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Suppose we role a dice and see what we get, the sample space is $\{1,2,3,4,5,6\}$ and each outcome occurs with probability 1/6. For example, if we look at the probability that 6 appears, it seems natural to interpret the probability of 1/6 as $plim \frac{n_6}{n}$ where $n$ is the total number of trials and $n_6$ is the number of times we get a 6. That is to say, the frequentist interpretation of 1/6 as the limiting frequency seems natural. This is also something that could be verified using repeated experiment.

My questions are:

  1. In this example, how to take a Bayesian interpretation of the probability 1/6? Is the Bayesian interpretation less appropriate/suitable in this example compared to the frequentist interpretation?

  2. In general, is it true that the Bayesian interpretation of probability makes more sense under certain scenarios, while the frequentist interpretation of probability is better in other scenarios?

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    $\begingroup$ We don't even need frequentism for this example where the probabilities are well defined without repeated sampling; the "classical probability" described in the following link suffices plato.stanford.edu/entries/probability-interpret/#ClaPro $\endgroup$ Commented Sep 14, 2022 at 12:39
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    $\begingroup$ See my answer at stats.stackexchange.com/questions/332026/… Also Bayesians need idealized frequencies to give meaning to probabilities! $\endgroup$ Commented Sep 14, 2022 at 13:08
  • $\begingroup$ @JohnMadden Thanks, John! This is extremely helpful! $\endgroup$ Commented Sep 15, 2022 at 15:59
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    $\begingroup$ @kjetilbhalvorsen Thanks! I've read it. Very insightful! $\endgroup$ Commented Sep 15, 2022 at 16:04

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If you know that the die is fair, then this can equally be viewed as a Bayesian interpretation with a dogmatic prior that puts all prior probability on the die being fair, so one could argue that the Bayesian interpretation does equally well.

If, on the other hand, the die is not fair, then, if you have prior knowledge on the sense in which it is not fair, you may improve your inference over the frequentist estimator $n_6/n$, especially when $n$ is "small".

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  • $\begingroup$ Thanks! Very interesting. So these two interpretations are not entirely exclusive for each other. $\endgroup$ Commented Sep 15, 2022 at 16:05
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Bayesians and modern "frequentists" both use the same underlying laws of probability theory, and in my view, both can reconcile their views on the "frequentist definition of probability". Both groups of practitioners agree with the strong law of large numbers, which says that if you have an exchangeable sequence $X_1,X_2,X_3,...$ then (with probability one):$^\dagger$

$$\mathbb{P}(X_i \in \mathcal{A}) = \lim_{n \rightarrow \infty} \frac{1}{n} \sum_{i=1}^n \mathbb{I}(X_i \in \mathcal{A}) \quad \quad \quad \text{for any measurable set } \mathcal{A}.$$

Frequentists take this as a "definition" of probability, whereas Bayesians use an alternative interpretation of probability (usually an epistemic interpretation), but also agree that it corresponds to limiting frequency in this case. Bayesians don't disagree with the strong law of large numbers, and they tend to view it as something that clarifies the notion of "repeated trials" in the frequentist viewpoint ---i.e., the idea of infinitely repeated trials of a stable experiment is provided by the concept of exchangeability.

If you go back to the older probabilists like Venn, Reichenbach and Von Mises, you will find that they made philosophical arguments about the frequentist meaning without the aid of Kolmogorov's strong law of large numbers. To do this, they tended to invoke the idea of repeated trials that could occur hypothetically infinitely often, under unchanging conditions. This is captured well by the modern condition of exchangeability of an infinite sequence, which is the basis for the strong law of large numbers.


$^\dagger$ One might wonder if it is legitimate to invoke laws of probability theory in answer to determining the meaning of probability --- i.e., whether this is circular logic. For this purpose we may treat $\mathbb{P}$ as purely a mathematical operator, and then imbut it as a "probability" only once we have satisfied ourselves that it meets the requirements of the meaning of this concept.

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