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Usual non-constructive mathematics leads to some paradoxes (e.g. the Banach-Tarski paradox), which are directly related to the axiom of choice. In non-constructive mathematics, the axiom of choice (as well as proofs by contradiction) are not accepted.

For statistics, the axiom of choice implies that for any $c$ there must be a model such that the BIC (or also the AIC) is below $c$ (see this answer, we can arbitrarily increase the likelihood without increasing $N$, the number of parameters), which kind of defeats the purpose of the BIC as a model selection criterion. This method would not work in constructive math, because the space-filling curves or bijections don't exists in constructive mathematics.

Is there a general treatment of statistics in constructive mathematics? AFAIK a lot of parts of statistics are non-constructive (e.g. Gaussian distribution, Beta distribution, etc). It would be interesting to see which parts of statistics are different when the constructive approach is taken.

EDIT:

It seems that my understanding of constructive mathematics is still to limited for this. So I may have been wrong about the existence of the distributions or the space-filling curves in constructive mathematics. I will edit this question, once I get a better understanding. Feel free to edit, if you can shed more light on this topic.

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  • $\begingroup$ Can you please explain what is non-constructive with the Gaussian distribution? Why do I need the axiom of choice to simulate random variates from a Gaussian? $\endgroup$ – kjetil b halvorsen Sep 12 '19 at 14:21
  • $\begingroup$ What makes you think that space-filling curves are not possible in constructive mathematics? From what I recall the theorem that for any infinite A, there exists bijection between A and AxA in fact relies on nonconstructive assumptions, but note that negation of that statement just means that there exist sets for which it doesn't hold, so doesn't exclude space-filling curves $\endgroup$ – Jakub Bartczuk Sep 12 '19 at 15:03
  • $\begingroup$ @JakubBartczuk According to reddit.com/r/askscience/comments/6opvae/… the existence of a bijection between $A$ and $A\times A$ is equivalent to the axiom of choice (see answer by midtek). However, I am still trying to understand constructive mathematics, so I can't really see how this happens. $\endgroup$ – LiKao Sep 13 '19 at 8:22
  • $\begingroup$ @JakubBartczuk Ok, I should have read the reddit further. It seems only the general case $|A|=|A\times A|$ is equivalent to the AOC, but $|\mathbb{R}|=|\mathbb{R}\times \mathbb{R}|$ can actually be shown without it... Interesting. $\endgroup$ – LiKao Sep 13 '19 at 8:24
  • $\begingroup$ @kjetilbhalvorsen Ok, I see that I should invest more time into grasping constructive mathematics. I thought that, because the integral over the gaussian distribution has no closed form, it would not be possible to show that it actually is a distribution. But the more I think about it, there may be a way around this. I will close this question and edit once I get a better understanding of constructive mathematics. $\endgroup$ – LiKao Sep 13 '19 at 8:26