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Let's say we have some binary variable of interest and fit a model to predict the probability of the two classes, say a logistic regression or a "classification" neural network. This model gives us predictions in the interval $[0, 1]$.

Must these predictions satisfy the Kolmogorov probability axioms, even if these predictions lack calibration?

We definitely get that the values are non-negative, and I think having an upper bound $1$ gives us unit measure (but I am not as confident about this). For $\sigma$-additivity, I have no idea.

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    $\begingroup$ The Kolmogoroff axioms are formal properties, calibration is not part of the axioms. $\endgroup$
    – kjetil b halvorsen ♦
    Commented Sep 12, 2023 at 20:27
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    $\begingroup$ Well, there are some other properties ... the logistic regression model you mention is based on the binomial distribution. Estimated probabilities you can calculate from the model is binomial probabilities, so you only need to check if the binomial distribution is consistent with Kolmogoroff! $\endgroup$
    – kjetil b halvorsen ♦
    Commented Sep 12, 2023 at 20:34
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    $\begingroup$ Sigma additivity of what space, specifically? The sample space seems to be $\{0,1\}$ with the discrete measure but surely you don't have any questions about additivity (much less sigma additivity) in that space! $\endgroup$
    – whuber ♦
    Commented Sep 12, 2023 at 20:41
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    $\begingroup$ So the Kolmogorov axioms are satisfied, since $p\ge 0$, $1-p\ge 0$ and $p+(1-p)=1$ $\endgroup$
    – Henry
    Commented Sep 12, 2023 at 22:08
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    $\begingroup$ @NuclearHoagie That makes it sound like the requirement is just that the values be in the $[0,1]$ interval. That seems to be the gist of the answer/comments by Kjetil, and perhaps that is what is meant by saying the Kolmogorov axioms are just formal properties. $\endgroup$
    – Dave
    Commented Sep 14, 2023 at 19:02

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Making comments into an answer:

  1. The Kolmogorov axioms are formal properties, calibration is not part of the axioms.

  2. The logistic regression model you mention is based on the binomial distribution. Estimated probabilities you can calculate from the model is binomial probabilities, so you only need to check if the binomial distribution is consistent with Kolmogorov!

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  • $\begingroup$ For the second point, is all you mean that $p,1-p\in[0,1]$, $p +(1-p)=1$, and $P(\{0\}\cup\{1\})=1=p+(1-p)?$ $\endgroup$
    – Dave
    Commented Sep 12, 2023 at 22:10
  • $\begingroup$ I guess I mean that the binomial distribution is developed using theory based on the axioms, so how could it not be consistent with the axioms? $\endgroup$
    – kjetil b halvorsen ♦
    Commented Sep 12, 2023 at 23:11
  • $\begingroup$ That makes sense. However, something is still troubling about this. Is the point of checking the Kolmogorov axioms that we check them for each Bernoulli distribution with fitted/predicted probability parameter $\hat p_i$, rather than checking across all predictions? Why? $\endgroup$
    – Dave
    Commented Sep 13, 2023 at 15:04
  • $\begingroup$ The point is maybe that no checking needs to be done, as it is already done! $\endgroup$
    – kjetil b halvorsen ♦
    Commented Sep 13, 2023 at 15:19
  • $\begingroup$ I agree that it is done if we only care about each individual Bernoulli distribution. However, why don’t we care about the entire collection of predictions? $\endgroup$
    – Dave
    Commented Sep 13, 2023 at 15:28

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