Suppose I observe a random variable $x$ drawn from a non-central t- or F- distribution, and would like to perform inference on the non-centrality parameter. How would a Bayesian approach this problem? Are there known conjugate priors for these?

For the t-distribution I have seen some reference to LeCoutre's 'lambda-prime' distribution, but I believe this is for a non-informative prior (and I know very little about Bayesian analysis). Is there a parallel result for the F-distribution? How are these used?

  • $\begingroup$ Are you desiring a closed-form solution, or would you be willing to use MCMC techniques to generate a (large) sample from the posterior distribution? $\endgroup$
    – jbowman
    Feb 17, 2012 at 13:48
  • $\begingroup$ I am looking for a closed-form solution. I was also curious if there was a well-developed Bayesian view of e.g. the frequentist t- and F-tests for regression, ANOVA, etc. $\endgroup$
    – shabbychef
    Feb 18, 2012 at 6:36

1 Answer 1


Both $t$ and $F$ distributions being outside exponential families, there is no conjugate distribution even in the central case (see Bayesian Choice, Chap. 3). You thus need to use non-conjugate priors and numerical methods like MCMC. For instance, using a flat prior on the non-centrality parameter.

As for your question on "How are these used?", I can only advise reading at least an introduction to Bayesian inference...

  • 1
    $\begingroup$ Note that t and F distributions are functions of random variables from exponential families. Thus there is hope of finding conditionally conjugate priors - andrew gelman's folded t distributions for variance components in multilevel modelling is one example. $\endgroup$ Feb 17, 2012 at 20:50
  • $\begingroup$ @probabilityislogic: truly, the hidden variable representations of the $t$ and $F$ distributions can rescind an exponential family setting. (I called this approach hidden mixtures in the early 90's.) For someone with little knowledge of Bayesian analysis, this may be too involved, though. $\endgroup$
    – Xi'an
    Feb 18, 2012 at 17:27

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