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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

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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.

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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).

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