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 $X\sim IG(\mu ,\lambda )$, $h(X)=\lambda (X-\mu {)}^2/({\mu }^{2}X)$ 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.