Past exam of the mathematics course of the University of Cambridge / 2012 / iii / Paper 38 / 3 / 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:One can also verify this chi-squared transform of an inverse Gaussian variable directly. The density contributions from the inverse branches areThe final equality follows from $x_{-}{x}_{+}={\mu }^2$. This confirms the transformed law independently of which reciprocal root was selected.