= Solution
For $x>\mu$, the corresponding smaller root is $\mu^2/x$. The <conditional probability> of selecting the larger root is
$$
1-\frac{\mu}{\mu+\mu^2/x}
=\frac{\mu}{\mu+x}.
$$
It is the same weight as a function of the actual selected value $x$. Hence
$$
p_+(x)=\frac{\mu}{\mu+x}\,g(h(x))\,|h'(x)|
=\sqrt{\frac{\lambda}{2\pi x^3}}
\exp\!\left(-\frac{\lambda(x-\mu)^2}{2\mu^2x}\right).
$$
Combining the two open intervals proves \b[the output has the <Inverse Gaussian distribution> $IG(\mu,\lambda)$]. The event $Y=0$ has probability zero, so no atom occurs at $\mu$; the limiting density there is finite. The algorithm always selects one of its positive roots, so these branch densities exhaust its <probability distribution>.
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