has mean $n\mu$ and shape $n^2\lambda$.

For the inverse Gaussian sum closure, take $x=s/n>0$. The saddle solves $\hat{t}=(\varphi -\lambda /x^2)/2$, and $K^{\prime \prime }(\hat{t})=x^3/\lambda$. Substitution in the saddlepoint density approximation givesThis is the exact sum density, so its leading saddlepoint expression needs no correction or renormalization.

For $X\sim IG(\mu ,\lambda )$, $h(X)=\lambda (X-\mu {)}^2/({\mu }^{2}X)$ has the chi-squared-one law. Its two inverse branches contribute weights $\mu /(\mu +x_{-})$ and $\mu /(\mu +x_{+})$ to the transformed density, and reciprocal roots make those weights sum to one.

For a chi-squared-one variable $Y$, the positive inverse branches of $h(x)=\lambda (x-\mu {)}^2/({\mu }^{2}x)$ have product ${\mu }^2$. Weight a branch value $x$ by $\mu /(\mu +x)$; the two weights sum to one. Multiplying the chi-squared density by this weight and the branch Jacobian gives the inverse Gaussian density on both sides of $\mu$.

With $t=\mu Y/(2\lambda )$, the lower root is $\mu /(1+t+\sqrt{t(t+2)})$ and the upper root is ${\mu }^2/x_{-}$. Rationalizing the lower-root formula avoids subtracting nearly equal large numbers when $Y$ is large.