= Solution
For the constructed output, its defining quadratic says $h(X)=Y$ identically. Since the preceding branch calculation proves that its law is $IG(\mu,\lambda)$, any variable with that law has the same pushforward distribution:
$$
\boxed{\frac{\lambda(X-\mu)^2}{\mu^2X}\sim\chi_1^2.}
$$
One can also verify this <chi-squared transform of an inverse Gaussian variable> directly. The density contributions from the inverse branches are
$$
f_X(x_-(y))|x_-'(y)|+f_X(x_+(y))|x_+'(y)|
=g(y)\left(\frac{\mu}{\mu+x_-(y)}+\frac{\mu}{\mu+x_+(y)}\right)=g(y).
$$
The final equality follows from $x_-x_+=\mu^2$. This confirms the transformed law independently of which reciprocal root was selected.
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