inverse Gaussian trick [or treat?]

When preparing my mid-term exam for my undergrad mathematical statistics course, I wanted to use the inverse Gaussian distribution IG(μ,λ) as an example of exponential family and include a random generator question. As shown above by a Fortran computer code from Michael, Schucany and Haas, a simple version can be based on simulating a χ²(1) variate and solving in x the following second degree polynomial equation

\dfrac{\lambda(x-\mu)^2}{\mu^2 x} = v

since the left-hand side transform is distributed as a χ²(1) random variable. The smallest root x¹, less than μ, is then chosen with probability μ/(μ+x¹) and the largest one, x²=μ²/x¹ with probability x¹/(μ+x¹). A relatively easy question then, except when one considers asking for the proof of the χ²(1) result, which proved itself to be a harder cookie than expected! The paper usually referred to for the result, Schuster (1968), is quite cryptic on the matter, essentially stating that the above can be expressed as the (bijective) transform of Y=min(X,μ²/X) and that V~χ²(1) follows immediately. I eventually worked out a proof by the “law of the unconscious statistician” [a name I do not find particularly amusing!], but did not include the question in the exam. But I found it fairly interesting that the inverse Gaussian can be generating by “inverting” the above equation, i.e. going from a (squared) Gaussian variate V to the inverse Gaussian variate X. (Even though the name stems from the two cumulant generating functions being inverses of one another.)

4 Responses to “inverse Gaussian trick [or treat?]”

  1. Gerard Letac Says:

    Le livre de Seshadri (1927-2020) Oxford University Press ‘(1993) ‘The inverse Gaussian distribution’ contient tous les details de ta question pages 81-83.

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