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Looking through the literature dedicated to Bayes factors I noticed that given formulae are quite unclear from the point of Kolmogorov probability axioms.

E.g. consider the following expression from Wikipedia: $$ \Pr (M \mid D) = \dfrac{\Pr (D \mid M) \Pr(M)}{\Pr(D)} \ , $$ where $D$ stands for our sampled data, $M$ - for the model.

Well, $D$ is a random variable. But how should I interpret $M$? In applications, e.g. solving a hierarchial probabilistic model I can consider $\Pr (D \mid M)$ as the conditional density under given parameter. But the concept of $\Pr(M)$ seems rather misleading to me.

Are there any papers with formal definitions based on axiomatic approach?

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  • $\begingroup$ $Pr(M|D)$ is interpreted essentially the same as with RV's. This says, "what is the probability that the model is true (or false) given the collected experimental data." $\endgroup$ Commented Jun 21, 2017 at 12:26
  • $\begingroup$ but what does it mean speaking formally? What is the probabilty of model? Axioms speak about the probability as a measure on some $\sigma$-algebra. How to deal with models in this manner? $\endgroup$ Commented Jun 21, 2017 at 12:31

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Since a model of the distribution of a data set isn't in any strict sence an event, then $\mathsf P(\mathcal M)$ isn't strictly the "probabity that the model is true" in terms of a sigma-algebra interpretation.   (The development of Bayes' probability theory preceeded measure theory.)   The prior is more sensibly interpreted as a measure of "our expectation for the truth of the model", or "our confidence in the model."

Assigning values to the prior is somewhat problematic, which is why Bayes' factor is a comparator that cancels the term.

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  • $\begingroup$ Considering Bayes' factors is definitely self-worth. But I wonder if any strict probabilistic model model may be applied to Bayesian reasoning. $\endgroup$ Commented Jun 22, 2017 at 6:46
  • $\begingroup$ However, I've read this answer and this one and now I feel like it's a case of two different axiomatic approaches. But it's still unclear if Cox's plausibility may be modeled by Kolmogorov probability $\endgroup$ Commented Jun 22, 2017 at 8:08

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