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.


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.

  • $\begingroup$ 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