The inverse Gaussian distribution $IG(\mu,\lambda)$ is associated with the density $$f(x;\mu,\lambda) = \sqrt{\frac{\lambda}{2\pi x^3}}\,\exp\left\{-\frac{\lambda(x-\mu)^2}{2\mu^2x}\right\}\qquad \lambda,\mu,x>0$$ In Schuster (1968), the following connection with the $\chi^2(1)$ distribution is made: if $X\sim IG(\mu,\lambda)$ then$$Z=\frac{\lambda(X-\mu)^2}{2\mu^2X}\sim\chi^2(1)$$ When looking at the proof

enter image description here

I cannot fill the gap between the definition of $Z$ [as a one-to-one transform of $Y$] and the "immediate" conclusion that it is a $\chi^2(1)$ variate. The 1978 review by Folks and Chhikara does not provide further enlightenment.


1 Answer 1


The proof is not exactly standard, although it relates to the "law of the unconscious statistician" [an expression I cannot fathom and do not find amusing] :

First, define $Y=\min\{X,\mu^2/X\}$ which belongs to $(0,\mu)$. The density of $Y$ can be derived from $(y<\mu)$ $$\mathbb P(Y\le y) = \mathbb P(X\le y)+\mathbb P(\mu^2/X \le y\,,\,X>\mu)$$ as $$f_Y(y;\mu,\lambda)=\left\{f_X(y)+\frac{\mu^2}{y^2}f_X(\mu^2/y)\right\}\mathbb I_{(0,\mu)}(y)$$ And if we notice that $$\dfrac{(\mu-\mu^2/y)^2}{\mu^2\,\mu^2/y}=\dfrac{(\mu-\mu^2/y)^2}{\mu^2\,\mu^2/y}=\dfrac{(\mu-y)^2}{\mu^2\,y}$$ which is also why $Z=\frac{(X-\mu)^2}{\mu^2X}$, then \begin{align}f_Y(y;\mu,\lambda)&=\sqrt{\frac{\lambda}{2\pi}}\,e^{-\frac{\lambda(\mu-y)^2}{2\mu^2\,y}}\left\{y^{-3/2}+\mu^{-1}\,y^{-1/2} \right\}\\ &=\sqrt{\frac{\lambda}{2\pi}}\,e^{-\frac{\lambda(\mu-y)^2}{2\mu^2\,y}}\,y^{-3/2}\mu^{-1}\,(\mu+y)\end{align} If we consider the transform$$H(y) = \dfrac{\lambda(\mu-y)^2}{\mu^2\,y}$$ then \begin{align}\left\vert\dfrac{\text{d}H(y)}{\text{d}y}\right\vert &=\frac{\lambda}{\mu^2} \frac{(\mu-y)}{y}\left\{\frac{\mu-y}{y}+2 \right\}\\ &=\frac{\lambda}{\mu^2}\frac{(\mu-y)(\mu+y)}{y^2}\\ &=\frac{\sqrt{\lambda}}{\mu}H(y)^{1/2}\frac{(\mu+y)}{y^{3/2}} \end{align} Which leads to $$\require{enclose} f_Y(y;\mu,\lambda)\text{d}y=\frac{1}{\sqrt{2\pi}}\,e^{-z/2}\,z^{-1/2}\frac{\text{d}z}{\enclose{horizontalstrike}{\text{d}y}}\,\enclose{horizontalstrike}{\text{d}y}=f_Z(z;\mu,\lambda)\text{d}z$$ i.e., a chi-square $\chi^2(1)$ density.

Note that a proof of the above using the moment generating function of $Z$ is straightforward (communication of Éric Marchand from Sherbrooke) and that Seshadri's 1994 book The Inverse Gaussian Distribution is the ultimate reference in the matter (communication of Gérard Letac).


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.